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Theorem 1. If a function \( f\left( z\right) = u\left( {x, y}\right) + {iv}\left( {x, y}\right) \) is analytic in a domain \( D \), then its component functions \( u \) and \( v \) are harmonic in \( D \) . | To show this, we need a result that is to be proved in Chap. 4 (Sec. 48). Namely, if a function of a complex variable is analytic at a point, then its real and imaginary components have continuous partial derivatives of all orders at that point.\n\nAssuming that \( f \) is analytic in \( D \), we start with the observa... | No |
Theorem 2. A function \( f\left( z\right) = u\left( {x, y}\right) + {iv}\left( {x, y}\right) \) is analytic in a domain \( D \) if and only if \( v \) is a harmonic conjugate of \( u \) . | The proof is easy. If \( v \) is a harmonic conjugate of \( u \) in \( D \), the theorem in Sec. 21 tells us that \( f \) is analytic in \( D \) . Conversely, if \( f \) is analytic in \( D \), we know from Theorem 1 above that \( u \) and \( v \) are harmonic in \( D \) ; and, in view of the theorem in Sec. 20, the Ca... | No |
Theorem 1. If a function \( f \) is analytic throughout a simply connected domain \( D \) , then\n\n(1)\n\n\[ \n{\int }_{C}f\left( z\right) {dz} = 0 \n\] \n\nfor every closed contour \( C \) lying in \( D \) . | The proof is easy if \( C \) is a simple closed contour or if it is a closed contour that intersects itself a finite number of times. For, if \( C \) is simple and lies in \( D \), the function \( f \) is analytic at each point interior to and on \( C \) ; and the Cauchy-Goursat theorem ensures that equation (1) holds.... | Yes |
Corollary 1. A function \( f \) that is analytic throughout a simply connected domain \( D \) must have an antiderivative everywhere in \( D \) . | This corollary follows immediately from Theorem 1 because of the theorem in Sec. 42, which tells us that a continuous function \( f \) always has an antiderivative in a given domain when equation (1) holds for each closed contour \( C \) in that domain. Note that, since the finite plane is simply connected, Corollary 1... | Yes |
Theorem 2. Suppose that\n\n(i) \( C \) is a simple closed contour, described in the counterclockwise direction;\n\n(ii) \( {C}_{k}\left( {k = 1,2,\ldots, n}\right) \) are simple closed contours interior to \( C \), all described in the clockwise direction, that are disjoint and whose interiors have no points in common ... | To prove the theorem, we introduce a polygonal path \( {L}_{1} \), consisting of a finite number of line segments joined end to end, to connect the outer contour \( C \) to the inner contour \( {C}_{1} \) . We introduce another polygonal path \( {L}_{2} \) which connects \( {C}_{1} \) to \( {C}_{2} \) ; and we continue... | Yes |
Corollary 2. Let \( {C}_{1} \) and \( {C}_{2} \) denote positively oriented simple closed contours, where \( {C}_{2} \) is interior to \( {C}_{1} \) (Fig. 59). If a function \( f \) is analytic in the closed region consisting of those contours and all points between them, then\n\n(3)\n\n\[{\int }_{{C}_{1}}f\left( z\rig... | For a verification, we use Theorem 2 to write\n\n\[{\int }_{{C}_{1}}f\left( z\right) {dz} + {\int }_{-{C}_{2}}f\left( z\right) {dz} = 0\]\n\nand we note that this is just a different form of equation (3). | Yes |
Theorem 1. If a function is analytic at a point, then its derivatives of all orders exist at that point. Those derivatives are, moreover, all analytic there. | To prove this remarkable theorem, we assume that a function \( f \) is analytic at a point \( {z}_{0} \) . There must, then, be a neighborhood \( \left| {z - {z}_{0}}\right| < \varepsilon \) of \( {z}_{0} \) throughout which \( f \) is analytic (see Sec. 23). Consequently, there is a positively oriented circle \( {C}_{... | Yes |
Theorem 2. Let \( f \) be continuous on a domain D. If\n\n(6)\n\n\[{\int }_{C}f\left( z\right) {dz} = 0\]\n\nfor every closed contour \( C \) lying in \( D \), then \( f \) is analytic throughout \( D \) . | To prove the theorem here, we observe that when its hypothesis is satisfied, the theorem in Sec. 42 ensures that \( f \) has an antiderivative in \( D \) ; that is, there exists an analytic function \( F \) such that \( {F}^{\prime }\left( z\right) = f\left( z\right) \) at each point in \( D \) . Since \( f \) is the d... | Yes |
Theorem 1. If \( f \) is entire and bounded in the complex plane, then \( f\left( z\right) \) is constant throughout the plane. | To start the proof, we assume that \( f \) is as stated in the theorem and note that, since \( f \) is entire, Cauchy’s inequality (1) with \( n = 1 \) holds for any choices of \( {z}_{0} \) and \( R \) :\n\n(2)\n\n\[ \left| {{f}^{\prime }\left( {z}_{0}\right) }\right| \leq \frac{{M}_{R}}{R} \]\n\nMoreover, the bounded... | Yes |
Theorem 2. Any polynomial\n\n\\[ \nP\\left( z\\right) = {a}_{0} + {a}_{1}z + {a}_{2}{z}^{2} + \\cdots + {a}_{n}{z}^{n}\\;\\left( {{a}_{n} \\neq 0}\\right) \n\\]\n\nof degree \\( n\\left( {n \\geq 1}\\right) \\) has at least one zero. That is, there exists at least one point \\( {z}_{0} \\) such that \\( P\\left( {z}_{0... | The proof here is by contradiction. Suppose that \\( P\\left( z\\right) \\) is not zero for any value of \\( z \\) . Then the reciprocal\n\n\\[ \nf\\left( z\\right) = \\frac{1}{P\\left( z\\right) } \n\\]\n\nis clearly entire, and it is also bounded in the complex plane.\n\nTo show that it is bounded, we first write\n\n... | Yes |
Theorem 1. If a power series\n\n(1)\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{\left( z - {z}_{0}\right) }^{n} \]\n\nconverges when \( z = {z}_{1}\left( {{z}_{1} \neq {z}_{0}}\right) \), then it is absolutely convergent at each point \( z \) in the open disk \( \left| {z - {z}_{0}}\right| < {R}_{1} \), wher... | We first prove the theorem when \( {z}_{0} = 0 \), and we assume that the series\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{z}_{1}^{n}\;\left( {{z}_{1} \neq 0}\right) \]\n\nconverges. The terms \( {a}_{n}{z}_{1}^{n} \) are thus bounded; that is,\n\n\[ \left| {{a}_{n}{z}_{1}^{n}}\right| \leq M\;\left( {n = 0... | Yes |
Theorem 2. If \( {z}_{1} \) is a point inside the circle of convergence \( \left| {z - {z}_{0}}\right| = R \) of a power series\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{\left( z - {z}_{0}\right) }^{n} \]\n\nthen that series must be uniformly convergent in the closed disk \( \left| {z - {z}_{0}}\right| \le... | As in the proof of Theorem 1, we first treat the case in which \( {z}_{0} = 0 \) . Given that \( {z}_{1} \) is a point lying inside the circle of convergence of the series\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{z}^{n} \]\n\nwe note that there are points with modulus greater than \( \left| {z}_{1}\right|... | Yes |
Theorem 1. Let \( C \) denote any contour interior to the circle of convergence of the power series (1), and let \( g\left( z\right) \) be any function that is continuous on \( C \) . The series formed by multiplying each term of the power series by \( g\left( z\right) \) can be integrated term by term over \( C \) ; t... | To prove this theorem, we note that since both \( g\left( z\right) \) and the sum \( S\left( z\right) \) of the power series are continuous on \( C \), the integral over \( C \) of the product\n\n\[g\left( z\right) S\left( z\right) = \mathop{\sum }\limits_{{n = 0}}^{{N - 1}}{a}_{n}g\left( z\right) {\left( z - {z}_{0}\r... | Yes |
Theorem 2. The power series (1) can be differentiated term by term. That is, at each point \( z \) interior to the circle of convergence of that series,\n\n(6)\n\n\[ \n{S}^{\prime }\left( z\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }n{a}_{n}{\left( z - {z}_{0}\right) }^{n - 1}.\n\] | To prove this, let \( z \) denote any point interior to the circle of convergence of series (1); and let \( C \) be some positively oriented simple closed contour surrounding \( z \) and interior to that circle. Also, define the function\n\n(7)\n\n\[ \ng\left( s\right) = \frac{1}{2\pi i} \cdot \frac{1}{{\left( s - z\ri... | Yes |
Theorem 1. If a series\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{\left( z - {z}_{0}\right) }^{n} \]\n\nconverges to \( f\left( z\right) \) at all points interior to some circle \( \left| {z - {z}_{0}}\right| = R \), then it is the Taylor series expansion for \( f \) in powers of \( z - {z}_{0} \) . | To prove this, we write the series representation\n\n\[ f\left( z\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{\left( z - {z}_{0}\right) }^{n}\;\left( {\left| {z - {z}_{0}}\right| < R}\right) \]\n\nin the hypothesis of the theorem using the index of summation \( m \) :\n\n\[ f\left( z\right) = \mathop{\sum... | Yes |
Theorem 2. If a series\n\n(8)\n\n\[ \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{c}_{n}{\left( z - {z}_{0}\right) }^{n} = \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{\left( z - {z}_{0}\right) }^{n} + \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{b}_{n}}{{\left( z - {z}_{0}\right) }^{n}} \]\n\nconverges to \... | The method of proof here is similar to the one used in proving Theorem 1. The hypothesis of this theorem tells us that there is an annular domain about \( {z}_{0} \) such that\n\n\[ f\left( z\right) = \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{c}_{n}{\left( z - {z}_{0}\right) }^{n} \]\n\nfor each point \( z \) i... | Yes |
Theorem 2. Given a function \( f \) and a point \( {z}_{0} \), suppose that\n\n(i) \( f \) is analytic at \( {z}_{0} \) ;\n\n(ii) \( f\left( {z}_{0}\right) = 0 \) but \( f\left( z\right) \) is not identically equal to zero in any neighborhood of \( {z}_{0} \) .\n\nThen \( f\left( z\right) \neq 0 \) throughout some dele... | To prove this, let \( f \) be as stated and observe that not all of the derivatives of \( f \) at \( {z}_{0} \) are zero. For, if they were, all of the coefficients in the Taylor series for \( f \) about \( {z}_{0} \) would be zero; and that would mean that \( f\left( z\right) \) is identically equal to zero in some ne... | Yes |
Theorem 3. Given a function \( f \) and a point \( {z}_{0} \), suppose that\n\n(i) \( f \) is analytic throughout a neighborhood \( {N}_{0} \) of \( {z}_{0} \) ;\n\n(ii) \( f\left( {z}_{0}\right) = 0 \) and \( f\left( z\right) = 0 \) at each point \( z \) of a domain or line segment containing \( {z}_{0} \) (Fig. 87).\... | We begin the proof with the observation that, under the stated conditions, \( f\left( z\right) \equiv 0 \) in some neighborhood \( N \) of \( {z}_{0} \) . For, otherwise, there would be a deleted neighborhood of \( {z}_{0} \) throughout which \( f\left( z\right) \neq 0 \), according to Theorem 2 above; and that would b... | Yes |
Theorem 1. Suppose that\n\n(i) two functions \( p \) and \( q \) are analytic at a point \( {z}_{0} \) ;\n\n(ii) \( p\left( {z}_{0}\right) \neq 0 \) and \( q \) has a zero of order \( m \) at \( {z}_{0} \) .\n\nThen the quotient \( p\left( z\right) /q\left( z\right) \) has a pole of order \( m \) at \( {z}_{0} \) . | The proof is easy. Let \( p \) and \( q \) be as in the statement of the theorem. Since \( q \) has a zero of order \( m \) at \( {z}_{0} \), we know from Theorem 2 in Sec. 68 that there is a deleted neighborhood of \( {z}_{0} \) in which \( q\left( z\right) \neq 0 \) ; and so \( {z}_{0} \) is an isolated singular poin... | Yes |
Theorem 2. Let two functions \( p \) and \( q \) be analytic at a point \( {z}_{0} \) . If\n\n\[ p\left( {z}_{0}\right) \neq 0,\;q\left( {z}_{0}\right) = 0,\;\text{ and }\;{q}^{\prime }\left( {z}_{0}\right) \neq 0, \]\n\nthen \( {z}_{0} \) is a simple pole of the quotient \( p\left( z\right) /q\left( z\right) \) and\n\... | To show this, we assume that \( p \) and \( q \) are as stated and observe that, because of the conditions on \( q \), the point \( {z}_{0} \) is a zero of order \( m = 1 \) of that function. According to Theorem 1 in Sec. 68, then,\n\n(3)\n\n\[ q\left( z\right) = \left( {z - {z}_{0}}\right) g\left( z\right) \]\n\nwher... | Yes |
Theorem 1. If \( {z}_{0} \) is a pole of a function \( f \), then\n\n(1)\n\n\[ \mathop{\lim }\limits_{{z \rightarrow {z}_{0}}}f\left( z\right) = \infty \] | To verify limit (1), we assume that \( f \) has a pole of order \( m \) at \( {z}_{0} \) and use the theorem in Sec. 66. It tells us that\n\n\[ f\left( z\right) = \frac{\phi \left( z\right) }{{\left( z - {z}_{0}\right) }^{m}} \]\n\nwhere \( \phi \left( z\right) \) is analytic and nonzero at \( {z}_{0} \) . Since\n\n\[ ... | Yes |
Theorem 2. If \( {z}_{0} \) is a removable singular point of a function \( f \), then \( f \) is analytic and bounded in some deleted neighborhood \( 0 < \left| {z - {z}_{0}}\right| < \varepsilon \) of \( {z}_{0} \) . | The proof is easy and is based on the fact that the function \( f \) here is analytic in a disk \( \left| {z - {z}_{0}}\right| < {R}_{2} \) when \( f\left( {z}_{0}\right) \) is properly defined; and \( f \) is then continuous in any closed disk \( \left| {z - {z}_{0}}\right| \leq \varepsilon \) where \( \varepsilon < {... | Yes |
Theorem 3. Suppose that \( {z}_{0} \) is an essential singularity of a function \( f \), and let \( {w}_{0} \) be any complex number. Then, for any positive number \( \varepsilon \), the inequality\n\n(4)\n\n\[ \left| {f\left( z\right) - {w}_{0}}\right| < \varepsilon \]\n\nis satisfied at some point \( z \) in each del... | The proof is by contradiction. Since \( {z}_{0} \) is an isolated singularity of \( f \), there is a deleted neighborhood \( 0 < \left| {z - {z}_{0}}\right| < \delta \) throughout which \( f \) is analytic; and we assume that condition (4) is not satisfied for any point \( z \) there. Thus \( \left| {f\left( z\right) -... | Yes |
The circle group \( {\mathbb{S}}^{1} = \mathrm{{SO}}\left( 2\right) \). | Viewed as a set of points in \( \mathbb{C} \) or \( {\mathbb{R}}^{2} \), the unit circle is a closed set because its complement (the set of points not on the circle) is clearly open. Figure 8.1 shows a typical point \( P \) not on the circle and an \( \varepsilon \) -neighborhood of \( P \) that lies in the complement ... | No |
The groups \( \mathrm{O}\left( n\right) \) and \( \mathrm{{SO}}\left( n\right) \). | We view \( \mathrm{O}\left( n\right) \) as a subset of the space \( {\mathbb{R}}^{{n}^{2}} \) of \( n \times n \) real matrices, which we also call \( {M}_{n}\left( \mathbb{R}\right) \). The complement of \( \mathrm{O}\left( n\right) \) is\n\n\[ \n{M}_{n}\left( \mathbb{R}\right) - \mathrm{O}\left( n\right) = \left\{ {A... | Yes |
Example 3. The group \( \operatorname{Aff}\left( 1\right) \). We view \( \operatorname{Aff}\left( 1\right) \) as in Section 4.6, namely, as the group of real matrices of the form \( A = \left( \begin{array}{ll} a & b \\ 0 & 1 \end{array}\right) \), where \( a, b \in \mathbb{R} \) and \( a > 0 \). It is now easy to see ... | However, \( \operatorname{Aff}\left( 1\right) \) is closed in the \ | No |
The general linear group \( \mathrm{{GL}}\left( {n,\mathbb{C}}\right) \) is the set of all invertible \( n \times n \) complex matrices. This set is a group because it is closed under products (since \( {A}^{-1}{B}^{-1} = \) \( {\left( BA\right) }^{-1} \) ) and under inverses (obviously). | It follows that every group of real or complex matrices is a subgroup of some \( \mathrm{{GL}}\left( {n,\mathbb{C}}\right) ,{}^{8} \) which is why we bring it up now. We are about to define what a \ | No |
Theorem 2. Suppose that\n\n(7)\n\[ \mathop{\lim }\limits_{{z \rightarrow {z}_{0}}}f\left( z\right) = {w}_{0}\;\text{ and }\;\mathop{\lim }\limits_{{z \rightarrow {z}_{0}}}F\left( z\right) = {W}_{0}. \]\n\nThen\n\n(8)\n\[ \mathop{\lim }\limits_{{z \rightarrow {z}_{0}}}\left\lbrack {f\left( z\right) + F\left( z\right) }\... | This important theorem can be proved directly by using the definition of the limit of a function of a complex variable. But, with the aid of Theorem 1, it follows almost immediately from theorems on limits of real-valued functions of two real variables.\n\nTo verify property (9), for example, we write\n\n\[ f\left( z\r... | Yes |
Theorem 2. If a function \( f\left( z\right) \) is continuous and nonzero at a point \( {z}_{0} \), then \( f\left( z\right) \neq 0 \) throughout some neighborhood of that point. | Assuming that \( f\left( z\right) \) is, in fact, continuous and nonzero at \( {z}_{0} \), we can prove Theorem 2 by assigning the positive value \( \left| {f\left( {z}_{0}\right) }\right| /2 \) to the number \( \varepsilon \) in statement (4). This tells us that there is a positive number \( \delta \) such that\n\n\[ ... | Yes |
Theorem 1. If a function \( f\left( z\right) = u\left( {x, y}\right) + {iv}\left( {x, y}\right) \) is analytic in a domain \( D \), then its component functions \( u \) and \( v \) are harmonic in \( D \) . | To show this, we need a result that is to be proved in Chap. 4 (Sec. 48). Namely, if a function of a complex variable is analytic at a point, then its real and imaginary components have continuous partial derivatives of all orders at that point.\n\nAssuming that \( f \) is analytic in \( D \), we start with the observa... | No |
Theorem 2. A function \( f\left( z\right) = u\left( {x, y}\right) + {iv}\left( {x, y}\right) \) is analytic in a domain \( D \) if and only if \( v \) is a harmonic conjugate of \( u \) . | The proof is easy. If \( v \) is a harmonic conjugate of \( u \) in \( D \), the theorem in Sec. 21 tells us that \( f \) is analytic in \( D \) . Conversely, if \( f \) is analytic in \( D \), we know from Theorem 1 above that \( u \) and \( v \) are harmonic in \( D \) ; and, in view of the theorem in Sec. 20, the Ca... | No |
Theorem 1. If a function \( f \) is analytic throughout a simply connected domain \( D \) , then\n\n\[{\int }_{C}f\left( z\right) {dz} = 0\]\n\nfor every closed contour \( C \) lying in \( D \) . | The proof is easy if \( C \) is a simple closed contour or if it is a closed contour that intersects itself a finite number of times. For, if \( C \) is simple and lies in \( D \), the function \( f \) is analytic at each point interior to and on \( C \) ; and the Cauchy-Goursat theorem ensures that equation (1) holds.... | Yes |
Corollary 1. A function \( f \) that is analytic throughout a simply connected domain \( D \) must have an antiderivative everywhere in \( D \) . | This corollary follows immediately from Theorem 1 because of the theorem in Sec. 42, which tells us that a continuous function \( f \) always has an antiderivative in a given domain when equation (1) holds for each closed contour \( C \) in that domain. Note that, since the finite plane is simply connected, Corollary 1... | Yes |
Theorem 2. Suppose that\n\n(i) \( C \) is a simple closed contour, described in the counterclockwise direction;\n\n(ii) \( {C}_{k}\left( {k = 1,2,\ldots, n}\right) \) are simple closed contours interior to \( C \), all described in the clockwise direction, that are disjoint and whose interiors have no points in common ... | To prove the theorem, we introduce a polygonal path \( {L}_{1} \), consisting of a finite number of line segments joined end to end, to connect the outer contour \( C \) to the inner contour \( {C}_{1} \) . We introduce another polygonal path \( {L}_{2} \) which connects \( {C}_{1} \) to \( {C}_{2} \) ; and we continue... | Yes |
Corollary 2. Let \( {C}_{1} \) and \( {C}_{2} \) denote positively oriented simple closed contours, where \( {C}_{2} \) is interior to \( {C}_{1} \) (Fig. 59). If a function \( f \) is analytic in the closed region consisting of those contours and all points between them, then\n\n(3)\n\n\[ \n{\int }_{{C}_{1}}f\left( z\... | For a verification, we use Theorem 2 to write\n\n\[ \n{\int }_{{C}_{1}}f\left( z\right) {dz} + {\int }_{-{C}_{2}}f\left( z\right) {dz} = 0 \n\]\n\nand we note that this is just a different form of equation (3). | Yes |
Theorem 1. If a function is analytic at a point, then its derivatives of all orders exist at that point. Those derivatives are, moreover, all analytic there. | To prove this remarkable theorem, we assume that a function \( f \) is analytic at a point \( {z}_{0} \) . There must, then, be a neighborhood \( \left| {z - {z}_{0}}\right| < \varepsilon \) of \( {z}_{0} \) throughout which \( f \) is analytic (see Sec. 23). Consequently, there is a positively oriented circle \( {C}_{... | Yes |
Theorem 2. Let \( f \) be continuous on a domain D. If\n\n(6)\n\n\[{\int }_{C}f\left( z\right) {dz} = 0\]\n\nfor every closed contour \( C \) lying in \( D \), then \( f \) is analytic throughout \( D \) . | To prove the theorem here, we observe that when its hypothesis is satisfied, the theorem in Sec. 42 ensures that \( f \) has an antiderivative in \( D \) ; that is, there exists an analytic function \( F \) such that \( {F}^{\prime }\left( z\right) = f\left( z\right) \) at each point in \( D \) . Since \( f \) is the d... | Yes |
Theorem 1. If \( f \) is entire and bounded in the complex plane, then \( f\left( z\right) \) is constant throughout the plane. | To start the proof, we assume that \( f \) is as stated in the theorem and note that, since \( f \) is entire, Cauchy’s inequality (1) with \( n = 1 \) holds for any choices of \( {z}_{0} \) and \( R \) :\n\n(2)\n\n\[ \left| {{f}^{\prime }\left( {z}_{0}\right) }\right| \leq \frac{{M}_{R}}{R} \]\n\nMoreover, the bounded... | Yes |
Theorem 2. Any polynomial\n\n\\[ \nP\\left( z\\right) = {a}_{0} + {a}_{1}z + {a}_{2}{z}^{2} + \\cdots + {a}_{n}{z}^{n}\\;\\left( {{a}_{n} \\neq 0}\\right) \n\\]\n\nof degree \\( n\\left( {n \\geq 1}\\right) \\) has at least one zero. That is, there exists at least one point \\( {z}_{0} \\) such that \\( P\\left( {z}_{0... | The proof here is by contradiction. Suppose that \\( P\\left( z\\right) \\) is not zero for any value of \\( z \\) . Then the reciprocal\n\n\\[ \nf\\left( z\\right) = \\frac{1}{P\\left( z\\right) } \n\\]\n\nis clearly entire, and it is also bounded in the complex plane.\n\nTo show that it is bounded, we first write\n\n... | Yes |
Theorem 1. If a power series\n\n(1)\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{\left( z - {z}_{0}\right) }^{n} \]\n\nconverges when \( z = {z}_{1}\left( {{z}_{1} \neq {z}_{0}}\right) \), then it is absolutely convergent at each point \( z \) in the open disk \( \left| {z - {z}_{0}}\right| < {R}_{1} \), wher... | We first prove the theorem when \( {z}_{0} = 0 \), and we assume that the series\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{z}_{1}^{n}\;\left( {{z}_{1} \neq 0}\right) \]\n\nconverges. The terms \( {a}_{n}{z}_{1}^{n} \) are thus bounded; that is,\n\n\[ \left| {{a}_{n}{z}_{1}^{n}}\right| \leq M\;\left( {n = 0... | Yes |
Theorem 2. If \( {z}_{1} \) is a point inside the circle of convergence \( \left| {z - {z}_{0}}\right| = R \) of a power series\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{\left( z - {z}_{0}\right) }^{n} \]\n\nthen that series must be uniformly convergent in the closed disk \( \left| {z - {z}_{0}}\right| \le... | As in the proof of Theorem 1, we first treat the case in which \( {z}_{0} = 0 \) . Given that \( {z}_{1} \) is a point lying inside the circle of convergence of the series\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{z}^{n} \]\n\nwe note that there are points with modulus greater than \( \left| {z}_{1}\right|... | Yes |
Theorem 1. Let \( C \) denote any contour interior to the circle of convergence of the power series (1), and let \( g\left( z\right) \) be any function that is continuous on \( C \) . The series formed by multiplying each term of the power series by \( g\left( z\right) \) can be integrated term by term over \( C \) ; t... | To prove this theorem, we note that since both \( g\left( z\right) \) and the sum \( S\left( z\right) \) of the power series are continuous on \( C \), the integral over \( C \) of the product\n\n\[ \ng\left( z\right) S\left( z\right) = \mathop{\sum }\limits_{{n = 0}}^{{N - 1}}{a}_{n}g\left( z\right) {\left( z - {z}_{0... | Yes |
Theorem 2. The power series (1) can be differentiated term by term. That is, at each point \( z \) interior to the circle of convergence of that series,\n\n(6)\n\n\[ \n{S}^{\prime }\left( z\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }n{a}_{n}{\left( z - {z}_{0}\right) }^{n - 1}.\n\] | To prove this, let \( z \) denote any point interior to the circle of convergence of series (1); and let \( C \) be some positively oriented simple closed contour surrounding \( z \) and interior to that circle. Also, define the function\n\n(7)\n\n\[ \ng\left( s\right) = \frac{1}{2\pi i} \cdot \frac{1}{{\left( s - z\ri... | Yes |
If a series\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{\left( z - {z}_{0}\right) }^{n} \]\n\nconverges to \( f\left( z\right) \) at all points interior to some circle \( \left| {z - {z}_{0}}\right| = R \), then it is the Taylor series expansion for \( f \) in powers of \( z - {z}_{0} \) . | To prove this, we write the series representation\n\n\[ f\left( z\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{\left( z - {z}_{0}\right) }^{n}\;\left( {\left| {z - {z}_{0}}\right| < R}\right) \]\n\nin the hypothesis of the theorem using the index of summation \( m \) :\n\n\[ f\left( z\right) = \mathop{\sum... | Yes |
Theorem 2. If a series\n\n(8)\n\n\[ \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{c}_{n}{\left( z - {z}_{0}\right) }^{n} = \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{\left( z - {z}_{0}\right) }^{n} + \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{b}_{n}}{{\left( z - {z}_{0}\right) }^{n}} \]\n\nconverges to \... | The method of proof here is similar to the one used in proving Theorem 1. The hypothesis of this theorem tells us that there is an annular domain about \( {z}_{0} \) such that\n\n\[ f\left( z\right) = \mathop{\sum }\limits_{{n = - \infty }}^{\infty }{c}_{n}{\left( z - {z}_{0}\right) }^{n} \]\n\nfor each point \( z \) i... | Yes |
Theorem 2. Given a function \( f \) and a point \( {z}_{0} \), suppose that\n\n(i) \( f \) is analytic at \( {z}_{0} \) ;\n\n(ii) \( f\left( {z}_{0}\right) = 0 \) but \( f\left( z\right) \) is not identically equal to zero in any neighborhood of \( {z}_{0} \) .\n\nThen \( f\left( z\right) \neq 0 \) throughout some dele... | To prove this, let \( f \) be as stated and observe that not all of the derivatives of \( f \) at \( {z}_{0} \) are zero. For, if they were, all of the coefficients in the Taylor series for \( f \) about \( {z}_{0} \) would be zero; and that would mean that \( f\left( z\right) \) is identically equal to zero in some ne... | Yes |
Theorem 3. Given a function \( f \) and a point \( {z}_{0} \), suppose that\n\n(i) \( f \) is analytic throughout a neighborhood \( {N}_{0} \) of \( {z}_{0} \) ;\n\n(ii) \( f\left( {z}_{0}\right) = 0 \) and \( f\left( z\right) = 0 \) at each point \( z \) of a domain or line segment containing \( {z}_{0} \) (Fig. 87).\... | We begin the proof with the observation that, under the stated conditions, \( f\left( z\right) \equiv 0 \) in some neighborhood \( N \) of \( {z}_{0} \) . For, otherwise, there would be a deleted neighborhood of \( {z}_{0} \) throughout which \( f\left( z\right) \neq 0 \), according to Theorem 2 above; and that would b... | Yes |
Theorem 1. Suppose that\n\n(i) two functions \( p \) and \( q \) are analytic at a point \( {z}_{0} \) ;\n\n(ii) \( p\left( {z}_{0}\right) \neq 0 \) and \( q \) has a zero of order \( m \) at \( {z}_{0} \) .\n\nThen the quotient \( p\left( z\right) /q\left( z\right) \) has a pole of order \( m \) at \( {z}_{0} \) . | The proof is easy. Let \( p \) and \( q \) be as in the statement of the theorem. Since \( q \) has a zero of order \( m \) at \( {z}_{0} \), we know from Theorem 2 in Sec. 68 that there is a deleted neighborhood of \( {z}_{0} \) in which \( q\left( z\right) \neq 0 \) ; and so \( {z}_{0} \) is an isolated singular poin... | Yes |
Theorem 2. Let two functions \( p \) and \( q \) be analytic at a point \( {z}_{0} \) . If\n\n\[ p\left( {z}_{0}\right) \neq 0,\;q\left( {z}_{0}\right) = 0,\;\text{ and }\;{q}^{\prime }\left( {z}_{0}\right) \neq 0, \]\n\nthen \( {z}_{0} \) is a simple pole of the quotient \( p\left( z\right) /q\left( z\right) \) and\n\... | To show this, we assume that \( p \) and \( q \) are as stated and observe that, because of the conditions on \( q \), the point \( {z}_{0} \) is a zero of order \( m = 1 \) of that function. According to Theorem 1 in Sec. 68, then,\n\n(3)\n\n\[ q\left( z\right) = \left( {z - {z}_{0}}\right) g\left( z\right) \]\n\nwher... | Yes |
Theorem 1. If \( {z}_{0} \) is a pole of a function \( f \), then\n\n(1)\n\n\[ \mathop{\lim }\limits_{{z \rightarrow {z}_{0}}}f\left( z\right) = \infty \] | To verify limit (1), we assume that \( f \) has a pole of order \( m \) at \( {z}_{0} \) and use the theorem in Sec. 66. It tells us that\n\n\[ f\left( z\right) = \frac{\phi \left( z\right) }{{\left( z - {z}_{0}\right) }^{m}} \]\n\nwhere \( \phi \left( z\right) \) is analytic and nonzero at \( {z}_{0} \) . Since\n\n\[ ... | Yes |
Theorem 2. If \( {z}_{0} \) is a removable singular point of a function \( f \), then \( f \) is analytic and bounded in some deleted neighborhood \( 0 < \left| {z - {z}_{0}}\right| < \varepsilon \) of \( {z}_{0} \) . | The proof is easy and is based on the fact that the function \( f \) here is analytic in a disk \( \left| {z - {z}_{0}}\right| < {R}_{2} \) when \( f\left( {z}_{0}\right) \) is properly defined; and \( f \) is then continuous in any closed disk \( \left| {z - {z}_{0}}\right| \leq \varepsilon \) where \( \varepsilon < {... | Yes |
Theorem 3. Suppose that \( {z}_{0} \) is an essential singularity of a function \( f \), and let \( {w}_{0} \) be any complex number. Then, for any positive number \( \varepsilon \), the inequality\n\n(4)\n\n\[ \left| {f\left( z\right) - {w}_{0}}\right| < \varepsilon \]\n\nis satisfied at some point \( z \) in each del... | The proof is by contradiction. Since \( {z}_{0} \) is an isolated singularity of \( f \), there is a deleted neighborhood \( 0 < \left| {z - {z}_{0}}\right| < \delta \) throughout which \( f \) is analytic; and we assume that condition (4) is not satisfied for any point \( z \) there. Thus \( \left| {f\left( z\right) -... | Yes |
Theorem 1. If all zeros of a polynomial \( P\left( z\right) \) lie in a half plane, then all zeros of the derivative \( {P}^{\prime }\left( z\right) \) lie in the same half plane. | From (8) we obtain\n\n(9)\n\n\[ \frac{{P}^{\prime }\left( z\right) }{P\left( z\right) } = \frac{1}{z - {\alpha }_{1}} + \cdots + \frac{1}{z - {\alpha }_{n}}. \]\n\nSuppose that the half plane \( H \) is defined as the part of the plane where \( \operatorname{Im}\left( {z - a}\right) /b < 0 \) (see Chap. 1, Sec. 2.3). I... | Yes |
For every power series (17) there exists a number \( R,0 \leqq \) \( R \leqq \infty \), called the radius of convergence, with the following properties:\n\n(i) The series converges absolutely for every \( z \) with \( \left| z\right| < R \) . If \( 0 \leqq \) \( \rho < R \) the convergence is uniform for \( \left| z\ri... | This is known as Hadamard’s formula for the radius of convergence.\n\nIf \( \left| z\right| < R \) we can find \( \rho \) so that \( \left| z\right| < \rho < R \) . Then \( 1/\rho > 1/R \), and by the definition of limes superior there exists an \( {n}_{0} \) such that \( {\left| {a}_{n}\right| }^{1/n} < 1/\rho , \) \(... | Yes |
Theorem 3. If \( \mathop{\sum }\limits_{0}^{\infty }{a}_{n} \) converges, then \( f\left( z\right) = \mathop{\sum }\limits_{0}^{\infty }{a}_{n}{z}^{n} \) tends to \( f\left( 1\right) \) as \( z \) approaches 1 in such a way that \( \left| {1 - z}\right| /\left( {1 - \left| z\right| }\right) \) remains bounded. | Proof. We may assume that \( \mathop{\sum }\limits_{0}^{\infty }{a}_{n} = 0 \), for this can be attained by adding\na constant to \( {a}_{0} \) . We write \( {s}_{n} = {a}_{0} + {a}_{1} + \cdots + {a}_{n} \) and make use of the identity (summation by parts)\n\n\[{s}_{n}\left( z\right) = {a}_{0} + {a}_{1}z + \cdots + {a... | Yes |
Theorem 1. The nonempty connected subsets of the real line are the intervals. | We reproduce one of the classical proofs, based on the fact that any monotone sequence has a finite or infinite limit.\n\nSuppose that the real line \( \mathbf{R} \) is represented as the union \( \mathbf{R} = A \cup B \) of two disjoint closed sets. If neither is empty we can find \( {a}_{1}{\epsilon A} \) and \( {b}_... | Yes |
Theorem 3. A nonempty open set in the plane is connected if and only if any two of its points can be joined by a polygon which lies in the set. | We prove first that the condition is necessary. Let \( A \) be an open connected set, and choose a point \( a \in A \) . We denote by \( {A}_{1} \) the subset of \( A \) whose points can be joined to \( a \) by polygons in \( A \), and by \( {A}_{2} \) the subset whose points cannot be so joined. Let us prove that \( {... | Yes |
Theorem 6. A set is compact if and only if it is complete and totally bounded. | To prove the other part, assume that the metric space \( S \) is complete and totally bounded. Suppose that there exists an open covering which does not contain any finite subcovering. Write \( {\varepsilon }_{n} = {2}^{-n} \) . We know that \( S \) can be covered by finitely many \( B\left( {x,{\varepsilon }_{1}}\righ... | Yes |
Theorem 7. A metric space is compact if and only if every infinite sequence has a limit point. | The first part of the proof is a repetition of an earlier argument. If \( y \) is not a limit point of \( \left\{ {x}_{n}\right\} \) it has a neighborhood which contains only finitely many \( {x}_{n} \) (abbreviated version of the correct phrase). If there were no limit points the open sets containing only finitely man... | Yes |
Theorem 8. Under a continuous mapping the image of every compact set is compact, and consequently closed. | Suppose that \( f \) is defined and continuous on the compact set \( X \) . Consider a covering of \( f\left( X\right) \) by open sets \( U \) . The inverse images \( {f}^{-1}\left( U\right) \) are open and form a covering of \( X \) . Because \( X \) is compact we can select a finite subcovering: \( X \subset {f}^{-1}... | Yes |
Theorem 9. Under a continuous mapping the image of any connected set is connected. | We may assume that \( f \) is defined and continuous on the whole space \( S \), and that \( f\left( S\right) \) is all of \( {S}^{\prime } \) . Suppose that \( {S}^{\prime } = A \cup B \) where \( A \) and \( B \) are open and disjoint. Then \( S = {f}^{-1}\left( A\right) \cup {f}^{-1}\left( B\right) \) is a represent... | Yes |
Theorem 10. On a compact set every continuous function is uniformly continuous. | The proof is typical of the way the Heine-Borel property can be used. Suppose that \( f \) is continuous on a compact set \( X \) . For every \( y \in X \) there is a ball \( B\left( {y,\rho }\right) \) such that \( {d}^{\prime }\left( {f\left( x\right), f\left( y\right) }\right) < \epsilon /2 \) for \( x \in B\left( {... | Yes |
Theorem 11. An analytic function in a region \( \Omega \) whose derivative vanishes identically must reduce to a constant. The same is true if either the real part, the imaginary part, the modulus, or the argument is constant. | The vanishing of the derivative implies that \( \partial u/\partial x,\partial u/\partial y,\partial v/\partial x \) , \( \partial v/\partial y \) are all zero. It follows that \( u \) and \( v \) are constant on any line segment in \( \Omega \) which is parallel to one of the coordinate axes. In Sec. 1.3 we remarked, ... | Yes |
Theorem 12. If \( {z}_{1},{z}_{2},{z}_{3},{z}_{4} \) are distinct points in the extended plane and \( T \) any linear transformation, then \( \left( {T{z}_{1}, T{z}_{2}, T{z}_{3}, T{z}_{4}}\right) = \left( {{z}_{1},{z}_{2},{z}_{3},{z}_{4}}\right) \) . | The proof is immediate, for if \( {Sz} = \left( {z,{z}_{2},{z}_{3},{z}_{4}}\right) \), then \( S{T}^{-1} \) carries \( T{z}_{2}, T{z}_{3}, T{z}_{4} \) into \( 1,0,\infty \) . By definition we have hence\n\n\[ \left( {T{z}_{1}, T{z}_{2}, T{z}_{3}, T{z}_{4}}\right) = S{T}^{-1}\left( {T{z}_{1}}\right) = S{z}_{1} = \left( ... | Yes |
Theorem 13. The cross ratio \( \left( {{z}_{1},{z}_{2},{z}_{3},{z}_{4}}\right) \) is real if and only if the four points lie on a circle or on a straight line. | This is evident by elementary geometry, for we obtain\n\n\[ \arg \left( {{z}_{1},{z}_{2},{z}_{3},{z}_{4}}\right) = \arg \frac{{z}_{1} - {z}_{3}}{{z}_{1} - {z}_{4}} - \arg \frac{{z}_{2} - {z}_{3}}{{z}_{2} - {z}_{4}} \]\n\nand if the points lie on a circle this difference of angles is either 0 or \( \pm \pi \) , dependin... | Yes |
Theorem 1. The line integral \( {\int }_{\gamma }{pdx} + {qdy} \), defined in \( \Omega \), depends only on the end points of \( \gamma \) if und only if there exists a function \( U\left( {x, y}\right) \) in \( \Omega \) with the partial derivatives \( \partial U/\partial x = p,\partial U/\partial y = q \) . | The sufficiency follows at once, for if the condition is fulfilled we can write, with the usual notations,\n\n\[ \n{\int }_{\gamma }{pdx} + {qdy} = {\int }_{a}^{b}\left( {\frac{\partial U}{\partial x}{x}^{\prime }\left( t\right) + \frac{\partial U}{\partial y}{y}^{\prime }\left( t\right) }\right) {dt} = {\int }_{a}^{b}... | Yes |
Theorem 3. Let \( f\left( z\right) \) be analytic on the set \( {R}^{\prime } \) obtained from a rectangle \( R \) by omitting a finite number of interior points \( {\zeta }_{j} \) . If it is true that\n\n\[ \mathop{\lim }\limits_{{z \rightarrow {\zeta }_{j}}}\left( {z - {\zeta }_{j}}\right) f\left( z\right) = 0 \]\n\n... | It is sufficient to consider the case of a single exceptional point \( \zeta \), for evidently \( R \) can be divided into smaller rectangles which contain at most one \( {\zeta }_{j} \) .\n\nWe divide \( R \) into nine rectangles, as shown in Fig. 4-3, and apply\n\n \) is analytic in an open disk \( \Delta \), then\n\n(18)\n\n\[{\int }_{\gamma }f\left( z\right) {dz} = 0\]\n\nfor every closed curve \( \gamma \) in \( \Delta \) . | The proof is a repetition of the argument used in proving the second half of Theorem 1. We define a function \( F\left( z\right) \) by\n\n(19)\n\n\[F\left( z\right) = {\int }_{\sigma }{fdz}\]\n\nwhere \( \sigma \) consists of the horizontal line segment from the center \( \left( {{x}_{0},{y}_{0}}\right) \) to \( \left(... | Yes |
Theorem 5. Let \( f\left( z\right) \) be analytic in the region \( {\Delta }^{\prime } \) obtained by omitting a finite number of points \( {\zeta }_{j} \) from an open disk \( \Delta \) . If \( f\left( z\right) \) satisfies the condition \( \mathop{\lim }\limits_{{z \rightarrow {\zeta }_{j}}}\left( {z - {\zeta }_{j}}\... | The proof must be modified, for we cannot let \( \sigma \) pass through the exceptional points. Assume first that no \( {\zeta }_{j} \) lies on the lines \( x = {x}_{0} \) and \( y = {y}_{0} \) . It is then possible to avoid the exceptional points by letting \( \sigma \) consist of three segments (Fig. 4-4). By an obvi... | Yes |
Lemma 1. If the piecewise differentiable closed curve \( \gamma \) does not pass through the point \( a \), then the value of the integral\n\n\[ \n{\int }_{\gamma }\frac{dz}{z - a}\n\]\n\nis a multiple of \( {2\pi i} \) . | The simplest proof is computational. If the equation of \( \gamma \) is \( z\; = \;z\left( t\right) , \) \( \alpha \leqq t \leqq \beta \), let us consider the function\n\n\[ \nh\left( t\right) = {\int }_{\alpha }^{t}\frac{{z}^{\prime }\left( t\right) }{z\left( t\right) - a}{dt}\n\]\n\nIt is defined and continuous on th... | Yes |
Lemma 2. Let \( {z}_{1},{z}_{2} \) be two points on a closed curve \( \gamma \) which does not pass through the origin. Denote the subarc from \( {z}_{1} \) to \( {z}_{2} \) in the direction of the curve by \( {\gamma }_{1} \), and the subarc from \( {z}_{2} \) to \( {z}_{1} \) by \( {\gamma }_{2} \) . Suppose that \( ... | For the proof we draw the half lines \( {L}_{1} \) and \( {L}_{2} \) from the origin through \( {z}_{1} \) and \( {z}_{2} \) (Fig. 4-5). Let \( {\zeta }_{1},{\zeta }_{2} \) be the points in which \( {L}_{1},{L}_{2} \) intersect a circle \( C \) about the origin. If \( C \) is described in the positive sense, the arc \(... | Yes |
Theorem 6. Suppose that \( f\left( z\right) \) is analytic in an open disk \( \Delta \), and let \( \gamma \) be a closed curve in \( \Delta \) . For any point a not on \( \gamma \)\n\n\[ n\left( {\gamma, a}\right) \cdot f\left( a\right) = \frac{1}{2\pi i}{\int }_{\gamma }\frac{f\left( z\right) {dz}}{z - a} \]\n\nwhere... | In this statement we have suppressed the requirement that \( a \) be a point in \( \Delta \) . We have done so in view of the obvious interpretation of the formula (20) for the case that \( a \) is not in \( \Delta \) . Indeed, in this case \( n\left( {\gamma, a}\right) \) and the integral in the right-hand member are ... | Yes |
Lemma 3. Suppose that \( \varphi \left( \zeta \right) \) is continuous on the arc \( \gamma \) . Then the function\n\n\[ \n{F}_{n}\left( z\right) = {\int }_{\gamma }\frac{\varphi \left( \zeta \right) }{{\left( \zeta - z\right) }^{n}}\n\]\n\nis analytic in each of the regions determined by \( \gamma \), and its derivati... | We prove first that \( {F}_{1}\left( z\right) \) is continuous. Let \( {z}_{0} \) be a point not on \( \gamma \) , and choose the neighborhood \( \left| {z - {z}_{0}}\right| < \delta \) so that it does not meet \( \gamma \) . By restricting \( z \) to the smaller neighborhood \( \left| {z - {z}_{0}}\right| < \delta /2 ... | Yes |
Theorem 7. Suppose that \( f\left( z\right) \) is analytic in the region \( {\Omega }^{\prime } \) obtained by omitting a point a from a region \( \Omega \) . A necessary and sufficient condition that there exist an analytic function in \( \Omega \) which coincides with \( f\left( z\right) \) in \( {\Omega }^{\prime } ... | The necessity and the uniqueness are trivial since the extended function must be continuous at \( a \) . To prove the sufficiency we draw a circle \( C \) about \( a \) so that \( C \) and its inside are contained in \( \Omega \) . Cauchy’s formula is valid, and we can write\n\n\[ f\left( z\right) = \frac{1}{2\pi i}{\i... | Yes |
Theorem 8. If \( f\left( z\right) \) is analytic in a region \( \Omega \), containing \( a \), it is possible to write\n\n\[ f\left( z\right) = f\left( a\right) + \frac{{f}^{\prime }\left( a\right) }{1!}\left( {z - a}\right) + \frac{{f}^{\prime \prime }\left( a\right) }{2!}{\left( z - a}\right) }^{2} + \cdots \]\n\n\[ ... | This finite development must be well distinguished from the infinite Taylor series which we will study later. It is, however, the finite development (28) which is the most useful for the study of the local properties of \( f\left( z\right) \) . Its usefulness is enhanced by the fact that \( {f}_{n}\left( z\right) \) ha... | Yes |
Theorem 9. An analytic function comes arbitrarily close to any complex value in every neighborhood of an essential singularity. | If the assertion were not true, we could find a complex number \( A \) and a \( \delta > 0 \) such that \( \left| {f\left( z\right) - A}\right| > \delta \) in a neighborhood of \( a \) (except for \( z = a) \) . For any \( \alpha < 0 \) we have then \( \mathop{\lim }\limits_{{z \rightarrow a}}{\left| z - a\right| }^{\a... | Yes |
Theorem 10. Let \( {z}_{j} \) be the zeros of a function \( f\left( z\right) \) which is analytic in a disk \( \Delta \) and does not vanish identically, each zero being counted as many times as its order indicates. For every closed curve \( \gamma \) in \( \Delta \) which does not pass through a zero\n\n\[ \mathop{\su... | The function \( w = f\left( z\right) \) maps \( \gamma \) onto a closed curve \( \Gamma \) in the \( w \) -plane, and we find\n\n\[ {\int }_{\Gamma }\frac{dw}{w} = {\int }_{\gamma }\frac{{f}^{\prime }\left( z\right) }{f\left( z\right) }{dz} \]\n\nThe formula (32) has thus the following interpretation:\n\n\[ n\left( {\G... | Yes |
Theorem 11. Suppose that \( f\left( z\right) \) is analytic at \( {z}_{0}, f\left( {z}_{0}\right) = {w}_{0} \), and that \( f\left( z\right) - {w}_{0} \) has a zero of order \( n \) at \( {z}_{0} \) . If \( \varepsilon > 0 \) is sufficiently small, there exists a corresponding \( \delta > 0 \) such that for all a with ... | We can choose \( \varepsilon \) so that \( f\left( z\right) \) is defined and analytic for \( \left| {z - {z}_{0}}\right| \leqq \varepsilon \) and so that \( {z}_{0} \) is the only zero of \( f\left( z\right) - {w}_{0} \) in this disk. Let \( \gamma \) be the circle \( \left| {z - {z}_{0}}\right| = \varepsilon \) and \... | Yes |
Corollary 1. A nonconstant analytic function maps open sets onto open sets. | This is merely another way of saying that the image of every sufficiently small disk \( \left| {z - {z}_{0}}\right| < \varepsilon \) contains a neighborhood \( \left| {w - {w}_{0}}\right| < \delta \) . | No |
Corollary 2. If \( f\left( z\right) \) is analytic at \( {z}_{0} \) with \( {f}^{\prime }\left( {z}_{0}\right) \neq 0 \), it maps a neighborhood of \( {z}_{0} \) conformally and topologically onto a region. | From the continuity of the inverse function it follows in the usual way that the inverse function is analytic, and hence the inverse mapping is\n\n\n\nFIG. 4-7. Local correspondence.\n\n\n\n![28926e29-52cf-461e-ad56-... | Yes |
Theorem 12. (The maximum principle.) If \( f\left( z\right) \) is analytic and nonconstant in a region \( \Omega \), then its absolute value \( \left| {f\left( z\right) }\right| \) has no maximum in \( \Omega \) . | The proof is clear. If \( {w}_{0} = f\left( {z}_{0}\right) \) is any value taken in \( \Omega \), there exists a neighborhood \( \left| {w - {w}_{0}}\right| < \varepsilon \) contained in the image of \( \Omega \). In this neighborhood there are points of modulus \( > \left| {w}_{0}\right| \), and hence \( \left| {f\lef... | Yes |
Theorem 13. If \( f\left( z\right) \) is analytic for \( \left| z\right| < 1 \) and satisfies the conditions \( \left| {f\left( z\right) }\right| \; \leqq \;1,\;f\left( 0\right) \; = \;0,\; \) then \( \;\left| {f\left( z\right) \;}\right| \; \leqq \;\left| z\right| \; \) and \( \;\left| {{f}^{\prime }\left( 0\right) \;... | We apply the maximum principle to the function \( {f}_{1}\left( z\right) \) which is equal to \( f\left( z\right) /z \) for \( z \neq 0 \) and to \( {f}^{\prime }\left( 0\right) \) for \( z = 0 \) . On the circle \( \left| z\right| = r < 1 \) it is of absolute value \( \leqq 1/r \), and hence \( \left| {{f}_{1}\left( z... | Yes |
Theorem 14. A region \( \Omega \) is simply connected if and only if \( n\left( {\gamma, a}\right) = 0 \) for all cycles \( \gamma \) in \( \Omega \) and all points a which do not belong to \( \Omega \) . | The necessity of the condition is almost trivial. Let \( \gamma \) be any cycle in \( \Omega \) . If the complement of \( \Omega \) is connected, it must be contained in one of the regions determined by \( \gamma \), and inasmuch as \( \infty \) belongs to the complement this must be the unbounded region. Consequently ... | Yes |
Corollary 1. If \( f\left( z\right) \) is analytic in a simply connected region \( \Omega \), then (40) holds for all cycles \( \gamma \) in \( \Omega \) . | Before proving the theorem, we make an observation which ties up with the considerations in Section 1.3. As pointed out in that connection, the validity of (40) for all closed curves \( \gamma \) in a region means that the line integral of \( {fdz} \) is independent of the path, or that \( {fdz} \) is an exact differen... | No |
Corollary 2. If \( f\left( z\right) \) is analytic and \( \neq 0 \) in a simply connected region \( \Omega \) , then it is possible to define single-valued analytic branches of \( \log f\left( z\right) \) and \( \sqrt[n]{f\left( z\right) } \) in \( \Omega \) . | In fact, we know that there exists an analytic function \( F\left( z\right) \) in \( \Omega \) such that \( {F}^{\prime }\left( z\right) = {f}^{\prime }\left( z\right) /f\left( z\right) \) . The function \( f\left( z\right) {e}^{-F\left( z\right) } \) has the derivative zero and is therefore a constant. Choosing a poin... | Yes |
Theorem 17. Let \( f\left( z\right) \) be analytic except for isolated singularities \( {a}_{j} \) in a region \( \Omega \) . Then\n\n\[ \frac{1}{2\pi i}{\int }_{\gamma }f\left( z\right) {dz} = \mathop{\sum }\limits_{j}n\left( {\gamma ,{a}_{j}}\right) {\operatorname{Res}}_{z = {a}_{j}}f\left( z\right) \]\n\nfor any cyc... | In the applications it is frequently the case that each \( n\left( {\gamma ,{a}_{\jmath }}\right) \) is either 0 or 1. Then we have simply\n\n\[ \frac{1}{2\pi i}{\int }_{\gamma }f\left( z\right) {dz} = \mathop{\sum }\limits_{j}{\operatorname{Res}}_{z = {a}_{j}}f\left( z\right) \]\n\nwhere the sum is extended over all s... | No |
Theorem 19. If \( {u}_{1} \) and \( {u}_{2} \) are harmonic in a region \( \Omega \), then\n\n\[{\int }_{\gamma }{u}_{1} * d{u}_{2} - {u}_{2} * d{u}_{1} = 0\]\nfor every cycle \( \gamma \) which is homologous to zero in \( \Omega \) . | For \( {u}_{1} = 1,{u}_{2} = u \) the formula reduces to (58). In the classical notation (60) would be written as\n\n\[{\int }_{\gamma }\left( {{u}_{1}\frac{\partial {u}_{2}}{\partial n} - {u}_{2}\frac{\partial {u}_{1}}{\partial n}}\right) \left| {dz}\right| = 0.\] | No |
Theorem 20. The arithmetic mean of a harmonic function over concentric circles \( \left| z\right| = r \) is a linear function of \( \log r \) , | \[ \frac{1}{2\pi }{\int }_{\left| z\right| = r}{ud\theta } = \alpha \log r + \beta \] and if \( u \) is harmonic in a disk \( \alpha = 0 \) and the arithmetic mean is constant. In the latter case \( \beta = u\left( 0\right) \), by continuity, and changing to a new origin we find \[ u\left( {z}_{0}\right) = \frac{1}{2\p... | No |
Theorem 21. A nonconstant harmonic function has neither a maximum nor a minimum in its region of definition. Consequently, the maximum and the minimum on a closed bounded set \( E \) are taken on the boundary of \( E \) . | The proof is the same as for the maximum principle of analytic functions and will not be repeated. It applies also to the minimum for the reason that \( - u \) is harmonic together with \( u \) . In the case of analytic functions the corresponding procedure would have been to apply the maximum principle to \( 1/f\left(... | No |
Theorem 23. The function \( {P}_{U}\left( z\right) \) is harmonic for \( \left| z\right| < 1 \), and\n\n\[ \mathop{\lim }\limits_{{z \rightarrow {e}^{i{\theta }_{0}}}}{P}_{U}\left( z\right) = U\left( {\theta }_{0}\right) \]\n\nprovided that \( U \) is continuous at \( {\theta }_{0} \) . | We have already remarked that \( {P}_{U} \) is harmonic. To study the boundary behavior, let \( {C}_{1} \) and \( {C}_{2} \) be complementary arcs of the unit circle, and denote by \( {U}_{1} \) the function which coincides with \( U \) on \( {C}_{1} \) and vanishes on \( {C}_{2} \), by \( {U}_{2} \) the corresponding ... | Yes |
Theorem 24. Let \( {\Omega }^{ + } \) be the part in the upper half plane of a symmetric region \( \Omega \), and let \( \sigma \) be the part of the real axis in \( \Omega \) . Suppose that \( v\left( x\right) \) is continuous in \( {\Omega }^{ + } \cup \sigma \), harmonic in \( {\Omega }^{ + } \), and zero on \( \sig... | For the proof we construct the function \( V\left( z\right) \) which is equal to \( v\left( z\right) \) in \( {\Omega }^{ + },0 \) on \( \sigma \), and equal to \( - v\left( \bar{z}\right) \) in the mirror image of \( {\Omega }^{ + } \) . We have to show that \( V \) is harmonic on \( \sigma \) . For a point \( {x}_{0}... | Yes |
Theorem 1. Suppose that \( {f}_{n}\left( z\right) \) is analytic in the region \( {\Omega }_{n} \), and that the sequence \( \left\{ {{f}_{n}\left( z\right) }\right\} \) converges to a limit function \( f\left( z\right) \) in a region \( \Omega \), uniformly on every compact subset of \( \Omega \) . Then \( f\left( z\r... | The analyticity of \( f\left( z\right) \) follows most easily by use of Morera’s theorem (Chap. 4, Sec. 2.3). Let \( \left| {z - a}\right| \leqq r \) be a closed disk contained in \( \Omega \) ; the assumption implies that this disk lies in \( {\Omega }_{n} \) for all \( n \) greater than a certain \( {n}_{0} + \) If \... | Yes |
Theorem 2. If the functions \( {f}_{n}\left( z\right) \) are analytic and \( \neq 0 \) in a region \( \Omega \) , and if \( {f}_{n}\left( z\right) \) converges to \( f\left( z\right) \), uniformly on every compact subset of \( \Omega \), then \( f\left( z\right) \) is either identically zero or never equal to zero in \... | Suppose that \( f\left( z\right) \) is not identically zero. The zeros of \( f\left( z\right) \) are in any case isolated. For any point \( {z}_{0} \in \Omega \) there is therefore a number \( r > 0 \) such that \( f\left( z\right) \) is defined and \( \neq 0 \) for \( 0 < \left| {z - {z}_{0}}\right| \leqq r \) . In pa... | Yes |
Theorem 3. If \( f\left( z\right) \) is analytic in the region \( \Omega \), containing \( {z}_{0} \), then the representation\n\n\[ f\left( z\right) = f\left( {z}_{0}\right) + \frac{{f}^{\prime }\left( {z}_{0}\right) }{1!}\left( {z - {z}_{0}}\right) + \cdots + \frac{f{(}^{\left( n\right) }\left( {z}_{0}\right) }{n!}{\... | The radius of convergence of the Taylor series is thus at least equal to the shortest distance from \( {z}_{0} \) to the boundary of \( \Omega \) . It may well be larger, but if it is there is no guarantee that the series still represents \( f\left( z\right) \) at all points which are simultaneously in \( \Omega \) and... | Yes |
Theorem 4. Let \( \left\{ {b}_{\nu }\right\} \) be a sequence of complex numbers with \( \mathop{\lim }\limits_{{\nu \rightarrow \infty }}{b}_{\nu } = \infty \) , and let \( {P}_{\nu }\left( \zeta \right) \) be polynomials without constant term. Then there are functions which are meromorphic in the whole plane with pol... | We may suppose that no \( {b}_{\nu } \) is zero. The function \( {P}_{\nu }\left( {1/\left( {z - {b}_{\nu }}\right) }\right) \) is analytic for \( \left| z\right| < \left| {b}_{\nu }\right| \) and can thus be expanded in a Taylor series about the origin. We choose for \( {\mathbf{p}}_{\nu }\left( z\right) \) a partial ... | Yes |
Theorem 5. The infinite product \( \mathop{\prod }\limits_{1}^{\infty }\left( {1 + {a}_{n}}\right) \) with \( 1 + {a}_{n} \neq 0 \) converges simultaneously with the series \( \mathop{\sum }\limits_{1}^{\infty }\log \left( {1 + {a}_{n}}\right) \) whose terms represent the values of the principal branch of the logarithm... | The question of convergence of a product can thus be reduced to the more familiar question concerning the convergence of a series. It can be further reduced by observing that the series (16) converges absolutely at the same time as the simpler series \( \sum \left| {a}_{n}\right| \) . This is an immediate consequence o... | Yes |
Theorem 7. There exists an entire function with arbitrarily prescribed zeros \( {a}_{n} \) provided that, in the case of infinitely many zeros, \( {a}_{n} \rightarrow \infty \) . Every entire function with these and no other zeros can be written in the form\n\n\[ f\left( z\right) = {z}^{m}{e}^{g\left( z\right) }\mathop... | This theorem is due to Weierstrass. | No |
Theorem 9. For \( \sigma = \operatorname{Re}s > 1 \) ,\n\n\[ \frac{1}{\zeta \left( s\right) } = \mathop{\prod }\limits_{{n = 1}}^{\infty }\left( {1 - {p}_{n}^{-s}}\right) \] | According to Theorem 6 the infinite product converges uniformly for \( \sigma \geqq {\sigma }_{0} > 1 \) if the same is true of the series \( \mathop{\sum }\limits_{1}^{\infty }\left| {p}_{n}^{-s}\right| = \mathop{\sum }\limits_{1}^{\infty }{p}_{n}^{-\sigma } \) . Since the latter is obtained by omitting terms of \( \m... | Yes |
Theorem 10. For \( \sigma > 1 \) ,\n\n\[ \zeta \left( s\right) = - \frac{\Gamma \left( {1 - s}\right) }{2\pi i}{\int }_{C}\frac{{\left( -z\right) }^{s - 1}}{{e}^{z} - 1}{dz} \]\n\nwhere \( {\left( -z\right) }^{s - 1} \) is defined on the complement of the positive real axis as \( {e}^{\left( {z - 1}\right) \log \left( ... | The integral is obviously convergent. By Cauchy's theorem its value does not depend on the shape of \( C \) as long as \( C \) does not enclose any multiples of \( {2\pi i} \) . In particular, we are free to let \( r \) tend to zero. It is readily seen that the integral over the circle tends to zero with \( r \) . In t... | Yes |
Theorem 11.\n\n\[ \zeta \left( s\right) = {2}^{s}{\pi }^{s - 1}\sin \frac{\pi s}{2}\Gamma \left( {1 - s}\right) \zeta \left( {1 - s}\right) . \] | For the proof we make use of the path \( {C}_{n} \) in Fig. 5-1; we assume that the square part lies on the lines \( t = \pm \left( {{2n} + 1}\right) \pi \) and \( \sigma = \pm \left( {{2n} + 1}\right) \pi \) . The cycle \( {C}_{n} - C \) has winding number one about the points \( \pm {2m\pi i} \) with \( m = 1,\ldots,... | Yes |
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