Statement:
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If $g$ is holomorphic on an open set $S$ and $g(z) = (z - a)f(z)$ for all $z \in S - \{a\}$, then $f$ is holomorphic on $S$.
Suppose $g$ is holomorphic on a set $S$ and $a$ is an interior point of $S$. Suppose that for all $z \in S - \{a\}$, we have $g(z) = (z - a)f(z)$. Suppose that $f(a) = g'(a)$ and $g(a) = 0$. Then $f$ is holomorphic on $S$.
If $g$ is holomorphic on an open set $S$ and $g(z) = (z - a)f(z)$ for all $z \in S - \{a\}$, then $f$ is holomorphic on $S$.
If $g$ is analytic on $S$ and for every $z \in S$, there exists a $d > 0$ such that for all $w \in B(z, d) - \{a\}$, we have $g(w) = (w - a)f(w)$, then $F$ is analytic on $S$, where $F(z) = f(z) - g(a)/(z - a)$ if $z \neq a$ and $F(a) = g'(a)$.
If $g$ is analytic on $S$ and for all $z \in S$, there exists a $d > 0$ such that for all $w \in B(z, d) - \{a\}$, $g(w) = (w - a)f(w)$, then $f$ is analytic on $S$.
If $g$ is analytic on $S$ and $S \subseteq T$ where $T$ is open, and if $g$ has a pole at $a$ with residue $g(a)$, then $f(z) = \frac{g(z)}{z - a}$ is analytic on $S$.
If $g$ is an analytic function on a set $S$ and $S$ is a subset of an open set $T$ such that $g$ has a pole of order $1$ at $a$ and $g(a) = 0$, then $f$ is analytic on $S$.
If $F$ is a continuous function from $U \times U$ to $\mathbb{C}$ and $F(w)$ is contour integrable for all $w \in U$, then the function $w \mapsto \int_{a}^{b} F(w)$ is continuous on $U$.
If $f$ is holomorphic on an open set $U$, and $\gamma$ is a polynomial function such that the image of $\gamma$ is contained in $U - \{z\}$, then the contour integral of $f(w)/(w-z)$ along $\gamma$ is equal to $2\pi i$ times the winding number of $\gamma$ around $z$ times $f(z)$.
If $f$ is holomorphic on an open set $S$, and $\gamma$ is a closed path in $S$ that does not pass through $z$, then the integral of $f(w)/(w-z)$ along $\gamma$ is $2\pi i$ times the winding number of $\gamma$ around $z$ times $f(z)$.
If $f$ is holomorphic on an open set $S$, and $\gamma$ is a closed path in $S$, then $\int_\gamma f(z) dz = 0$.
If $f$ is holomorphic on an open set $S$ and $\gamma$ is a closed path in $S$, then $\int_\gamma f(z) dz = 0$.
If $S$ is a simply connected set and $g$ is a closed path in $S$ that does not pass through $z$, then the winding number of $g$ about $z$ is zero.
If $S$ is an open simply connected set, $f$ is a holomorphic function on $S$, and $g$ is a closed path in $S$, then $\int_g f(z) dz = 0$.
If $f$ is a holomorphic function on a convex open set $A$ such that $f(x) \neq 0$ for all $x \in A$, then there exists a holomorphic function $g$ on $A$ such that $e^{g(x)} = f(x)$ for all $x \in A$.
If $f$ is holomorphic on a ball $B(z,r)$ and continuous on the closed ball $\overline{B}(z,r)$, and if $f(B(z,r)) \subset B(y,B_0)$, then the $n$th derivative of $f$ at $z$ is bounded by $(n!)B_0/r^n$.
If $f$ is holomorphic on a ball $B(\xi, r)$ and continuous on the closed ball $\overline{B}(\xi, r)$, and if $f$ is bounded on the boundary of the ball, then the $n$th derivative of $f$ at $\xi$ is bounded by a constant times $B/r^n$.
If $f$ is a holomorphic function on $\mathbb{C}$ and there exist constants $A$ and $B$ such that $|f(z)| \leq B|z|^n$ for all $z$ with $|z| \geq A$, then $f$ is a polynomial of degree at most $n$.
If $f$ is a bounded holomorphic function on the complex plane, then $f$ is constant.
If $f$ is a power series with radius of convergence $r > 0$ and $f(\xi) = 0$, then there exists $s > 0$ such that $f(z) \neq 0$ for all $z \<in> \mathbb{C}$ with $|z - \xi| < s$ and $z \neq \xi$.
Suppose $f$ is holomorphic on the ball $B(z_0, r)$. Then the radius of convergence of the power series expansion of $f$ at $z_0$ is at least $r$. Moreover, the power series expansion of $f$ at $z_0$ converges to $f$ on the ball $B(z_0, r)$.
If $f$ is a complex power series such that $f(z) \neq 0$ for all $z$ with $|z| < r$, then the radius of convergence of $f^{-1}$ is at least $\min(r, \text{radius of convergence of } f)$. Moreover, if $|z| < \text{radius of convergence of } f$ and $|z| < r$, then $f^{-1}(z) = 1/f(z)$.
If $f$ and $g$ are two complex power series with positive convergence radii, and $g$ is nonzero on the ball of radius $r$, then the convergence radius of $f/g$ is at least $r$.
If $f$ and $g$ are power series with $g$ having a lower degree than $f$, and $g$ has a nonzero constant term, then the power series $f/g$ has a radius of convergence at least as large as the minimum of the radii of convergence of $f$ and $g$.
If $f$ is holomorphic on an open set $A$ containing $0$, then $f$ has a power series expansion about $0$.
The radius of convergence of the Taylor series of $\tan(z)$ at $c$ is at least $\pi / (2 \|c\|)$.
If $|z| < \pi / (2 |c|)$, then the Taylor series for $\tan(cz)$ at $0$ converges to $\tan(cz)$.
If $f$ is a function from a convex set $U$ to the space of functions from $[a, b]$ to $\mathbb{C}$ such that $f$ is differentiable in $x$ and $f(x)$ is integrable for all $x \in U$, then the function $g(x) = \int_a^b f(x)(t) dt$ is holomorphic on $U$.
The natural logarithm function is measurable.
If $f$ and $g$ are measurable functions from a measurable space $M$ to the real numbers, then the function $x \mapsto f(x)^{g(x)}$ is measurable.
If $f$ has a contour integral along the line segment from $a$ to $b$, then $f$ has a contour integral along the line segment from $a$ to the midpoint of $a$ and $b$, and $f$ has a contour integral along the line segment from the midpoint of $a$ and $b$ to $b$, then $f$ has a contour integral along the line segment from...
If $f$ is continuous on the closed line segment from $a$ to $b$, then the contour integral of $f$ along the line segment from $a$ to $b$ is equal to the sum of the contour integrals of $f$ along the line segments from $a$ to the midpoint of $a$ and $b$, and from the midpoint of $a$ and $b$ to $b$.
If $m$ is a real number and $a, b, c$ are points in the plane, then the integral of the function $f(x) = mx + d$ over the triangle with vertices $a, b, c$ is zero.
If $f$ has a contour integral along the path $a \rightarrow b \rightarrow c \rightarrow d$, then the contour integral of $f$ along the path $a \rightarrow b$ plus the contour integral of $f$ along the path $b \rightarrow c$ plus the contour integral of $f$ along the path $c \rightarrow d$ is equal to the contour integr...
If $f$ has a contour integral along the path $a \rightarrow b \rightarrow c \rightarrow d \rightarrow e$, then the sum of the contour integrals of $f$ along the paths $a \rightarrow b$, $b \rightarrow c$, $c \rightarrow d$, and $d \rightarrow e$ is equal to the contour integral of $f$ along the path $a \rightarrow b \r...
If the norm of the sum of two numbers is at least $e$, then the norm of one of the numbers is at least $e/2$.
If $e \leq \|a + b + c + d\|$, then at least one of $\|a\|$, $\|b\|$, $\|c\|$, or $\|d\|$ is at least $e/4$.
Suppose $f$ is a continuous function defined on the convex hull of three points $a$, $b$, and $c$. If the distances between these points are bounded by $K$, and if the integral of $f$ around the triangle formed by these points is at least $eK^2$, then there exists a triangle with vertices $a'$, $b'$, and $c'$ such that...
If $x$ and $y$ are points in the convex hull of $a$, $b$, and $c$, then the distance between $x$ and $y$ is less than or equal to the distance between any two of $a$, $b$, and $c$.
If $f$ is a continuous function on a set $S$ and $f$ is differentiable at a point $x \in S$, then there exists a positive number $k$ such that for any triangle with vertices $a$, $b$, and $c$ and with $x$ in the interior of the triangle, if the triangle is contained in $S$ and the sides of the triangle are less than $k...
$f$ is at $n$.
If $f(n+1)$ follows $f(n)$, then $f(n+2)$ follows $f(n+1)$, and so on.
Suppose that we have a predicate $At(x,y,z,n)$ that holds for some $x_0, y_0, z_0$ and $n=0$, and that for any $x,y,z,n$ such that $At(x,y,z,n)$ holds, there exists $x',y',z'$ such that $At(x',y',z',n+1)$ holds and $Follows(x',y',z',x,y,z)$ holds. Then there exist functions $f,g,h$ such that $f(0) = x_0$, $g(0) = y_0$,...
If $f$ is holomorphic on the triangle with vertices $a$, $b$, and $c$, then $\int_a^b f(z) dz + \int_b^c f(z) dz + \int_c^a f(z) dz = 0$.
If $f$ is a continuous function on the convex hull of three points $a$, $b$, and $c$, and $c$ is a convex combination of $a$ and $b$, then the sum of the contour integrals of $f$ along the three sides of the triangle is zero.
If $f$ is a continuous function on the triangle with vertices $a$, $b$, and $c$, and $c$ is a multiple of $b - a$, then the sum of the integrals of $f$ around the three sides of the triangle is zero.
If $f$ is a continuous function defined on the triangle with vertices $a$, $b$, and $c$, and $f$ is holomorphic on the interior of the triangle, then $\int_\gamma f(z) dz = 0$, where $\gamma$ is the boundary of the triangle.
If $f$ is continuous on the triangle with vertices $a$, $b$, and $c$, and if $f$ is differentiable at all points of the interior of the triangle except for a finite set $S$, then $\int_\gamma f(z) dz = 0$, where $\gamma$ is the triangle with vertices $a$, $b$, and $c$.
If $a \in S$, $b,c \in S$, and $S$ is starlike with respect to $a$, then the convex hull of $a,b,c$ is contained in $S$.
If $f$ is a continuous function on an open set $S$ and $a \in S$, and if the integral of $f$ around any triangle with vertices in $S$ is zero, then $f$ is a primitive of $f$ on $S$.
Suppose $f$ is a continuous function defined on a starlike set $S$ and $f$ is complex differentiable on $S$ except for a finite number of points. Then there exists a function $g$ such that $g'(x) = f(x)$ for all $x \<in> S$.
If $S$ is an open, starlike set, $f$ is a continuous function on $S$ that is differentiable on $S$ except for a finite set $k$, $g$ is a closed path in $S$, and $f$ is holomorphic on $S - k$, then $\int_g f(z) dz = 0$.
If $f$ is holomorphic on an open, starlike set $S$, and $g$ is a closed path in $S$, then $\int_g f(z) dz = 0$.
If $f$ is a continuous function on a convex set $S$ and $a \in S$, then the function $g(x) = \int_a^x f(t) dt$ is differentiable at every point $x \in S$ and $g'(x) = f(x)$.
If $f$ is a continuous function on a convex set $S$ and $f$ has a contour integral of $0$ around every triangle in $S$, then $f$ is the derivative of some function $g$ on $S$.
Suppose $f$ is a continuous function defined on a convex set $S$ and $f$ is differentiable at every point of $S$ except for a finite set $K$. Then there exists a function $g$ such that $g$ is differentiable at every point of $S$ and $g'(x) = f(x)$ for all $x \in S$.
If $f$ is a holomorphic function on a convex open set $S$, then there exists a function $g$ such that $g'(x) = f(x)$ for all $x \in S$.
If $f$ is a continuous function on a convex set $S$, and $f$ is differentiable on the interior of $S$ except for a finite number of points, then the integral of $f$ along any closed path in $S$ is zero.
If $f$ is holomorphic on a convex set $S$ and $g$ is a closed path in $S$, then $\int_g f(z) dz = 0$.
If $f$ is a continuous function on the closed disc of radius $e$ centered at $a$, and $f$ is differentiable on the open disc of radius $e$ centered at $a$ except at finitely many points, then the integral of $f$ around the boundary of the closed disc is zero.
If $f$ is holomorphic on a disc, and $g$ is a closed path in the disc, then $\int_g f(z) dz = 0$.
If $g$ is piecewise differentiable on $[a,b]$ and $f$ is differentiable on $g([a,b])$, then the function $x \mapsto f'(g(x)) \cdot g'(x)$ is integrable on $[a,b]$.
If $g$ is piecewise differentiable on $[a,b]$ and $f$ is differentiable at every point in the image of $g$, then the composition $f \circ g$ is integrable on $[a,b]$.
If $f$ is locally integrable on a set $S$ and $g$ is a path in $S$, then $f$ is integrable along $g$.
If $f$ is a continuous function on an open set $S$ and $f$ is holomorphic on $S$ except for a finite number of points, then $f$ is contour integrable on any path in $S$.
If $f$ is holomorphic on an open set $S$ and $g$ is a path in $S$, then $f$ is contour integrable along $g$.
If $z$ is not in $S$, then the function $f(w) = 1/(w-z)$ is continuous on $S$.
If $g$ is a valid path and $z$ is not in the image of $g$, then the function $1/(w-z)$ is contour-integrable along $g$.
If $p$ is a path in an open set $S$, then there is a $\delta > 0$ such that if $g$ and $h$ are paths with $g(0) = h(0)$, $g(1) = h(1)$, and $g(t)$ and $h(t)$ are within $\delta$ of $p(t)$ for all $t \in [0,1]$, then $g$ and $h$ are also in $S$, and the contour integrals of $g$ and $h$ are equal for any holomorphic func...
If $p$ is a path in an open set $S$, then there is a positive number $d$ such that if $g$ and $h$ are paths with $g(0) = h(0)$ and $g(1) = h(1)$ and $g(t)$ and $h(t)$ are within $d$ of $p(t)$ for all $t$, then $g$ and $h$ are paths in $S$ and $\int_g f = \int_h f$ for all holomorphic functions $f$ on $S$. Similarly, if...
If $S$ is an open set and $g$ is a valid path with image contained in $S$, then there exists a constant $L$ such that for any holomorphic function $f$ on $S$ with $|f(z)| \leq B$ for all $z \in S$, we have $|\int_g f(z) dz| \leq LB$.
If $g$ and $h$ are homotopic paths or loops in an open set $S$, then $\int_g f = \int_h f$ for any holomorphic function $f$ on $S$.
If two paths are homotopic in an open set, then the contour integrals of a holomorphic function along the paths are equal.
If two paths are homotopic, then their contour integrals are equal.
If $f$ has a contour integral along a path $h$, and $f$ is contour integrable along a path $g$, and the contour integrals of $f$ along $g$ and $h$ are equal, then $f$ has a contour integral along $g$.
If $f$ is holomorphic on an open set $S$, $g$ is a valid path in $S$, and $g$ is homotopic to a constant path, then $f$ has contour integral $0$ along $g$.
The complex number $z$ is equal to the complex number with real part $Re(z)$ and imaginary part $Im(z)$.
If the real and imaginary parts of two complex numbers are equal, then the complex numbers are equal.
Two complex numbers are equal if and only if their real and imaginary parts are equal.
The real part of the quotient of two complex numbers is equal to the quotient of the real parts of the two complex numbers.
The imaginary part of the quotient of two complex numbers is given by the formula $\frac{\text{Im}(x) \cdot \text{Re}(y) - \text{Re}(x) \cdot \text{Im}(y)}{\text{Re}(y)^2 + \text{Im}(y)^2}$.
The quotient of two complex numbers is given by the following formula.
The real part of $x^2$ is equal to $(\text{Re}(x))^2 - (\text{Im}(x))^2$.
The imaginary part of $x^2$ is twice the product of the real and imaginary parts of $x$.
If $x$ is a real number, then $\Re(x^n) = x^n$.
If $x$ is a real number, then $x^n$ is a real number.
The real part of a natural number is the natural number itself.
The imaginary part of a natural number is zero.
The real part of an integer is the integer itself.
The imaginary part of an integer is zero.
The real part of a numeral is the numeral itself.
The imaginary part of a numeral is zero.
The real part of a real number is the number itself.
The imaginary part of a real number is zero.
The real part of a complex number divided by a numeral is the real part of the complex number divided by the numeral.
The imaginary part of a complex number divided by a numeral is the imaginary part of the complex number divided by the numeral.
The real part of a complex number divided by a natural number is equal to the real part of the complex number divided by the natural number.
The imaginary part of a complex number divided by a natural number is equal to the imaginary part of the complex number divided by the natural number.
If $z$ is a real number, then $\operatorname{Re}(z) = z$.
The real part of the factorial of a natural number is the factorial of that natural number.