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Example 1 Decide which, if any, of the matrices\n\n\[ \n{\mathbf{A}}_{1} = \left( \begin{array}{ll} 0 & 4 \\ i & 0 \end{array}\right) ,\;{\mathbf{A}}_{2} = \left( \begin{matrix} 8 & 0 \\ - {2i} & - 8 \end{matrix}\right) ,\;{\mathbf{A}}_{3} = \left( \begin{matrix} - {4i} & 0 \\ 0 & 1 \end{matrix}\right) \n\]\n\nare asso... | ## Solution\n\n(a) Every matrix associated with this \( M \) is a non-zero multiple of the matrix\n\n\[ \n\left( \begin{matrix} - {3i} & 2 \\ 1 & - {3i} \end{matrix}\right) \n\]\n\nHence none of the three given matrices is associated with \( M \) .\n\n(b) Every matrix associated with this \( M \) is a non-zero multiple... | Yes |
Determine the composite \( {M}_{1} \circ {M}_{2} \), where \( {M}_{1} \) and \( {M}_{2} \) are the Möbius transformations defined by\n\n\[ \n{M}_{1}\left( z\right) = \frac{{iz} + 1}{{2z} - 2}\;\text{ and }\;{M}_{2}\left( z\right) = \frac{z + i}{{2z} - 1}.\n\] | Solution Since \( {M}_{1} \) and \( {M}_{2} \) are one-one mappings of \( \widehat{\mathbb{C}} \) onto \( \widehat{\mathbb{C}} \), the same\n\n---\n\nStrictly speaking, we should check our convention for the exceptional points separately. For example, by convention \( {M}_{2}\left( \infty \right) = \frac{1}{2} \), so t... | Yes |
Problem 3 Write down matrices \( {\mathbf{A}}_{1} \) and \( {\mathbf{A}}_{2} \) associated with the Möbius transformations \( {M}_{1} \) and \( {M}_{2} \) defined in Example 2. Calculate the product \( {\mathbf{A}}_{1}{\mathbf{A}}_{2} \), and compare it with the Möbius transformation \( {M}_{1} \circ {M}_{2} \) . | In Example 2, the composite of two Möbius transformations \( {M}_{1} \) and \( {M}_{2} \) turned out to be another Möbius transformation. The solution to Problem 3 demonstrates that the product of matrices associated with \( {M}_{1} \) and \( {M}_{2} \) is a matrix associated with \( {M}_{1} \circ {M}_{2} \). | No |
Theorem 6 Composition of Möbius Transformations\n\nLet \( {M}_{1} \) and \( {M}_{2} \) be Möbius transformations with associated matrices \( {\mathbf{A}}_{1} \) and \( {\mathbf{A}}_{2} \), respectively. Then \( {M}_{1} \circ {M}_{2} \) is a Möbius transformation with an associated matrix \( {\mathbf{A}}_{1}{\mathbf{A}}... | Proof Let \( {M}_{1} \) and \( {M}_{2} \) be defined by\n\n\[ \n{M}_{1}\left( z\right) = \frac{{az} + b}{{cz} + d}\;\text{ and }\;{M}_{2}\left( z\right) = \frac{{ez} + f}{{gz} + h}.\n\]\n\nSince \( {M}_{1} \) and \( {M}_{2} \) are one-one mappings of \( \widehat{\mathbb{C}} \) onto \( \widehat{\mathbb{C}} \), the same ... | Yes |
Example 3 Use the strategy to determine the composite \( {M}_{1} \circ {M}_{2} \) of the Möbius transformations\n\n\[ \n{M}_{1}\left( z\right) = \frac{{3z} + 1}{{iz} - 2}\;\text{ and }\;{M}_{2}\left( z\right) = \frac{{2iz} + 3}{z - 2}.\n\] | Solution The Möbius transformations \( {M}_{1} \) and \( {M}_{2} \) have associated matrices\n\n\[ \n{\mathbf{A}}_{1} = \left( \begin{array}{rr} 3 & 1 \\ i & - 2 \end{array}\right) \;\text{ and }\;{\mathbf{A}}_{2} = \left( \begin{array}{rr} {2i} & 3 \\ 1 & - 2 \end{array}\right) ,\n\]\n\nrespectively. It follows that a... | Yes |
Problem 5 Let \( {M}_{1} \) and \( {M}_{2} \) be the Möbius transformations defined by\n\n\[ \n{M}_{1}\left( z\right) = \frac{z - i}{{iz} + 2}\;\text{ and }\;{M}_{2}\left( z\right) = \frac{{2z} + i}{-{iz} + 1}.\n\]\n\nUse the strategy to determine the composite \( {M}_{1} \circ {M}_{2} \) . | In the solution to Problem 5 you saw that the composite \( {M}_{1} \circ {M}_{2} \) of the Möbius transformations\n\n\[ \n{M}_{1}\left( z\right) = \frac{z - i}{{iz} + 2}\;\text{ and }\;{M}_{2}\left( z\right) = \frac{{2z} + i}{-{iz} + 1}\n\]\n\nis the identity function on \( \widehat{\mathbb{C}} \) . This shows that \( ... | Yes |
The inverse of the Möbius transformation\n\n\[ M\left( z\right) = \frac{{az} + b}{{cz} + d} \]\n\nis also a Möbius transformation, and it may be written in the form\n\n\[ {M}^{-1}\left( z\right) = \frac{{dz} - b}{-{cz} + a}. \] | For example, the inverse of the Möbius transformation\n\n\[ M\left( z\right) = \frac{{iz} + 1}{{2z} - 2} \]\n\nis given by\n\n\[ {M}^{-1}\left( z\right) = \frac{-{2z} - 1}{-{2z} + i} = \frac{{2z} + 1}{{2z} - i}. \] | Yes |
Theorem 8 The set of all Möbius transformations forms a group under composition of functions. | Proof We show that the four group axioms hold.\n\nGl closure By Theorem 6, the composite of two Möbius transformations is itself a Möbius transformation.\n\nG2 IDENTITY The identity is the Möbius transformation given by \( M\left( z\right) = \frac{{1z} + 0}{{0z} + 1} \) .\n\nG3 INVERSES By Theorem 7, every Möbius trans... | Yes |
Theorem 9 Every inversion \( t \) has the form \( t\left( z\right) = M\left( \bar{z}\right) \), where \( M \) is a Möbius transformation. | Proof If \( t \) is an inversion of \( \widehat{\mathbb{C}} \) in an extended line, then by Theorem 1 of Section 5.2 it must have the form \( t\left( z\right) = a\bar{z} + b \), with \( t\left( \infty \right) = \infty \) . It follows that \( t\left( z\right) = M\left( \bar{z}\right) \), where \( M \) is the Möbius tran... | Yes |
Theorem 10 Every inversive transformation \( t \) can be represented in \( \widehat{\mathbb{C}} \) by one of the formulas\n\n\[ \nt\left( z\right) = \frac{{az} + b}{{cz} + d}\;\text{ or }\;t\left( z\right) = \frac{a\bar{z} + b}{c\bar{z} + d},\n\]\n\nwhere \( a, b, c, d \in \mathbb{C} \) and \( {ad} - {bc} \neq 0 \) . | Proof We first show that the composite of two inversions \( {t}_{1} \) and \( {t}_{2} \) is a Möbius transformation. By Theorem 9 above, we can write \( {t}_{1}\left( z\right) = {M}_{1}\left( \bar{z}\right) \) and \( {t}_{2}\left( z\right) = \) \( {M}_{2}\left( \bar{z}\right) \), where \( {M}_{1} \) and \( {M}_{2} \) a... | Yes |
Example 4 Use the strategy to find the image of the unit circle \( \mathcal{C} \) under the inversive transformation defined by\n\n\[ t\left( z\right) = \frac{\bar{z} + i}{\bar{z} - 1} \] | Solution We first pick three distinct points on the unit circle. There is no definite rule about which points should be chosen, so to keep the calculations simple, we pick the points \( 1, i \) and -1, as shown below. Now\n\n\[ t\left( 1\right) = \infty, t\left( i\right) = \frac{\bar{i} + i}{\bar{i} - 1} = 0\;\text{ an... | Yes |
For each set of three points given below, determine a Möbius transformation \( M \) that maps the points to \( 0,1,\infty \), respectively.\n\n(a) \( \frac{1}{2}, - 1,3 \) | (a) To ensure that \( M\left( \frac{1}{2}\right) = 0 \) and \( M\left( 3\right) = \infty \) we let \( M \) have the form\n\n\[ M\left( z\right) = K\frac{z - \frac{1}{2}}{z - 3} \]\n\nThis has the form\n\nfor some complex number \( K \) . Since \( M\left( {-1}\right) = 1 \), we must have\n\n\[ 1 = K\frac{-1 - \frac{1}{2... | Yes |
Theorem 1 The Fundamental Theorem of Inversive Geometry\n\nLet \( {z}_{1},{z}_{2},{z}_{3} \) and \( {w}_{1},{w}_{2},{w}_{3} \) be two sets of three points in the extended complex plane \( \widehat{\mathbb{C}} \) . Then there is a unique Möbius transformation \( M \) which maps \( {z}_{1} \) to \( {w}_{1},{z}_{2} \) to ... | Proof According to the above strategy there is a Möbius transformation \( {M}_{1} \) which maps the points \( {z}_{1},{z}_{2},{z}_{3} \) to the points \( 0,1,\infty \), respectively. Similarly, there is a Möbius transformation \( {M}_{2} \) which maps the points \( {w}_{1},{w}_{2},{w}_{3} \) to the points \( 0,1,\infty... | Yes |
Example 2 Find the Möbius transformation \( M \) which maps the points \( i,\infty ,3 \) to the points \( \frac{1}{2}, - 1,3 \), respectively. | Solution We follow the steps in the above strategy.\n\n1. From part (c) of Example 1, we know that the Möbius transformation \( {M}_{1} \) which maps the points \( i,\infty ,3 \) to the points \( 0,1,\infty \), respectively, is given by\n\n\[ \n{M}_{1}\left( z\right) = \frac{z - i}{z - 3}.\n\]\n\n2. Also, from part (a)... | No |
Example 3 Determine whether the four points \( i,1 + {4i},3,4 + {3i} \) lie on a generalized circle. | Solution First, we determine the Möbius transformation \( M \) that maps \( i \) , \( 1 + {4i},3 \) to \( 0,1,\infty \), respectively. Following the strategy in the previous subsection, we observe that this transformation must be of the form\n\n\[ M\left( z\right) = K\frac{z - i}{z - 3} \]\n\nfor some complex number \(... | Yes |
Theorem 2 Let \( {C}_{1} \) and \( {C}_{2} \) be generalized circles in the extended complex plane. Then there is a Möbius transformation that maps \( {C}_{1} \) onto \( {C}_{2} \) . | Proof Let \( a, b, c \) be any three points on \( {C}_{1} \) and let \( d, e, f \) be any three points on \( {C}_{2} \) . By the Fundamental Theorem of Inversive Geometry, there is a Möbius transformation \( M \) that maps \( a, b, c \) to \( d, e, f \), respectively.\n\nSince \( M \) maps generalized circles to genera... | Yes |
Theorem 1 Apollonian Circles Theorem\n\nLet \( A \) and \( B \) be two distinct points in the plane, and let \( k \) be a positive real\n\nnumber other than 1 . Then the locus of points \( P \) that satisfy \( {PA} : {PB} = k : 1 \) is a circle whose centre lies on the line through \( A \) and \( B \) . | ## First Proof\n\nTo keep the algebra simple, we introduce \( x \) - and \( y \) -axes into the plane such that \( A \) and \( B \) have coordinates \( \\left( {-a,0}\\right) \) and \( \\left( {a,0}\\right) \), respectively, where \( a > 0 \) .\n\nNow fix a value of \( k > 0, k \\neq 1 \), and let \( C \) be the locus ... | Yes |
For the Apollonian family defined by the point circles \( \left( {-1,0}\right) \) and \( \left( {1,0}\right) \), determine the equation of the circle in the family that passes through the point \( \left( {2,1}\right) \) . | Let \( P \) be a point \( \left( {x, y}\right) \) in the plane whose distance from the point \( \left( {-1,0}\right) \) is \( k \) times its distance from the point \( \left( {1,0}\right) \) . Then, if we use the Euclidean formula for distance between points in the plane, we obtain\n\n\[ \n{\left( x + 1\right) }^{2} + ... | Yes |
Theorem 2 Let \( A \) and \( B \) be distinct points in the plane, and let \( t \) be the inversion in the circle with centre \( A \) and radius 1 . Then the Apollonian family of circles defined by the point circles \( A \) and \( B \) is mapped by \( t \) to the family of all concentric circles with centre \( t\left( ... | For \( t \) is self-inverse. | No |
Theorem 3 Coaxal Circles Theorem\n\nLet \( A \) and \( B \) be distinct points in the plane. Let \( \mathcal{F} \) be the Apollonian family defined by the point circles \( A \) and \( B \), and let \( \mathcal{G} \) be the family of all generalized circles through \( A \) and \( B \) . Then every member of \( \mathcal{... | Proof Let \( t \) be the inversion in the unit circle with centre \( A \) . By Theorem 2, the Apollonian family \( \mathcal{F} \) is mapped by \( t \) to the family of concentric circles with centre \( t\left( B\right) \) .\n\nNow let \( C \) be an arbitrary member of \( \mathcal{G} \) . Since \( C \) is a generalized ... | Yes |
Theorem 4 Two points \( A \) and \( B \) in the extended complex plane are inverse points with respect to a generalized circle \( C \) if and only if every generalized circle through \( A \) and \( B \) meets \( C \) at right angles. | Proof We first show that \( A \) and \( B \) are inverse points if every generalized circle through \( A \) and \( B \) meets \( C \) at right angles.\n\nLet \( {C}_{1} \) be any generalized circle through \( A \) and \( B \) that meets \( C \) at right\n\nangles, at points \( R \) and \( S \) . Now invert the figure \... | Yes |
Corollary 1 If \( C \) is an Apollonian circle defined by the point circles \( A \) and \( B \), then \( A \) and \( B \) are inverse points with respect to \( C \) . | Proof Let \( C \) be an Apollonian circle with respect to the point circles \( A \) and \( B \) . Then by the Coaxal Circles Theorem, every generalized circle through \( A \) and \( B \) meets \( C \) at right angles. By Theorem 4, \( A \) and \( B \) are inverse points with respect to \( C \) . | Yes |
Corollary 2 Let \( A \) and \( B \) be inverse points with respect to a generalized circle \( C \), and let \( t \) be an inversive transformation. Then \( t\left( A\right) \) and \( t\left( B\right) \) are inverse points with respect to the generalized circle \( t\left( C\right) \) . | \n\nLet \( {C}_{1} \) be an arbitrary generalized circle through \( t\left( A\right) \) and \( t\left( B\right) \) . Then \( {t}^{-l}\left( {C}_{1}\right) \) is a generalized circle that passes through \( A \) and \(... | Yes |
Theorem 5 Let \( \mathcal{F} \) be the family of all generalized circles that have \( A \) and \( B \) as inverse points. Then \( \mathcal{F} \) is either a concentric family of circles with centres \( A \) or \( B \), or the Apollonian family of circles with point circles \( A \) and \( B \) . | Proof First, suppose that either \( A \) or \( B \) is the point \( \infty \) ; to be definite, assume \( A = \infty \) . Then each circle in \( \mathcal{F} \) has \( \infty \) and \( B \) as inverse points. But this can happen only if \( B \) is the centre of each circle in \( \mathcal{F} \) ; in other words \( \mathc... | Yes |
Example 2 Let \( {C}_{1},{C}_{2} \) and \( {C}_{3} \) be circles in the plane such that \( {C}_{2} \) and \( {C}_{3} \) touch at a point \( P,{C}_{1} \) and \( {C}_{3} \) touch at a point \( Q \), and \( {C}_{1} \) and \( {C}_{2} \) touch at a point \( R \) . Let \( C \) be the circle that passes through \( P, Q \) and... | Solution Let \( t \) be an inversion in a circle with centre \( R \) . Then \( t\left( {C}_{1}\right) \) and \( t\left( {C}_{2}\right) \) are straight lines. Moreover, since \( {C}_{1} \) and \( {C}_{2} \) do not meet at any point other than \( P \), it follows that \( t\left( {C}_{1}\right) \) and \( t\left( {C}_{2}\r... | Yes |
Theorem 6 The Concentricity Theorem\n\nLet \( {C}_{1} \) and \( {C}_{2} \) be any two non-intersecting circles in the plane. Then there is a inversive transformation that maps \( {C}_{1} \) and \( {C}_{2} \) onto a pair of concentric circles. | Proof If the circles are concentric then there is nothing to prove, so we may assume they are not concentric.\n\nStep 1 Pick a point \( O \) on the circle \( {C}_{1} \), and invert both circles in the circle of unit radius with centre \( O \) . Under this inversion \( {C}_{1} \) maps to a line, \( {C}_{1}^{\prime } \),... | Yes |
Theorem 7 Two Apollonian Circles Theorem\n\nLet \( {C}_{1} \) and \( {C}_{2} \) be two non-concentric circles that do not intersect. Then there is a unique Apollonian family of circles that contains \( {C}_{1} \) and \( {C}_{2} \) . | Proof By the Concentricity Theorem there is an inversive transformation \( t \) Theorem 6 above that maps \( {C}_{1} \) and \( {C}_{2} \) onto a pair of concentric circles \( t\left( {C}_{1}\right) \) and \( t\left( {C}_{2}\right) \) . Let \( O \) be the common centre of these circles, and let \( \mathcal{G} \) be the ... | Yes |
Lemma 1 Let \( a \) and \( b \) be points in \( \mathbb{C} - \{ O\} \) . Then, under inversion in a circle \( C \) with centre \( O \) and radius \( r \) the distance between the images of \( a \) and \( b \) is \( \frac{{r}^{2}}{\left| a\right| \cdot \left| b\right| }\left| {a - b}\right| \) . | Proof Inversion in \( C \) may be represented in the complex plane by the transformation \[ t\left( z\right) = \frac{{r}^{2}}{\bar{z}},\;\text{ where }z \in \mathbb{C} - \{ O\} . \] It follows from this formula that \[ t\left( a\right) - t\left( b\right) = \frac{{r}^{2}}{\bar{a}} - \frac{{r}^{2}}{\bar{b}} \] \[ = \frac... | Yes |
Theorem 9 Ptolemy's Theorem\n\nLet \( A, B, C, D \) be the vertices (in order round a circle) of a quadrilateral \( {ABCD} \) inscribed in a circle. Then\n\n\[ {AD} \cdot {BC} + {AB} \cdot {CD} = {AC} \cdot {BD}. \] | Proof Let the transformation \( t \) be inversion of the figure \( {ABCD} \) in the circle \( \mathcal{C} \) with centre \( A \) and radius 1 . This inversion maps \( A \) to \( \infty \), the circle \( \mathcal{C} \) to an extended line (which we will denote by \( \ell \) ), and the points \( B, C, D \) on \( \mathcal... | Yes |
Problem 3 Prove that the height above the \( x \) -axis of the centre of the \( n \) -th circle, \( {C}_{n} \) say, in the chain of circles constructed in Theorem 10 is \( n \) times the diameter of that circle. | Hint: Invert the figure in a circle with centre \( A \) that intersects \( {C}_{n} \) at right angles. | No |
Problem 1 Sketch the following parts of generalized circles, and determine which of them are \( d \) -lines. | \[ {\ell }_{1} = \{ \left( {x, y}\right) : y = {3x}\} \cap \mathcal{D} \] \[ {\ell }_{2} = \{ \left( {x, y}\right) : {3x} + y = 1\} \cap \mathcal{D} \] \[ {\ell }_{3} = \left\{ {\left( {x, y}\right) : {x}^{2} + {y}^{2} + {2x} + {2y} + 1 = 0}\right\} \cap \mathcal{D} \] | No |
Lemma 1 The equation of a \( d \) -line \( \ell \) is of one of the following forms:\n\n\( {ax} + {by} = 0,\; \) where \( a \) and \( b \) are not both zero;\n\n\[ \n{x}^{2} + {y}^{2} + {fx} + {gy} + 1 = 0,\;\text{ where }{f}^{2} + {g}^{2} > 4. \n\] | Proof If a \( d \) -line \( \ell \) is (part of) a Euclidean line through the origin, then certainly its equation is of the form \( {ax} + {by} = 0 \), where \( a \) and \( b \) are not both zero.\n\nThe other possibility is that the \( d \) -line \( \ell \) is (part of) a Euclidean circle \( C \) that intersects the b... | Yes |
Theorem 1 Let \( \ell \) be a \( d \) -line that is part of a Euclidean circle \( C \) . Then inversion in \( C \) maps \( \mathcal{C} \) onto \( \mathcal{C} \), and \( \mathcal{D} \) onto \( \mathcal{D} \) . | Proof Let \( C \) meet \( \mathcal{C} \) at the points \( A \) and \( B \) . Under inversion in \( C, A \) and \( B \) map to themselves. Since inversion preserves angles, the circle \( \mathcal{C} \) maps onto some circle that meets \( C \) at right angles at the points \( A \) and \( B \) . There is only one such cir... | Yes |
Let \( A \) be a point of \( \mathcal{D} \) other than the origin \( O \) . Then there exists a \( d \) -line \( \ell \) such that hyperbolic reflection in \( \ell \) maps \( A \) to \( O \) . | Proof We seek a \( d \) -line \( \ell \) which is part of a Euclidean circle with centre \( R \) , say, such that inversion in this circle maps \( A \) to \( O \) . Suppose that this circle meets \( \mathcal{C} \) at the point \( P \) .\n\nThe condition that this inversion maps \( A \) to \( O \) is\n\n\[ \n{RO} \cdot ... | Yes |
Theorem 3 Let \( A \) be a point of \( \mathcal{D} \). Then there exist infinitely many \( d \)-lines through \( A \). | Proof Let \( A \) be the origin. As we said above, each of the infinitely many diameters of \( \mathcal{D} \) passes through the origin, and each of these diameters is a \( d \)-line.\n\nIf \( A \) is not the origin, then by the Origin Lemma, there is a hyperbolic transformation, \( r \) say, that maps \( A \) to the o... | Yes |
Theorem 4 Let \( A \) and \( B \) be any two distinct points of \( \mathcal{D} \). Then there exists a unique \( d \)-line \( \ell \) through \( A \) and \( B \). | Proof (Existence) By the Origin Lemma, there is a hyperbolic transformation \( r \) that maps \( A \) to the origin \( O \); let the image of \( B \) under \( r \) be the point \( {B}^{\prime } \) of \( \mathcal{D} \).\n\n and \( {A}_{2} \) be any two points of \( \mathcal{D} \), and let \( {\ell }_{1} \) and \( {\ell }_{2} \) be \( d \) -lines through \( {A}_{1} \) and \( {A}_{2} \), respectively. Then there is a hyperbolic transformation which maps \( {A}_{1} \) to \( {A}_{2} \) and \( {\ell }_{1} \) to \( {... | Proof By the Origin Lemma, there are hyperbolic transformations \( {r}_{1} \) and \( {r}_{2} \) which map \( {A}_{1} \) and \( {A}_{2} \), respectively, to the origin \( O \) . Let the images of \( {\ell }_{1} \) and \( {\ell }_{2} \) under \( {r}_{1} \) and \( {r}_{2} \) be the \( d \) -lines \( {\ell }_{1}^{\prime } ... | Yes |
Let \( A, B \) and \( C \) be points of \( \mathcal{C} \) such that the \( d \) -lines \( {AB} \) and \( {BC} \) are parallel. Let \( r \) be a hyperbolic transformation under which the images of \( {AB} \) and \( {BC} \) are the \( d \) -lines \( {A}^{\prime }{B}^{\prime } \) and \( {B}^{\prime }{C}^{\prime } \), resp... | Solution First, we show that \( {A}^{\prime }{B}^{\prime } \) and \( {B}^{\prime }{C}^{\prime } \) do not meet in \( \mathcal{D} \) . If they do, let them meet at the point \( {P}^{\prime } \), say. Then the point \( {P}^{\prime } \) corresponds to a point \( P \), say, on the \( d \) -lines \( {AB} \) and \( {BC} \) ;... | Yes |
Show that inversion in \( C \) maps \( {\ell }^{\prime } \) to itself. | Inversion in \( C \) exchanges \( P \) and \( Q \) . So it maps \( {\ell }^{\prime } \), the unique \( d \) -line through \( P \) and \( Q \), to itself. | Yes |
Using the fact that \( {a}^{2} + {b}^{2} = \alpha \bar{\alpha } \), we may rewrite this equation as\n\n\[ \n{r}^{2} - \alpha \bar{\alpha } = - 1 \n\] | We now use equation (1), and the fact that inversion \( t \) in the circle \( C \) has the form\n\n\[ \nt\left( z\right) = \frac{{r}^{2}}{\overline{z - \alpha }} + \alpha \;\left( {z \in \mathbb{C}-\{ \alpha \} }\right) ,\n\]\n\nto obtain the form of the hyperbolic reflection \( \rho \) in the \( d \) -line \( \ell \) ... | Yes |
Lemma 1 The hyperbolic reflection \( \rho \) in the \( d \) -line \( \ell \) that is part of a Euclidean circle with centre \( \alpha \) is given by the hyperbolic transformation | \[ \rho \left( z\right) = \frac{\alpha \bar{z} - 1}{\bar{z} - \bar{\alpha }}\;\left( {z \in \mathcal{D}}\right) . \] Recall that \( \left| \alpha \right| > 1 \) . Notice that we may write \[ \rho \left( z\right) = \left( {M \circ B}\right) \left( z\right) ,\;\text{ for }z \in \mathcal{D}, \] where \( M\left( z\right) =... | Yes |
Problem 1 Find the point which has image 0 under reflection in the \( d \) -line obtained from \( \alpha \) . Hence obtain a second proof of the Origin Lemma. | Reflection in a \( d \) -line which is a diameter of \( \mathcal{D} \) is simply (Euclidean) reflection in that line. Recall that Subsection 5.2.1 \[ {t}_{1}\left( z\right) = \bar{z} \] is reflection in the \( x \) -axis, and \[ {t}_{2}\left( z\right) = {\alpha z} \] where \( \alpha = \cos \theta + i\sin \theta \) with... | No |
Find the composite \( {\sigma }_{2} \circ {\sigma }_{1} \) of the hyperbolic reflections\n\n\[ \n{\sigma }_{1}\left( z\right) = {\alpha }^{2}\bar{z}\;\text{ and }\;{\sigma }_{2}\left( z\right) = {\beta }^{2}\bar{z}, \n\]\n\nwhere \( \alpha = \cos {\theta }_{1} + i\sin {\theta }_{1} \) and \( \beta = \cos {\theta }_{2} ... | Solution We have\n\n\[ \n\left( {{\sigma }_{2} \circ {\sigma }_{1}}\right) \left( z\right) = {\beta }^{2}\overline{\left( {\alpha }^{2}\bar{z}\right) } = {\beta }^{2}{\bar{\alpha }}^{2}z, \n\]\n\nwhich is a Euclidean rotation about the origin of \( \mathcal{D} \) . | Yes |
Example 2 Show that the hyperbolic reflection \( \rho \) is its own inverse, by showing that \( \rho \left( {\rho \left( z\right) }\right) = z \) . | Solution From Theorem 1, we know that the composite of the reflection \( \rho \) with itself is given by\n\n\[ \left( {\rho \circ \rho }\right) \left( z\right) = \frac{\left( {\bar{\alpha }\alpha - 1}\right) z + \alpha - \alpha }{\left( {\bar{\alpha } - \bar{\alpha }}\right) z + \alpha \bar{\alpha } - 1} = \frac{\left(... | Yes |
Show that the composite \( {M}_{2} \circ {M}_{1} \) of the Möbius transformations\n\n\[ \n{M}_{1}\left( z\right) = \frac{{az} + b}{\bar{b}z + \bar{a}}\;\text{ and }\;{M}_{2}\left( z\right) = \frac{{cz} + d}{\bar{d}z + \bar{c}} \]\n\nis a Möbius transformation of the same form. That is, of the form \( \frac{{ez} + f}{\b... | Solution By the strategy for finding composites of Möbius transformations, for some \( e, f \in \mathbb{C} \) . a matrix associated with the composite \( {M}_{2} \circ {M}_{1} \) is the matrix product\n\n\[ \n\left( \begin{array}{ll} c & d \\ \bar{d} & \bar{c} \end{array}\right) \left( \begin{array}{ll} a & b \\ \bar{b... | Yes |
Example 4 Show that the composite \( {\sigma }_{2} \circ {\sigma }_{1} \) of the two reflections in lines through the origin given by \( {\sigma }_{1}\left( z\right) = {\alpha }^{2}\bar{z} \) and \( {\sigma }_{2}\left( z\right) = {\beta }^{2}\bar{z} \) can be written in the form \( M\left( z\right) = \frac{{az} + b}{\b... | Solution From Example 1, we know that\n\n\[ \left( {{\sigma }_{2} \circ {\sigma }_{1}}\right) \left( z\right) = {\beta }^{2}{\bar{\alpha }}^{2}z \]\n\nSince \( \left| \alpha \right| = 1 \) and \( \left| \beta \right| = 1 \),\n\n\[ {\alpha }^{-1} = \bar{\alpha }\;\text{ and }\;{\beta }^{-1} = \bar{\beta }, \]\n\nand so\... | Yes |
bolic reflection reverses the orientation of angles between \( d \) -lines. So a composite of two such transformations leaves the orientation unchanged, while a composite of three reverses it again. We call a hyperbolic transformation that leaves orientation unchanged a direct hyperbolic transformation, and one that re... | It is possible to say more about the direct transformations. Let \( {r}_{1} \) and \( {r}_{2} \) be reflections in the \( d \) -lines \( {\ell }_{1} \) and \( {\ell }_{2} \), respectively.\n\nFirst, suppose that \( {\ell }_{1} \) and \( {\ell }_{2} \) intersect at some point \( A \) . (Certainly they cannot intersect a... | Yes |
Theorem 4 A direct hyperbolic transformation \( M \) can be written in the form\n\n\[ M\left( z\right) = K\frac{z - m}{1 - \bar{m}z} \]\n\nwhere \( K \) and \( m \) are complex numbers with \( \left| K\right| = 1 \) and \( m \in \mathcal{D} \). | Proof We know from the previous subsection that a direct hyperbolic transformation can always be written in the form\n\n\[ M\left( z\right) = \frac{{az} + b}{\bar{b}z + \bar{a}},\;\text{ with }\left| b\right| < \left| a\right| . \]\n\n---\n\nSee the discussion before Theorem 2.\n\n---\n\nIndeed, on dividing the express... | Yes |
Problem 4 For each of the following points, determine all the direct hyperbolic transformations that map it to the origin.\n\n(a) \( \frac{1}{4}i\; \) (b) \( - \frac{1}{3} + \frac{2}{3}i \) | The general form of the inverse of a direct hyperbolic transformation \( M \) mapping the point \( m \) to the origin is a direct hyperbolic transformation sending the origin to \( m \) . To find this inverse explicitly, we write\n\n\[ M\left( z\right) = K\frac{z - m}{1 - \bar{m}z}\;\text{ (by Theorem 4) } \]\n\n\[ = \... | No |
Example 6 Determine the general form of the direct hyperbolic transformation that maps \( \frac{1}{2}i \) to \( \frac{3}{4} \) . | Solution We have already seen that the general form of the direct hyperbolic Example 5, part (a). transformation \( {M}_{1} \) that maps \( \frac{1}{2}i \) to 0 is\n\n\[ \n{M}_{1}\left( z\right) = K\frac{{2z} - i}{{iz} + 2},\;\text{ where }\left| K\right| = 1; \]\n\n\na matrix associated with \( {M}_{1} \) is\n\n\[ \n{... | Yes |
Example 1 Find the hyperbolic distance between the points 0.1 and 0.2 . | Solution Here we use Property 4, which states that distances along a\n\n\n\nThroughout Section 6.3 we shall work to the full accuracy of our calculator, but we shall record our results only to 3 or 4 decimal places.\... | Yes |
Problem 1 Determine the hyperbolic distances \( d\left( {0,\frac{1}{3}i}\right) \) and \( d\left( {{0.8},{0.9}}\right) \) . | By rearranging equation (1) we obtain the formula\n\n\[\n\left| z\right| = \tanh d\left( {0, z}\right) \text{.}\n\]\n\n(2)\n\nThis can be used to locate a point, given its hyperbolic distance along a radius of the disc \( \mathcal{D} \) . For example, the point \( a \) that is at a hyperbolic distance 0.1 from the orig... | No |
Find the two missing entries in the table. | <table><tr><td>\\( d\\left( {0, a}\\right), a > 0 \\)</td><td>0.2</td><td>0.4</td><td>0.8</td><td>1.6</td><td>3.2</td></tr><tr><td>\\( a \\)</td><td>\\( {0.197}\\ldots \\)</td><td>\\( {0.380}\\ldots \\)</td><td>\\( {0.664}\\ldots \\)</td><td></td><td></td></tr></table> | No |
Example 2 Find the hyperbolic distance between the points \( \frac{1}{2} \) and \( \frac{1}{3}i \) . | Solution From the definition of hyperbolic distance we have\n\n\[ d\left( {\frac{1}{2},\frac{1}{3}i}\right) = {\tanh }^{-1}\left( \left| \frac{\frac{1}{2} - \frac{1}{3}i}{1 - \overline{\frac{1}{3}i} \cdot \frac{1}{2}}\right| \right) \]\n\n\[ = {\tanh }^{-1}\left( \left| \frac{3 - {2i}}{6 + i}\right| \right) \]\n\n\[ = ... | Yes |
Example 3 Find the hyperbolic midpoint \( m \) of the line segment which joins each of the following pairs of points.\n\n(a) \( \frac{1}{4}i \) and \( \frac{3}{4}i \) (b) \( \frac{1}{4}i \) and \( - \frac{3}{4}i \) | Solution First observe that\n\n\[ d\left( {0,\frac{1}{4}i}\right) = {\tanh }^{-1}\left( \left| {\frac{1}{4}i}\right| \right) = {\tanh }^{-1}\left( {0.25}\right) = {0.255}\ldots ,\]\n\nand\n\n\[ d\left( {0,\frac{3}{4}i}\right) = {\tanh }^{-1}\left( \left| {\frac{3}{4}i}\right| \right) = {\tanh }^{-1}\left( {0.75}\right)... | Yes |
Problem 3 Find the hyperbolic midpoint \( m \) of the line segment which joins each of the following pairs of points:\n\n\[ \text{0.5 and 0.8 ;}\; - {0.2}\text{and 0.8 .} \] | ## Remark\n\nTo calculate the hyperbolic midpoint of a line segment which joins two arbitrary points \( p, q \) in \( \mathcal{D} \) we would use a Möbius transformation \( M \) to map \( p \) to 0 and \( q \) to \( M\left( q\right) \) . After calculating the hyperbolic midpoint \( {m}^{\prime } \) of the segment from ... | No |
Theorem 1 Every hyperbolic circle is a Euclidean circle in \( \mathcal{D} \), and vice versa. | Proof We have already established the result for circles centred at 0 , so let For the steps in the \( C \) be any hyperbolic circle with hyperbolic centre \( m \neq 0 \) . Let the diameter of argument before Theorem \( \mathcal{D} \) through \( m \) meet \( C \) at the points \( a \) and \( b \), and let \( K \) be th... | Yes |
Find the Euclidean centre and radius of the hyperbolic circle | Solution Here the hyperbolic centre \( m \) is the point \( \frac{1}{2} \), and so\n\n\[ d\left( {0, m}\right) = {\tanh }^{-1}\left( \frac{1}{2}\right) \simeq {0.549}.\n\]\n\nSince the hyperbolic radius of \( C \) is equal to \( \frac{1}{2} \), it follows that \( {Om} \) meets \( C \) at the points \( a, b \), where\n\... | Yes |
Example 5 Find the hyperbolic centre and radius of the Euclidean circle\n\n\\[ \nK = \\left\\{ {z : \\left| {z + \\frac{1}{2}i}\\right| = \\frac{1}{4}}\\right\\} .\n\\] | Solution Here the Euclidean centre \\( p \\) is the point \\( - \\frac{1}{2}i \\), and the Euclidean\n\nradius is \\( \\frac{1}{4} \\), so \\( {Op} \\) meets \\( K \\) at the points \\( a = - \\frac{1}{4}i \\) and \\( b = - \\frac{3}{4}i \\) . Thus\n\n\\[\nd\\left( {0, a}\\right) = {\\tanh }^{-1}\\left( \\left| {-\\fra... | Yes |
Problem 5 Determine the hyperbolic centre and radius of the Euclidean circle\n\n\[ K = \\left\\{ {z : \\left| {z - \\frac{1}{4}}\\right| = \\frac{1}{2}}\\right\\} . \] | We now use the fact that hyperbolic circles are also Euclidean circles to prove the Triangle Inequality property of hyperbolic distance. | No |
Theorem 2 For all \( {z}_{1},{z}_{2},{z}_{3} \) in \( \mathcal{D} \) :\n\n\[ d\left( {{z}_{1},{z}_{3}}\right) + d\left( {{z}_{3},{z}_{2}}\right) \geq d\left( {{z}_{1},{z}_{2}}\right) . \] | Proof First let \( M \) be a hyperbolic transformation that maps \( {z}_{1} \) to 0, and let \( M\left( {z}_{2}\right) = b \) and \( M\left( {z}_{3}\right) = c \) . Then \( {Ob} \) is a straight line.\n\n\n\n## Dista... | Yes |
Theorem 3 Let \( A \) and \( {A}^{\prime } \) be points in the unit disc \( \mathcal{D} \) that are images of each other under reflection in a \( d \) -line \( \ell \) . Then \( \ell \) is the hyperbolic perpendicular bisector of the hyperbolic line segment \( A{A}^{\prime } \) . | Proof Let the \( d \) -line \( {\ell }^{\prime } \) through \( A \) and \( {A}^{\prime } \) meet \( \ell \) at \( P \) . Now \( A \) and \( {A}^{\prime } \) map to each other under hyperbolic reflection in \( \ell \), and \( P \) remains invariant. But hyperbolic reflection maps \( d \) -lines to \( d \) -lines, and th... | Yes |
Example 6 Determine the equation of the hyperbolic perpendicular bisector of \( \\left\\lbrack {{0.2},{0.9}}\\right\\rbrack \), the line segment from 0.2 to 0.9. | Solution Using the Reflection Lemma, with \( p = {0.9} \) and \( q = {0.2} \), we find that\n\n\[ \n\\alpha = \\frac{{0.7} + {0.9} \\cdot {0.2} \\cdot {0.7}}{{0.81} - {0.04}} \\simeq {1.0727}.\n\]\n\nSo the equation of the \( d \) -line which is the (hyperbolic) perpendicular bisector of \( \\left\\lbrack {{0.2},{0.9}}... | Yes |
Theorem 1 The sum of the angles of a \( d \) -triangle is less than \( \pi \) . | Proof By the Origin Lemma, we can map the \( d \) -triangle \( \bigtriangleup {ABC} \) onto a Subsection 6.1.2, \( d \) -triangle \( \bigtriangleup O{B}^{\prime }{C}^{\prime } \) by any hyperbolic transformation that sends \( A \) to the origin Lemma 2 \( O \) . Since hyperbolic transformations preserve angles, the sum... | Yes |
Theorem 2 The sum of the angles of a \( d \) -quadrilateral is less than \( {2\pi } \) . | Proof Any \( d \) -quadrilateral can be divided into two (non-overlapping) without proof; see the \( d \) -triangles by one or other of the \( d \) -lines joining alternate vertices. figures below.\n\n\n\nThe angles ... | No |
Theorem 3 Let \( \\bigtriangleup {ABC} \) be a \( d \) -triangle in which \( \\angle {ABC} = \\angle {ACB} \). Then the sides \( {AB} \) and \( {AC} \) are of equal length. | Proof Let \( D \) be the midpoint of the \( d \) -line segment \( {BC} \). By applying the Origin Lemma, if necessary, we may assume that \( D \) coincides with \( O \), the centre of the disc \( \\mathcal{D} \). (Although this is not strictly necessary for the proof, it simplifies the picture.) Then \( {BC} \) is part... | Yes |
Problem 3 Let \( \\bigtriangleup {ABC} \) be a \( d \) -triangle in which the sides \( {AB} \) and \( {AC} \) have equal hyperbolic length. Prove that \( \\angle {ABC} = \\angle {ACB} \) . | Hint: Consider reflection in the \( d \) -line that bisects angle \( \\angle {BAC} \) . | No |
Theorem 4 Similar \( d \) -triangles are \( d \) -congruent. | Proof We have to prove that if the \( d \) -triangles \( \bigtriangleup {ABC} \) and \( \bigtriangleup {PQR} \) have the angles at \( A, B \) and \( C \) and the angles at \( P, Q \) and \( R \) equal, respectively, then the two \( d \) -triangles are \( d \) -congruent.\n\nWe may apply a hyperbolic transformation to m... | Yes |
Theorem 7 Altitude Theorem\n\nLet the sides \( {AB} \) and \( {AC} \) of a \( d \) -triangle \( \bigtriangleup {ABC} \) be of equal hyperbolic length, and let the angle at \( A \) be \( \theta \) . Then the hyperbolic length of the altitude through \( A \) of the triangle is less than some number that depends only on \... | Proof Map the \( d \) -triangle \( \bigtriangleup {ABC} \) onto a \( d \) -triangle \( \bigtriangleup O{B}^{\prime }{C}^{\prime } \) by any hyperbolic\n\n\n\ntransformation that sends \( A \) to the origin \( O \) . ... | Yes |
Theorem 8 Pythagoras' Theorem | We prove Theorem 8 later in this subsection. | No |
Prove that, if \( \bigtriangleup {ABC} \) is an isosceles \( d \) -triangle in which \( {AB} \) and \( {AC} \) are of equal hyperbolic lengths and the angle at \( A \) is a right angle, then the hyperbolic length of the altitude \( {AD} \) is less than \( \frac{1}{2}{\cosh }^{-1}\left( \sqrt{2}\right) \simeq {0.4407} \... | Solution The given triangle is \( d \) -congruent to the \( d \) -triangle \( {\Delta O}{B}^{\prime }{C}^{\prime } \) with vertices at \( 0, r \) and \( {ir} \), for some \( r \) with \( 0 < r < 1 \) . Let \( O{D}^{\prime } \) be an altitude of this triangle. Let the hyperbolic lengths of the sides \( O{B}^{\prime } \)... | Yes |
Theorem 10 Sine Formula\n\nLet \( \\bigtriangleup {ABC} \) be a \( d \) -triangle right-angled at \( C \) . Let \( a \) and \( c \) be the hyperbolic\n\nlengths of \( {BC} \) and \( {AB} \) . Then\n\n\[ \sin A = \\frac{\\sinh {2a}}{\\sinh {2c}}. \] | A proof of this result will be found as an Exercise in Section 6.7. | No |
Theorem 1 All trebly asymptotic \( d \) -triangles are \( d \) -congruent to each other. | Proof It is sufficient to prove that trebly asymptotic \( d \) -triangles are \( d \) -congruent to the trebly asymptotic \( d \) -triangle with vertices -1,1 and \( i \) .\n\nSo, let \( \bigtriangleup {ABC} \) be any trebly asymptotic \( d \) -triangle, where (for convenience) we assume that the vertices \( A, B, C \)... | Yes |
Theorem 2 Let \( \bigtriangleup {ABC} \) be a \( d \) -triangle. Then there exists a trebly asymptotic \( d \) -triangle \( \bigtriangleup {DEF} \) that contains \( \bigtriangleup {ABC} \) . | Proof First, we construct a trebly asymptotic \( d \) -triangle \( \bigtriangleup {DEF} \) as follows: extend the segment \( {AB} \) beyond \( B \) to meet \( \mathcal{C} \) at \( D \), the segment \( {BC} \) beyond \( C \) to meet \( \mathcal{C} \) at \( E \), and the segment \( {CA} \) beyond \( A \) to meet \( \math... | Yes |
Lemma 1 Let \( P \) be a point on a \( d \) -line \( \ell \) of \( \mathcal{D} \), and let \( 0 < \beta < \frac{\pi }{2} \) . Then there are exactly two \( d \) -lines through \( P \) that make an acute angle \( \beta \) with \( \ell \) , and each is a (hyperbolic) reflection of the other in \( \ell \) . | Proof Since all the properties in the statement of the Lemma are hyperbolic properties, and there is a hyperbolic transformation (for example, a Möbius transformation) that maps the point \( P \) to the origin \( O \), it is sufficient to prove the result when \( P \) is the origin. Then there are (by Euclidean geometr... | Yes |
4. Let \( \bigtriangleup {ABC} \) be a \( d \) -triangle with \( \angle {ABC} = \pi /3 \), and let the sides \( {BA} \) and \( {BC} \) have the same hyperbolic length \( c \) . Let \( {BD} \) be the altitude from \( B \) to the side \( {AC} \) . Show that the hyperbolic length \( b \) of \( {AC} \) is greater than \( c... | Hint: Apply the Sine Formula to triangle \( \bigtriangleup {BDA} \) . | No |
Prove that any plane that cuts the sphere \( {S}^{2} \) in more than one point cuts it in a circle. | Solution Let \( {OT} \) be a radius of the sphere perpendicular to the given plane. A rotation of the sphere about this axis maps the plane to itself, and the sphere to itself; so it maps their intersection to itself. Each point on the intersection traces out a circle (of the same radius and centre) under the rotation,... | Yes |
Example 2 Let \( P \) and \( Q \) be any two points on \( {S}^{2} \) . Show that there is a rotation of \( {S}^{2} \) that maps \( P \) to \( Q \) . | Solution Consider the great circle through \( P \) and \( Q \), and let \( T \) be one of its poles. The rotation of \( {S}^{2} \) about the axis \( {OT} \) and through the angle \( \angle {POQ} \) (in the appropriate direction) sends \( P \) to \( Q \), as required. | Yes |
Show that \( R\left( {X,\frac{\pi }{2}}\right) \cdot R\left( {Y,\frac{\pi }{2}}\right) \neq R\left( {Y,\frac{\pi }{2}}\right) \cdot R\left( {X,\frac{\pi }{2}}\right) \). | Solution Here\n\n\[ R\left( {X,\frac{\pi }{2}}\right) \cdot R\left( {Y,\frac{\pi }{2}}\right) = \left( \begin{matrix} 1 & 0 & 0 \\ 0 & \cos \frac{\pi }{2} & - \sin \frac{\pi }{2} \\ 0 & \sin \frac{\pi }{2} & \cos \frac{\pi }{2} \end{matrix}\right) \]\n\n\[ \times \left( \begin{matrix} \cos \frac{\pi }{2} & 0 & \sin \fr... | Yes |
For the elementary rotation \( R\left( {X,\alpha }\right) \), verify that the transpose of its matrix is equal to the matrix of the elementary rotation \( R\left( {X, - \alpha }\right) \). | Notice that the inverse of an elementary rotation through a given angle is an elementary rotation through the same angle but in the opposite direction;\n\nThis shows that multiplication of the matrices of elementary rotations (and hence composition of elementary rotations) is not commutative. that is,\n\n\[ R{\left( X,... | No |
We now see how we can determine the matrix of such a rotation explicitly. | First, we see how to rotate \( {S}^{2} \) so as to send the point \( A\left( {1,0,0}\right) \) to any point \( P\left( {\cos \phi \sin \theta ,\sin \phi \sin \theta ,\cos \theta }\right) \) of \( {S}^{2} \) . First we rotate \( {S}^{2} \) to send \( A \) to the point \( {P}_{1}\left( {\cos \phi ,\sin \phi ,0}\right) \)... | Yes |
Theorem 2 A rotation of \( {S}^{2} \) that maps \( A\left( {1,0,0}\right) \) to \( P(\cos \phi \sin \theta ,\sin \phi \sin \theta ,\cos \theta ) \) | is given by the composition \( R\left( {Z,\phi }\right) R\left( {Y, - {\theta }^{\prime }}\right) \), where \( {\theta }^{\prime } = \frac{\pi }{2} - \theta \) | Yes |
Example 2 Determine a rotation that maps the point \( P(\cos \phi \sin \theta \) , \( \sin \phi \sin \theta ,\cos \theta ) \) of \( {S}^{2} \) to \( N\left( {0,0,1}\right) \), in terms of elementary rotations. | Solution By Theorem 2, one rotation that sends \( A \) to \( P \) is given by the composition \( R\left( {Z,\phi }\right) R\left( {Y, - {\theta }^{\prime }}\right) \), where \( {\theta }^{\prime } = \frac{\pi }{2} - \theta \) . It follows that the mapping\n\n\[ \n{\left( R\left( Z,\phi \right) R\left( Y, - {\theta }^{\... | Yes |
Theorem 3 Reflection of \( {S}^{2} \) in the plane \( \pi \) with equation \( {ax} + {by} + {cz} = 0 \) , where \( {a}^{2} + {b}^{2} + {c}^{2} = 1 \), is given by the mapping \( \mathbf{x} \mapsto \mathbf{A}\mathbf{x} \) where | \[ \mathbf{A} = \left( \begin{matrix} 1 - 2{a}^{2} & - {2ab} & - {2ac} \\ - {2ab} & 1 - 2{b}^{2} & - {2bc} \\ - {2ac} & - {2bc} & 1 - 2{c}^{2} \end{matrix}\right) . \] | Yes |
Theorem 4 The product of any two reflections of \( {\mathbb{R}}^{3} \) in planes through \( O \) that meet in a common line is a rotation about that common line. | Proof By choosing coordinate axes suitably, we may arrange that the common line is the \( z \) -axis. The planes are therefore vertical, and it follows that each reflection leaves the \( z \) -coordinate of each point unaltered. It is therefore sufficient to look at the effect of each reflection and composite on planes... | Yes |
Prove that the image of a circle \( C \) on \( {S}^{2} \) under an isometry \( t \) of \( {S}^{2} \) to itself is a circle. | Solution Let the circle \( C \) be cut out by a plane \( \pi \), and let \( \ell \) be the line through the origin perpendicular to \( \pi \) . Let \( \ell \) meet \( {S}^{2} \) at the points \( P \) and \( {P}^{\prime } \) . Then all points of \( C \) are the same distance from \( P \) .\n\nSince \( t \) is an isometr... | Yes |
Theorem 6 Isosceles Triangle Theorem\n\nLet a triangle \( \bigtriangleup {PQR} \) on \( {S}^{2} \) have sides \( {PQ} \) of length \( r,{QR} \) of length \( p \), and\n\n\n\n\( {RP} \) of length \( q \), and let \( p... | Proof Consider the reflection in the great circle that is the internal bisector of the angle \( \angle {PRQ} \) .\n\nIt maps the great circle through \( R \) and \( P \) to the great circle through \( R \) and \( Q \) ; and, since it is an isometry and \( p = q \), it therefore maps \( P \) to \( Q \), and \( Q \) to \... | Yes |
Example 1 Show that the triangles \( \bigtriangleup {PQR} \) and \( \bigtriangleup {P}^{\prime }{Q}^{\prime }{R}^{\prime } \) have their angles equal in pairs. | Solution The angle \( \angle {QPR} \) at \( P \) in \( \bigtriangleup {PQR} \) is formed by two great circles that meet again at \( {P}^{\prime } \), forming the angle \( \angle {Q}^{\prime }{P}^{\prime }{R}^{\prime } \) at \( {P}^{\prime } \) in \( \bigtriangleup {P}^{\prime }{Q}^{\prime }{R}^{\prime } \), so they mus... | Yes |
Example 2 Estimate the distance between New York and Sydney, taking the radius of the Earth as 4000 miles. | Solution The colatitudes of New York and Sydney are \( {90}^{ \circ } - {41}^{ \circ } = {49}^{ \circ } \) and \( {90}^{ \circ } + {34}^{ \circ } = {124}^{ \circ } \), so that the coordinates of the corresponding points \( P \) and \( Q \) on \( {S}^{2} \) are\n\n\[ \left( {\cos \left( {-{74}^{ \circ }}\right) \sin \le... | Yes |
Let \( \\bigtriangleup {ABC} \) be a triangle on \( {S}^{2} \) in which the angle at \( C \) is a right angle. If \( a \) , \( b \) and \( c \) are the lengths of \( {BC},{CA} \) and \( {AB} \), then\n\n\[ \n\\cos c = \\cos a \\times \\cos b.\n\] | Proof We may rotate \( {S}^{2} \) so that \( C \) is the North Pole \( N\\left( {0,0,1}\\right) \) of \( {S}^{2}, A \) lies on the Greenwich meridian and has coordinates \( \\left( {\\sin b,0,\\cos b}\\right) \), and \( B \) has coordinates \( \\left( {0,\\sin a,\\cos a}\\right) \) .\n\nWe then apply the rotation \( R\... | Yes |
Theorem 7 Let \( \\bigtriangleup {ABC} \) be a triangle on \( {S}^{2} \) in which the angle at \( C \) is a right angle. If \( a, b \) and \( c \) are the lengths of \( {BC},{CA} \) and \( {AB} \), and \( \\alpha \) denotes the angle \( \\angle {CAB} \), then\n\n\[ \n\\cos c = \\cos a \\times \\cos b, \n\]\n\n\[ \n\\si... | A similar discussion to that above gives that\n\n\[ \n\\cos \\beta \\sin c = \\sin a\\cos b\\;\\text{ and }\\;\\sin \\beta \\sin c = \\sin b, \n\]\n\nso that\n\n\[ \n\\sin \\beta = \\frac{\\sin b}{\\sin c}\\;\\text{ and }\\;\\tan \\beta = \\frac{\\tan b}{\\sin a}. \n\]\n\nFrom these formulas and those in Theorem 7 abov... | No |
Theorem 9 Cosine Rule for Sides of a Strict Spherical Triangle\n\nLet \( \\bigtriangleup {ABC} \) be a strict triangle on \( {S}^{2} \), in which the sides \( {AB},{BC} \) and \( {CA} \) have lengths \( c, a \) and \( b \), respectively, and the angles at \( A, B \) and \( C \) are \( \\alpha ,\\beta \) and \( \\gamma ... | \[ \cos c = \cos a\\cos b + \\sin a\\sin b\\cos \\gamma . \] | Yes |
Example 3 Let \( \\bigtriangleup {ABC} \) be a strict isosceles triangle on \( {S}^{2} \), in which the sides \( {AB},{BC} \) and \( {CA} \) have lengths \( c, a \) and \( a \), respectively, and the angles at \( A, B \) and \( C \) are \( \\alpha ,\\alpha \) and \( \\gamma \), respectively. Prove that\n\n\[ \n\\text{(... | Solution Putting \( b = a \) in the conclusion of Theorem 9, we obtain the formula (a).\n\nNext, by applying Theorem 9 to find \( \\cos a \), we get\n\n\[ \n\\cos a = \\cos a\\cos c + \\sin a\\sin c\\cos \\alpha .\n\]\n\nWe may then rearrange this formula to get\n\n\[ \n\\cos \\alpha = \\frac{\\cos a\\left( {1 - \\cos ... | Yes |
Let \( \bigtriangleup {ABC} \) be a strict equilateral triangle on \( {S}^{2} \), in which for \( x \in \mathbb{R} \) . the sides have length \( a \) and the angles are equal to \( \alpha \) . Prove that \[ \cos \alpha = \frac{\cos a}{1 + \cos a}. \] | We can now use any dual triangle \( \Delta {A}^{\prime }{B}^{\prime }{C}^{\prime } \) of \( \bigtriangleup {ABC} \) to prove a new trigonometric result. The dual triangle has sides of length \( \pi - \alpha ,\pi - \beta \) and \( \pi - \gamma \), and has angles of magnitude \( \pi - a,\pi - b \) and \( \pi - c \) . By ... | No |
Example 1 Let \( C \) denote a great circle on \( {S}^{2} \) that is the intersection of \( {S}^{2} \) with Since \( c \neq 0, C \) does not the plane \( {aX} + {bY} + {cZ} = 0 \), where \( {a}^{2} + {b}^{2} + {c}^{2} = 1, c \neq 0 \) . pass through the North Pole \( N \) of \( {S}^{2} \) .\n\n(a) Determine the point \... | (a) The point \( {N}^{\prime } \) has coordinates We follow the approach of\n\n\[ \left( {0,0,1}\right) - 2\{ \left( {0,0,1}\right) \cdot \left( {a, b, c}\right) \} \left( {a, b, c}\right) \]\nSubsection 7.2.2.\n\n\[ = \left( {0,0,1}\right) - {2c}\left( {a, b, c}\right) = \left( {-{2ac}, - {2bc},1 - 2{c}^{2}}\right) . ... | Yes |
Prove that the image \( P \) of \( {P}^{\prime } \) under stereographic projection has coordinates \( \left( {\frac{\cos \phi }{\tan \left( {\frac{1}{2}\theta }\right) },\frac{\sin \phi }{\tan \left( {\frac{1}{2}\theta }\right) }}\right) \). | Solution From the formula for \( \pi \), the image \( P \) of \( {P}^{\prime } \) under \( \pi \) is the point\n\n\[ \left( {\frac{\cos \phi \sin \theta }{1 - \cos \theta },\frac{\sin \phi \sin \theta }{1 - \cos \theta }}\right) \]\n\n\[ = \left( {\frac{\cos \phi \cdot 2\sin \left( {\frac{1}{2}\theta }\right) \cos \lef... | Yes |
Theorem 2 Let \( {P}^{\prime } \) and \( {Q}^{\prime } \) be points on \( {S}^{2} \) that are mirror images under reflection in a great circle \( {C}^{\prime } \), and let the images of \( {P}^{\prime },{Q}^{\prime } \) and \( {C}^{\prime } \) under stereographic projection \( \pi \) be \( P, Q \) and \( C \), respecti... | Proof The great circle lies in a plane, \( \Pi \) say, through the origin. \( \Pi \) consists of those points in \( {\mathbb{R}}^{3} \) that are equidistant from the points \( {P}^{\prime } \) and \( {Q}^{\prime } \) . So any sphere through \( {P}^{\prime } \) and \( {Q}^{\prime } \) has its centre somewhere in \( \Pi ... | Yes |
Prove that a Möbius transformation \( M \) that is conjugate to a rotation \( R\left( {Y,\beta }\right) \) of \( {S}^{2} \) is of the form \( M\left( z\right) = \frac{{az} - b}{{bz} + a} \) where \( a, b \) are real. | Solution A rotation \( R\left( {Y,\beta }\right) \) of \( {S}^{2} \) fixes the points \( \left( {0,1,0}\right) \) and \( \left( {0, - 1,0}\right) \) , which map under stereographic projection to the points \( i \) and \( - i \), respectively.\n\nBy Theorem 3, a conjugate Möbius transformation must be of the form\n\n\[ ... | Yes |
Problem 5 Determine the fixed points in \( \mathbb{C} \) and in \( \widehat{\mathbb{C}} \) of the following Möbius transformations: | \[ z \mapsto z + 1;\;z \mapsto - \frac{4}{z + 4};\;z \mapsto - \frac{4}{z + 5}. \] | No |
Prove that a Möbius transformation \( M \) with distinct fixed points \( a \) and \( b \) in \( \mathbb{C} \) maps the family \( \mathcal{A} \) of Apollonian circles defined by the point circles \( a \) and \( b \) to itself. | Solution Let \( \mathcal{B} \) be the family of generalized circles through \( a \) and \( b \) .\n\nSince \( M \) has fixed points \( a \) and \( b \), it maps the family \( \mathcal{B} \) onto itself. Also, A Möbius transformation since a Möbius transformation preserves angles, \( M \) thus maps the family of maps ge... | No |
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