Complex numbers and multi-valued functions

1. State the Euler’s formula (which connect exponential functions with trigonometric functions). Express $\cos3\theta$ in terms of $\cos\theta$ and $\sin\theta$ .

Euler's formula: $e^{i\theta}=\cos\theta+i\sin\theta, \theta\in\R$ .
$\cos 3\theta=\operatorname{Re}e^{3i\theta}=\operatorname{Re}(e^{i\theta})^3=\operatorname{Re}(\cos\theta+i\sin\theta)^3=\operatorname{Re}(\cos^3\theta+3i\cos^2\theta\sin\theta-3\cos\theta\sin^2\theta-i\sin^3\theta)=\cos^3\theta-3\cos\theta\sin^2\theta$ .

2. Find all values of $(-2+2i)^\frac{1}{3}$ (hint: you should use the polar representation.

Consider $-2+2i=2\sqrt 2 e^{i(\frac{3}{4}+2k)\pi}$ , therefore $(-2+2i)^{\frac{1}{3}}=\sqrt2e^{i(\frac{1}{4}+\frac{2}{3}k)\pi}$ , which means the set of all values is $\{1+i,-\frac{\sqrt3+1}{2}+\frac{\sqrt3-1}{2}i,\frac{\sqrt3-1}{2}-\frac{\sqrt3+1}{2}i\}$ .

3. Find all values of $(-i)^i$ (hint: an expression in the form $a^b$ should be considered as $\exp(b\log a))$ . What is the value of $(-i)^i$ if we use the principle branch of the $\log$ function. By the principle branch of $\log$ , we mean the domain of definition of $\log$ is restricted to $z = re^{iθ}$ such that $r > 0$ , $-\pi < \theta ≤ \pi$ .

As $(-i)^i=(e^{i(\frac{3}{2}+2k)\pi})^i=e^{-(\frac{3}{2}+2k)\pi}$ , we have the set of all values is $\{e^{-(\frac{3}{2}+2k)\pi};k\in\Z\}$ .
When we consider the principle branch, it means $k=-1$ . So $(-i)^i=e^{-\frac{1}{2}\pi}$ .

Partial fraction of a rational function

4. Put the following rational functions into a partial fraction form $\frac{1-z^2}{(z-\alpha)^3}$ , $\frac{1-z^2}{(z-\alpha_1)(z-\alpha_2)(z-\alpha_3)}$ , where $\alpha_1$ , $\alpha_2$ , $\alpha_3$ are simple poles.

Suppose $\frac{1-z^2}{(z-\alpha)^3}=\frac{A}{z-\alpha}+\frac{B}{(z-\alpha)^2}+\frac{C}{(z-\alpha)^3}$ . By solving $1-z^2=A(z-\alpha)^2+B(z-\alpha)+C$ , we can get $A=-1,B=-2\alpha,C=1-\alpha^2$ . So $\frac{1-z^2}{(z-\alpha)^3}=\frac{-1}{z-\alpha}+\frac{-2\alpha}{(z-\alpha)^2}+\frac{1-\alpha^2}{(z-\alpha)^3}$ .
Suppose $\frac{1-z^2}{(z-\alpha_1)(z-\alpha_2)(z-\alpha_3)}=\frac{A}{z-\alpha_1}+\frac{B}{z-\alpha_2}+\frac{C}{z-\alpha_3}$ . By solving $1-z^2=A(z-\alpha_2)(z-\alpha_3)+B(z-\alpha_1)(z-\alpha_3)+C(z-\alpha_1)(z-\alpha_2)$ , we can get $A=\frac{1-\alpha_1^2}{(\alpha_1-\alpha_2)(\alpha_1-\alpha_3)},B=\frac{1-\alpha_2^2}{(\alpha_1-\alpha_2)(\alpha_2-\alpha_3)},C=\frac{1-\alpha_3^2}{(\alpha_1-\alpha_3)(\alpha_2-\alpha_3)}$ . So $\frac{1-z^2}{(z-\alpha_1)(z-\alpha_2)(z-\alpha_3)}=\frac{\frac{1-\alpha_1^2}{(\alpha_1-\alpha_2)(\alpha_1-\alpha_3)}}{z-\alpha_1}+\frac{\frac{1-\alpha_2^2}{(\alpha_1-\alpha_2)(\alpha_2-\alpha_3)}}{z-\alpha_2}+\frac{\frac{1-\alpha_3^2}{(\alpha_1-\alpha_3)(\alpha_2-\alpha_3)}}{z-\alpha_3}$ .

Möbius transformations

Recall that a Möbius transformation is defined as $M:z\mapsto\frac{az+b}{cz+d}$ , where $a$ , $b$ , $c$ , $d$ are complex numbers such that $ad - bc\neq 0$ . Here, we consider the Möbius transformation as a map from the extended complex plane $\overline{\mathbb C} = \mathbb C\cup\{\infty\}$ to itself. For $\overline{\mathbb C}$ , we have two coordinate systems: let $\overline{\mathbb C} = \mathcal A \cup \mathcal B$ , where $\mathcal A := \mathbb C$ and $\mathcal B := \mathbb C\cup\{\infty\}−\{0\}$ . From $\mathcal A$ and $\mathcal B$ , one has the maps $f: \mathcal A\ni z\mapsto z\in\mathbb C$ , $g:\mathcal B\ni z\mapsto\frac{1}{z}\in\mathbb C$ .

5. Give the domain of definition and the range of the maps $g\circ f^{-1}$ and $f\circ g^{-1}$ . Are these maps bijective holomorphic functions?

Notice that $f:\mathcal A\leftrightarrow \mathbb C,g:\mathcal B\leftrightarrow\mathbb C$ . Therefore $\mathcal D(g)\cap\mathcal R(f^{-1})=\mathcal B\cap\mathcal A=\mathbb C-\{0\}$ , $\mathcal D(f)\cap\mathcal R(g^{-1})=\mathcal A\cap\mathcal B=\mathbb C-\{0\}$ . So $\mathcal D(g\circ f^{-1})=f^{-1}(\mathbb C-\{0\})=\mathbb C-\{0\},\mathcal D(f\circ g^{-1})=g^{-1}(\mathbb C-\{0\})=\mathbb C-\{0\}$ , and $\mathcal R(g\circ f^{-1})=g(\mathbb C-\{0\})=\mathbb C-\{0\},\mathcal R(f\circ g^{-1})=f(\mathbb C-\{0\})=\mathbb C-\{0\}$ .
Because $f,g$ are all bijective, $g\circ f^{-1},f\circ g^{-1}$ are all bijective. As $g\circ f^{-1}:z\mapsto\frac{1}{z},f\circ g^{-1}:z\mapsto\frac{1}{z}$ , and $h(z)=\frac{1}{z}$ is holomorphic on $\mathbb C-\{0\}$ , we know they both are holomorphic. So in conclusion, they are bijective holomorphic functions.

6. Prove that Möbius transformation is a conformal mapping.

Let $z=x+iy$ , then $\frac{\partial M}{\partial\overline z}=\frac{1}{2}(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y})M=\frac{1}{2}(\frac{(az+b)_x(cz+d)-(az+b)(cz+d)_x}{(cz+d)^2}+i\frac{(az+b)_y(cz+d)-(az+b)(cz+d)_y}{(cz+d)^2})=\frac{1}{2}(\frac{a(cz+d)-(az+b)c}{(cz+d)^2}+i\frac{ai(cz+d)-(az+b)ci}{(cz+d)^2})=0,\frac{\partial M}{\partial z}=\frac{ad-bc}{(cz+d)^2}$ . So $M(z)$ is holomorphic and nonzero on $\mathbb C\cup\{\infty\}-\{\frac{d}{c}\}$ . As holomorphic functions are conformal, we have proved our statement.

7. Construct a special Möbius transformation that maps three distinct points $\{\infty,1,−1\}$ to $\{1,−i,i\}$ respectively. Such map is called Cayley transformation.

By solving $M(\infty)=\frac{a}{c}=1,M(1)=\frac{a+b}{c+d}=-i,M(-1)=\frac{-a+b}{-c+d}=i$ and letting $a=1$ , we have $a=1,b=-i,c=1,d=i$ , which means $M(z)=\frac{z-i}{z+i}$ .

8. Prove that the Cayley transformation maps the real axis to the unit circle (that is circle of radius $1$ centered at $0$ ).

$M_{Cayley}(x)=\frac{x-i}{x+i}=\frac{x^2-2ix-1}{x^2+1}=\frac{x^2-1}{x^2+1}-\frac{2x}{x^2+1}i,x\in\R$ . Notice that $(\frac{x^2-1}{x^2+1})^2+(\frac{-2x}{x^2+1})^2=1$ . So the Cayley transformation maps the real axis to the unit circle.

9. Using the the Cayley transformation, prove that in general that a Möbius transformation will map a line to a circle in the extended complex plane, and construct such a particular map that maps a line (a line can be expressed as $z = ax+b,a,b\in\mathbb C$ by a real parameter $x \in \R$ ) to the unit circle. (hint: you should make compositions of Möbius transformations. Möbius transformation maps circles to circles, in this sense, a straight line is a circle in the extended complex plane).

Let $R_\theta(z)=e^{i\theta}z$ stands for rotation, $S_r(z)=rz$ stands for scale, $T_b(z)=z+b$ stands for translation and $C(z)=\frac{z-i}{z+i}$ stands for Cayley transformation. Notice that $R_\theta,S_r,P_b$ all map circles to circles, lines to lines. As $M(x)=|\frac{a}{2c}-\frac{b}{2d}|\exp(i\operatorname{Arg(\frac{a}{2c}-\frac{b}{2d})})\frac{|\frac{ci}{d}|\exp(i\operatorname{Arg}\frac{ci}{d})z-i}{|\frac{ci}{d}|\exp(i\operatorname{Arg}\frac{ci}{d})z+i}+\frac{a}{2c}+\frac{b}{2d}=T_{\frac{a}{2c}+\frac{b}{2d}}\circ S_{|\frac{a}{2c}-\frac{b}{2d}|}\circ R_{\operatorname{Arg}(\frac{a}{2c}-\frac{b}{2d})}\circ C\circ S_{|\frac{ci}{d}|}\circ R_{\operatorname{Arg}\frac{ci}{d}}(z)$ , we can know that $M$ will map a line to a circle.
For $z=ax+b$ , $C\circ R_{-\operatorname{Arg}a}\circ S_{\frac{1}{|a|}}\circ T_{-b}(z)=\frac{z-b-ai}{z-b+ai}$ will maps it to the unit circle.

Cross-ratio

A cross-ratio of an ordered four points $z_1$ , $z_2$ , $z_3$ , $z_4$ can be defined as $[z_1, z_2, z_3, z_4] := \frac{(z_1 − z_3)(z_2 − z_4)}{(z_1 −z_2)(z_3 − z_4)}$ . It is a map from $\overline{\mathbb{C}}\times\overline{\mathbb{C}}\times\overline{\mathbb{C}}\times\overline{\mathbb{C}}$ to $\overline{\mathbb{C}}$ .

10. Prove $[z_1, z_2, z_3, z_4] = [M(z_1),M(z_2),M(z_3),M(z_4)]$ , where $M$ is a Möbius transformation. You can check it by direct computations. Another way is to think the cross-ratio as a special Möbius transformation of one of the points say $z_1$ , and use the property that a unique Möbius transformation carries three distinct points (in our case $z_2$ , $z_3$ , $z_4$ ) to $1$ , $0$ , $\infty$ . Refer to [Ahlfors, page 78] for details.

$[M(z_1),M(z_2),M(z_3),M(z_4)]=(\frac{a}{c}+\frac{bc-ad}{c^2z_1+cd}-\frac{a}{c}-\frac{bc-ad}{c^2z_3+cd})(\frac{a}{c}+\frac{bc-ad}{c^2z_2+cd}-\frac{a}{c}-\frac{bc-ad}{c^2z_4+cd})/[(\frac{a}{c}+\frac{bc-ad}{c^2z_1+cd}-\frac{a}{c}-\frac{bc-ad}{c^2z_2+cd})(\frac{a}{c}+\frac{bc-ad}{c^2z_3+cd}-\frac{a}{c}-\frac{bc-ad}{c^2z_4+cd})]=\frac{z_3-z_1}{(cz_1+d)(cz_3+d)}\frac{z_4-z_2}{(cz_2+d)(cz_4+d)}/(\frac{z_2-z_1}{(cz_1+d)(cz_2+d)}\frac{z_4-z_3}{(cz_3+d)(cz_4+d)})=\frac{(z_3-z_1)(z_4-z_2)}{(z_2-z_1)(z_4-z_3)}=[z_1,z_2,z_3,z_4]$ .
(Another way: Let $L(z)=[z,z_2,z_3,z_4],S(z)=[z,M(z_2),M(z_3),M(z_4)]$ . Notice that $L(z_2)=1,L(z_3)=0,L(z_4)=\infty,S(z_2)=1,S(z_3)=0,S(z_4)=\infty$ . According to the property that a unique Möbius transformation carries three distinct points to $1$ , $0$ , $\infty$ , $M=S^{-1}\circ L$ . Therefore $S(M(z_1))=L(z_1)$ .)

11. Prove the following statement: the cross-ratio $[z_1,z_2,z_3,z_4]$ is real if and only if the four points $z_1$ , $z_2$ , $z_3$ , $z_4$ lie on a circle or on a straight line. Try to give a proof by yourself. But you can refer to [Ahlfors, page 78] for details.

Let $L(z)=[z,z_2,z_3,z_4]$ , we have $L(z_2)=1,L(z_3)=0,L(z_4)=\infty$ .
$P\Rightarrow Q$ : If $L(z_1)\in\R$ , it means that $L^{-1}$ maps the real axis into a circle or line that determined by $z_2$ , $z_3$ , $z_4$ . Therefore $z_1=L^{-1}(L(z_1))$ must on the circle or line too, which means $z_1$ , $z_2$ , $z_3$ , $z_4$ lie on the same circle or straight line.
$Q\Rightarrow P$ : If the four points $z_1$ , $z_2$ , $z_3$ , $z_4$ lie on a circle or on a straight line, it means that $L$ maps the circle or line that $z_2$ , $z_3$ , $z_4$ on into the real axis. Therefore $L(z_1)=[z_1,z_2,z_3,z_4]\in\R$ .
$Q.E.D.$

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Complex Functions Assignment NO.1

Complex numbers and multi-valued functions

1. State the Euler’s formula (which connect exponential functions with trigonometric functions). Express $\cos3\theta$ in terms of $\cos\theta$ and $\sin\theta$ .

2. Find all values of $(-2+2i)^\frac{1}{3}$ (hint: you should use the polar representation.

Partial fraction of a rational function

4. Put the following rational functions into a partial fraction form $\frac{1-z^2}{(z-\alpha)^3}$ , $\frac{1-z^2}{(z-\alpha_1)(z-\alpha_2)(z-\alpha_3)}$ , where $\alpha_1$ , $\alpha_2$ , $\alpha_3$ are simple poles.

Möbius transformations

5. Give the domain of definition and the range of the maps $g\circ f^{-1}$ and $f\circ g^{-1}$ . Are these maps bijective holomorphic functions?

6. Prove that Möbius transformation is a conformal mapping.

7. Construct a special Möbius transformation that maps three distinct points $\{\infty,1,−1\}$ to $\{1,−i,i\}$ respectively. Such map is called Cayley transformation.

8. Prove that the Cayley transformation maps the real axis to the unit circle (that is circle of radius $1$ centered at $0$ ).

Cross-ratio

11. Prove the following statement: the cross-ratio $[z_1,z_2,z_3,z_4]$ is real if and only if the four points $z_1$ , $z_2$ , $z_3$ , $z_4$ lie on a circle or on a straight line. Try to give a proof by yourself. But you can refer to [Ahlfors, page 78] for details.

Ordinary Differential Equation Assignment NO.2

Analytic Geometry Assignment No.3

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人は繋がっているのよ

Complex numbers and multi-valued functions

1. State the Euler’s formula (which connect exponential functions with trigonometric functions). Express \cos3\theta in terms of \cos\theta and \sin\theta.

2. Find all values of (-2+2i)^\frac{1}{3} (hint: you should use the polar representation.

Partial fraction of a rational function

4. Put the following rational functions into a partial fraction form \frac{1-z^2}{(z-\alpha)^3}, \frac{1-z^2}{(z-\alpha_1)(z-\alpha_2)(z-\alpha_3)}, where \alpha_1, \alpha_2, \alpha_3 are simple poles.

Möbius transformations

5. Give the domain of definition and the range of the maps g\circ f^{-1} and f\circ g^{-1}. Are these maps bijective holomorphic functions?

6. Prove that Möbius transformation is a conformal mapping.

7. Construct a special Möbius transformation that maps three distinct points \{\infty,1,−1\} to \{1,−i,i\} respectively. Such map is called Cayley transformation.

8. Prove that the Cayley transformation maps the real axis to the unit circle (that is circle of radius 1 centered at 0).

Cross-ratio

11. Prove the following statement: the cross-ratio [z_1,z_2,z_3,z_4] is real if and only if the four points z_1, z_2, z_3, z_4 lie on a circle or on a straight line. Try to give a proof by yourself. But you can refer to [Ahlfors, page 78] for details.

Ordinary Differential Equation Assignment NO.2

Analytic Geometry Assignment No.3

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1. State the Euler’s formula (which connect exponential functions with trigonometric functions). Express $\cos3\theta$ in terms of $\cos\theta$ and $\sin\theta$ .

2. Find all values of $(-2+2i)^\frac{1}{3}$ (hint: you should use the polar representation.

4. Put the following rational functions into a partial fraction form $\frac{1-z^2}{(z-\alpha)^3}$ , $\frac{1-z^2}{(z-\alpha_1)(z-\alpha_2)(z-\alpha_3)}$ , where $\alpha_1$ , $\alpha_2$ , $\alpha_3$ are simple poles.

5. Give the domain of definition and the range of the maps $g\circ f^{-1}$ and $f\circ g^{-1}$ . Are these maps bijective holomorphic functions?

7. Construct a special Möbius transformation that maps three distinct points $\{\infty,1,−1\}$ to $\{1,−i,i\}$ respectively. Such map is called Cayley transformation.

8. Prove that the Cayley transformation maps the real axis to the unit circle (that is circle of radius $1$ centered at $0$ ).

11. Prove the following statement: the cross-ratio $[z_1,z_2,z_3,z_4]$ is real if and only if the four points $z_1$ , $z_2$ , $z_3$ , $z_4$ lie on a circle or on a straight line. Try to give a proof by yourself. But you can refer to [Ahlfors, page 78] for details.