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Math416Complex Variables (Lecture 6)

Math416 Lecture 6

Review

Linear Fractional Transformations

Transformations of the form f(z)=az+bcz+df(z)=\frac{az+b}{cz+d},a,b,c,d∈Ca,b,c,d\in\mathbb{C} and ad−bc≠0ad-bc\neq 0 are called linear fractional transformations.

Theorem 3.8 Preservation of clircles

We defined clircle to be a circle or a line.

The circle equation is:

Let z=u+ivz=u+iv be the center of the circle, rr be the radius of the circle.

circle={z∈C:∣z−c∣=r}circle=\{z\in\mathbb{C}:|z-c|=r\}

This is:

∣z∣2−cz‾−c‾z+∣c∣2−r2=0|z|^2-c\overline{z}-\overline{c}z+|c|^2-r^2=0

If ϕ\phi is a non-constant linear fractional transformation, then ϕ\phi maps clircles to clircles.

We claim that a map is circle preserving if and only if for some α,β,γ,δ∈R\alpha,\beta,\gamma,\delta\in\mathbb{R}.

α∣z∣2+βRe(z)+γIm(z)+δ=0\alpha|z|^2+\beta Re(z)+\gamma Im(z)+\delta=0

when α=0\alpha=0, it is a line.

when α≠0\alpha\neq 0, it is a circle.

Proof

Let w=u+iv=1zw=u+iv=\frac{1}{z}, so 1w=uu2+v2−ivu2+v2\frac{1}{w}=\frac{u}{u^2+v^2}-i\frac{v}{u^2+v^2}.

Then the original equation becomes:

α(uu2+v2)2+β(uu2+v2)+γ(−vu2+v2)+δ=0\alpha\left(\frac{u}{u^2+v^2}\right)^2+\beta\left(\frac{u}{u^2+v^2}\right)+\gamma\left(-\frac{v}{u^2+v^2}\right)+\delta=0

Which is in the form of circle equation.

Chapter 4 Elementary functions

et=∑n=0∞tnn!e^t=\sum_{n=0}^{\infty}\frac{t^n}{n!}

So, following the definition of eze^z, we have:

ex+iy=exeiy=ex(∑n=0∞(iy)nn!)=ex(∑n=0∞(−1)nynn!)=ex(cos⁡y+isin⁡y)\begin{aligned} e^{x+iy}&=e^xe^{iy} \\ &=e^x\left(\sum_{n=0}^{\infty}\frac{(iy)^n}{n!}\right) \\ &=e^x\left(\sum_{n=0}^{\infty}\frac{(-1)^ny^n}{n!}\right) \\ &=e^x(\cos y+i\sin y) \end{aligned}

eze^z

The exponential of ez=x+iye^z=x+iy is defined as:

ez=exp(z)=ex(cos⁡y+isin⁡y)e^z=exp(z)=e^x(\cos y+i\sin y)

So,

∣ez∣=∣ex∣∣cos⁡y+isin⁡y∣=ex|e^z|=|e^x||\cos y+i\sin y|=e^x

Theorem 4.3 eze^z is holomorphic

eze^z is holomorphic on C\mathbb{C}.

Proof

∂∂zez=12(∂∂x+i∂y)ex(cos⁡y+isin⁡y)=12ex(cos⁡y+isin⁡y)+iex(−sin⁡y+icos⁡y)=0\begin{aligned} \frac{\partial}{\partial z}e^z&=\frac{1}{2}\left(\frac{\partial}{\partial x}+\frac{i}{\partial y}\right)e^x(\cos y+i\sin y) \\ &=\frac{1}{2}e^x(\cos y+i\sin y)+ie^x(-\sin y+i\cos y) \\ &=0 \end{aligned}

Theorem 4.4 eze^z is periodic

eze^z is periodic with period 2πi2\pi i.

Proof

ez+2πi=eze2πi=ez⋅1=eze^{z+2\pi i}=e^z e^{2\pi i}=e^z\cdot 1=e^z

Theorem 4.5 eze^z as a map

eze^z is a map from C\mathbb{C} to C\mathbb{C} with period 2πi2\pi i.

eπi+1=0e^{\pi i}+1=0

This is a map from cartesian coordinates to polar coordinates, where exe^x is the radius and yy is the angle.

This map attains every value in C∖{0}\mathbb{C}\setminus\{0\}.

Definition 4.6-8 cos⁡z\cos z and sin⁡z\sin z

cos⁡z=12(eiz+e−iz)\cos z=\frac{1}{2}(e^{iz}+e^{-iz}) sin⁡z=12i(eiz−e−iz)\sin z=\frac{1}{2i}(e^{iz}-e^{-iz}) cosh⁡z=12(ez+e−z)\cosh z=\frac{1}{2}(e^z+e^{-z}) sinh⁡z=12(ez−e−z)\sinh z=\frac{1}{2}(e^z-e^{-z})

From this definition, we can see that cos⁡z\cos z and sin⁡z\sin z are no longer bounded in the complex plane.

And this definition is still compatible with the previous definition of cos⁡\cos and sin⁡\sin when zz is real.

Moreover,

cosh⁡(iz)=cos⁡z\cosh(iz)=\cos z sinh⁡(iz)=isin⁡z\sinh(iz)=i\sin z

Logarithm

Definition 4.9 Logarithm

A logarithm of aa is any bb such that eb=ae^b=a.

If a=0a=0, then no logarithm exists.

If a≠0a\neq 0, then there exists infinitely many logarithms of aa.

Let a=reiθa=re^{i\theta}, b=x+iyb=x+iy be a logarithm of aa.

Then,

ex+iy=reiθe^{x+iy}=re^{i\theta}

Since logarithm is not unique, we can always add 2kπi2k\pi i to the angle.

If y∈(−π,π]y\in(-\pi,\pi], then log⁡a=b\log a=b means eb=ae^b=a and Im(b)∈(−π,π]Im(b)\in(-\pi,\pi].

If a=reiθa=re^{i\theta}, then log⁡a=log⁡r+i(θ0+2kπ)\log a=\log r+i(\theta_0+2k\pi).

Definition 4.10 of Branch of arg⁡z\arg z and log⁡z\log z

Let GG be an open connected subset of C∖{0}\mathbb{C}\setminus\{0\}.

A branch of arg⁡(z)\arg(z) in GG is a continuous function α:G→G\alpha:G\to G, such that α(z)\alpha(z) is a value of arg⁡(z)\arg(z).

A branch of log⁡(z)\log(z) in GG is a continuous function β\beta, such that eβ(z)=ze^{\beta(z)}=z.

Note: GG has a branch of arg⁡(z)\arg(z) if and only if it has a branch of log⁡(z)\log(z).

Proof

Suppose there exists α(z)\alpha(z) such that ∀z∈G\forall z\in G, α(z)∈G\alpha(z)\in G, then l(z)=ln⁡∣z∣+iα(z)l(z)=\ln|z|+i\alpha(z) is a branch of log⁡(z)\log(z).

Suppose there exists l(z)l(z) such that ∀z∈G\forall z\in G, l(z)∈Gl(z)\in G, then α(z)=Im(z)\alpha(z)=Im(z) is a branch of arg⁡(z)\arg(z).

If G=C∖{0}G=\mathbb{C}\setminus\{0\}, then not branch of arg⁡(z)\arg(z) exists.

Corollary of 4.10

Suppose α1\alpha_1 and α2\alpha_2 are two branches of arg⁡(z)\arg(z) in GG.

Then,

α1(z)−α2(z)=2kπ\alpha_1(z)-\alpha_2(z)=2k\pi

for some k∈Zk\in\mathbb{Z}.

Suppose l1l_1 and l2l_2 are two branches of log⁡(z)\log(z) in GG.

Then,

l1(z)−l2(z)=2kπil_1(z)-l_2(z)=2k\pi i

for some k∈Zk\in\mathbb{Z}.

Theorem 4.11

log⁡(z)\log(z) is holomorphic on C∖{0}\mathbb{C}\setminus\{0\}.

Proof (continue on next lecture)

Method 1: Use polar coordinates. (See in homework)

Method 2: Use the fact that log⁡(z)\log(z) is the inverse of eze^z.

Suppose h=s+ith=s+it, eh=es(cos⁡t+isin⁡t)e^h=e^s(\cos t+i\sin t), eh−1=es(cos⁡t−1)+isin⁡te^h-1=e^s(\cos t-1)+i\sin t. So

eh−1h=(s+it)es(cos⁡t−1)+isin⁡ts2+t2=es(cos⁡t−1)s2+t2+isin⁡ts2+t2\begin{aligned} \frac{e^h-1}{h}&=\frac{(s+it)e^s(\cos t-1)+i\sin t}{s^2+t^2} \\ &=\frac{e^s(\cos t-1)}{s^2+t^2}+i\frac{\sin t}{s^2+t^2} \end{aligned}

Continue next time.

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