202102190222 Homework 1 (Q2)

Show whether or not the function (a) $f(z)=\mathrm{Re}(z)=x$ ; (b) $f(z)=z^2$ is analytic.

Setup.

Theorem 1. Suppose $f:\Omega \rightarrow \mathbb{C}$ is analytic. Then, writing

$f(z)=f(x+\mathrm{i}y)=u(x,y)+\mathrm{i}v(x,y)$

with $u(x,y)=\mathrm{Re}(f(z))$ and $v(x,y)=\mathrm{Im}(f(z))$ , we have the Cauchy-Riemann equations

$\begin{cases} \displaystyle{\frac{\partial u}{\partial x}} & = \displaystyle{\frac{\partial v}{\partial y}} \\ \displaystyle{\frac{\partial u}{\partial y}} & = \displaystyle{-\frac{\partial v}{\partial x}} \end{cases}$

everywhere on $\Omega$ .

Theorem 2. If $f:\Omega\rightarrow \mathbb{C}$ is of class $\mathcal{C}^1$ , $f=u+\mathrm{i}v$ . Then $(u,v)$ satisfies the Cauchy-Riemann equations if and only if $f$ is analytic.

(a) Now that

$u(x,y)=x$ and $v(x,y)=0$ ,

the partial derivatives are as follows

$\begin{aligned} \frac{\partial u}{\partial x} & = 1 & \qquad \frac{\partial u}{\partial y} = & 0 \\ \frac{\partial v}{\partial y} & = 0 & \qquad -\frac{\partial v}{\partial x} = & 0 \\ \end{aligned}$

Since

$1=\displaystyle{\frac{\partial u}{\partial x}}\neq \displaystyle{\frac{\partial v}{\partial y}}=0$ ,

the Cauchy-Riemann equations are not satisfied, and $f$ is $\textrm{\scriptsize{NOT}}$ analytic.

(b)

$\begin{aligned} f(z) & = z^2 \\ f(x+\mathrm{i}y) & = (x+\mathrm{i}y)^2 \\ & = (x^2-y^2)+\mathrm{i}(2xy) \\ & = u(x,y)+\mathrm{i}v(x,y) \end{aligned}$

where $u(x,y)=x^2-y^2$ and $v(x,y)=2xy$ .

Then,

$\displaystyle{\frac{\partial u}{\partial x}} = 2x$ , $\displaystyle{\frac{\partial u}{\partial y}} = -2y$ , $\displaystyle{\frac{\partial v}{\partial y}} = 2x$ , $\displaystyle{-\frac{\partial v}{\partial x}} = -2y$

and the Cauchy-Riemann equations are satisfied. $f$ is thus analytic.

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