Question

prove that f(z)=z² is an analytic function​

Answer 1

$f(z) = {z}^{2}$

$z := x + yi$

$f(z) = {x}^{2} - {y}^{2} + 2xyi := u + vi$

$\begin{cases}u = {x}^{2} - {y}^{2} \\ v = 2xy\end{cases}$

Now, With the help of Cauchy Reimann system of Equations.

$\begin{cases} \frac{ \partial u}{ \partial x} = \frac{ \partial v}{ \partial y} \\ \frac{ \partial u}{ \partial y} + \frac{ \partial v}{ \partial x} = 0 \end{cases}$

$\begin{cases}\frac{ \partial u}{ \partial x} = 2x \\ \frac{ \partial v}{ \partial y} = 2x \end{cases} \: \: \: \: \implies \frac{ \partial u}{ \partial x} = \frac{ \partial v}{ \partial y}$

$\begin{cases} \frac{ \partial u}{ \partial y} = - 2y \\ \frac{ \partial v}{ \partial x} = 2y \end{cases} \: \: \: \: \implies \frac{ \partial u}{ \partial y} + \frac{ \partial v}{ \partial x}= 0$

Therefore, the Function f(z)=z² is analytic all over the complex plane.