Question

Show that the function fz = z is continuous but not differentiable at the

point z = 0.​

Answer 1

Given:

A function $f(z)=\bar z$

To show:

$f(z)=\bar z$ is continuous but not differentiable at the point $z=0$.

Solution:

$f(z)=\bar z=x-iy$           (∵ $z=x+iy$ is a complex number, where $x$ and $y$ are real valued functions )  

We know that real value functions are everywhere continuous and hence, $x$ and $y$ are continuous everywhere in $z$-plane.

∴ $f(z)=\bar z=x-iy$ is continuous at the point $z=0$.

Now we will check differentiability $z=0$

$f(z)=\bar z=x-iy$$=u(x,y)+iv(x,y)$

$u_{x} =1$, $v_{x} =0$

$u_{y}=0$,$v_y=-1$

$u_x\neq v_y$ for every $(x,y)$

Hence, Cauchy-Riemann's equations are nowhere satisfied at $z$-plane.

∴By necessary condition, $f(z)=\bar z$ is nowhere differentiable in $z$-plane

⇒ $f(z)=\bar z$ is not differentiable at $z=0$.

Hence proved.