Analytic function related

Question

The function $f(z)=y+ix$ is

(a) Differentiable everywhere

(b) Differentiable no-where

(c)Differentiable only when $y=x$

(d)Differentiable only at $0$

As $f(z)=y+ix$

Hence $u=y$ and $v=x$

According to C-R equation a function is analytic if $u_x=v_y$ and $u_y=-v_x$

But in this case $u_x=0,u_y=1,v_x=1,v_y=0$

Hence it does not satisfy C-R equation for any $x$ and $y$.

So my answer is $b$

Please guide me whether I am correct or not?

Answer 1

Use $f^{'}(z_0)=lim_{z\rightarrow z_0}\frac{f(z)-f(z_0)}{z-z_0}$

Answer 2

I think the answer key is right in saying that option c) is true. This is because, when $x=y$, $f(z)=f(x+iy)=f(x+ix)=x+ix$, which then reduces to an identity on the set $A=[z:z\in \mathbb{C},\Re(z)=\Im(z)]$, hence rendering it differentiable on the set $A$.

Answer 3

The question is vague. "Differentiable" can mean complex differentiable (holomorphic) as a function $\mathbb{C}\to\mathbb{C}$ or real differentiable as a function $\mathbb{R}^2\to\mathbb{R}^2$.

As a function $\mathbb{C}\to\mathbb{C}$, the function is nowhere holomorphic since it does not satisfy the Cauchy-Riemann equations, as you showed.

As a function $\mathbb{R}^2\to\mathbb{R}^2$ it is everywhere differentiable since it's just the map $f(x,y)=(y,x)$, which has continuous partial derivatives everywhere.