Prove that $f$ defined by $f(z) = \frac{z^5}{|z|^4} | (z\neq 0), f(0)=0$ satisfies the Cauchy-Riemann equations at $z=0$ but is not differentiable there.
So I know that $u_x = v_x$.
Therefore let $u(x,y) = |z|^4 = (x^2+y^2)^2$ and let $v(x,y)=
z^5 = (x+iy)^5$.
These are not equal. Where have I made my mistake?
Thank you for any guidance.
Best Answer
There is no reason to take $u(x,y)=\lvert x+yi\rvert^4$ and $v(x,y)=(x+yi)^5$. Note that it is not even true that $v$ is a function from $\mathbb{C}$ into $\mathbb R$.
Note that, if $x\in\mathbb{R}\setminus\{0\}$, then $f(x)=\frac{x^5}{x^4}=x$ and that therefore $u(x,0)=x$ and $v(x,0)=0$. So, $u_x(0,0)=1$ and $v_x(0,0)=0$. On the other hand, $y\in\mathbb{R}\setminus\{0\}\implies f(yi)=\frac{(yi)^5}{y^4}=yi$ . Therefore, $u(0,y)=0$ and $v(0,y)=y$. So, $u_y(0,0)=0$ and $v_y(0,0)=1$. So, $(0,0)$ is a solution of the Cauchy-Riemann equations.
However, $f$ is not differentiable at $0$. The limit$$\lim_{z\to0}\frac{z^5}{z|z|^4}=\lim_{z\to0}\frac{z^2}{\overline z^2}$$does not exist. Just see what happens when $z\in\mathbb R$ and when $z$ s of the form $(1+i)t$, for a real $t$.