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Let $f(z): \mathbb{C}\rightarrow \mathbb{C}$ be a complex function such that $f(z) = \begin{Bmatrix} \frac{z^5}{|z|^4},z\neq 0\\ 0, z=0 \end{Bmatrix}$.

Prove that Cauchy-Riemann equations(in x and y version) are met in $(0,0)$ but the function is not differentiable in $(0,0).$

In order to avoid using the binomial theorem on the expression $(x+yi)^5$ and getting a general expression for $u(x,y),v(x,y)$. I did as followed -

$u_x(0,0) = \lim_{x\rightarrow 0} \frac{u(x,0)-u(0,0)}{x}$

an expression for $u(x,0)$ would be -

$f(x+0i) = u(x,0)+v(x,0)i$ = $\frac{(x+0i)^5}{x^4}$ = $x$.

so $u_{x}(0,0)$ is just 1.

and getting $u_{y}(0,0),v_{x}(0,0),v_{y}(0,0)$ would be in a similiar way.

Is that legal mathematically? or a logical mistake was fallen somewhere? beacuse there are a lot of arguments around my fellows in class.

Thanks in advance.

mcr0yal
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  • This is correct: you know that there are real-valued functions $u$ and $v$ s.t. $f(x+iy)=u(x,y)+iv(x,y)$ for all $(x,y)\in\mathbb{R}^2$, and you remarked that you only needed to see what are the functions $x\mapsto u(x,0)$, $x\mapsto v(x,0)$, $y\mapsto u(0,y)$ and $y\mapsto v(0,y)$ in order to check if the Cauchy-Riemann equations are met. – Balloon Nov 17 '21 at 17:04

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