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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.

José Carlos Santos
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fynmnx
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    $u$ should be the real part of $f$ and $v$ the imaginary part of $f$. You chose them quite randomly to be the numerator and the denominator, that's why you get a wrong result. – Yaddle Sep 20 '18 at 16:03
  • Wouldn't $u$ then be denominator $|z|^4$? – fynmnx Sep 20 '18 at 16:09
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    I think you're confused about what "real part" and "imaginary part" means. It doesn't mean components which are real and components which are not real. What it means is: $f(z)$ is a complex number, so it splits into a real and imaginary part $f(z) = u(z) + i\cdot v(z)$ where $u$ and $v$ are real valued functions. $u$ is the real part and $v$ is the imaginary part. – Callus - Reinstate Monica Sep 20 '18 at 16:12
  • I think that might be my sticking point. So in this case would would be the real and imaginary? – fynmnx Sep 20 '18 at 16:14

1 Answers1

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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 $t$.

José Carlos Santos
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