complex analysis - Determining Points of $mathbb{C}$ where $f(z)$ is differentiable

Sorry for the question but i'm partially confused about the result and i'm hoping someone can help. here we go

Stating clearly any result you use, determine the set of points in $\mathbb{C}$ where the following functions are differentiable.

$$f(z) = \left\{ \begin{matrix} \frac{z^4}{|z|^2} ~ \text{if z}\neq 0\\ 0 ~\text{if z} = 0\end{matrix}\right.$$

so here's where i'm at so far. let $z\neq 0$, then we use euler's form giving

$$f(z) = \frac{|z|^4(\cos{4\theta}+i\sin{4\theta})}{|z|^2}$$ via De'moivres $\implies$ (subbing |z| = r) $$f(z) = r^2(\cos{4\theta}+i\sin{4\theta})$$

From here, we consider f as $f: \mathbb{R}^2 \longrightarrow \mathbb{R}^2$ where $f(u(r,\theta),v(r,\theta)) = u(r,\theta)+i v(r,\theta)$

for f to be holomorphic (this is one of the problems i've got already...the phrasing of the question says to determine the points in $\mathbb{C}$ where f is differentiable, and not necessarily holomorphic?)

either way...crack on: for f to be holomorphic, u,v must be Real-differentiable (which they are) and the CR-Equations must be satisfied.

Using

$$u(r,\theta)=r^2 \cos{4\theta}~\&~v(r,\theta) = r^2\sin{4\theta}$$

we need to satisfy

$$u_{r}= \frac{1}{r}v_{\theta}, ~\&~ v_r = -\frac{1}{r}u_{\theta}$$

Running the calculations we have

$$u_r = 2r \cos{4\theta}$$ $$u_{\theta} = -4r^2 \sin{4\theta}$$ $$v_{r} = 2r \sin{4\theta}$$ $$v_{\theta} = 4r^2 \cos{4\theta}$$

substituing into the above values gives

$$u_{r}= \frac{1}{r}v_{\theta} \implies 2r \cos{4\theta} = 4r \cos{4\theta} \implies 1 = 2$$ and $$v_{r}= -\frac{1}{r}u_{\theta} \implies 2r \sin{4\theta} = 4r \sin{4\theta} \implies 1 = 2$$

what have i done wrong? if it's nothing...does this mean that the function isn't differentiable at any point $(z \neq 0 )\in \mathbb{C}$?

cheers for reading.

i think it may be a bit of a long day on my behalf, and i think i'm just doing something stupid at this point...but tbh i just can't see it right now

If i minus one side from the other and then compare the two i get

$$(r,\theta) = (1/2,\theta)$$ satisfies both equations, otherwise nothing else does as Cos and sin are never zero at the same time.

does this mean that f(z) is holomorphic on the disk of radius 1/2?

Cheers for reading.

Answer

The function is real differentiable, indeed smooth, for $z\ne 0$, being the quotient of two smooth functions. Now, note that for $z\ne 0$ we can rewrite

$$f(z) = \frac{z^4}{|z|^2} = \frac{z^4}{z\bar z} = \frac{z^3}{\bar z};$$ thus, $\dfrac{\partial f}{\partial\bar z} \ne 0$, and so $f$ is not holomorphic.

Now, what about at $z=0$? Write down the difference quotient

$$f'(0) = \lim_{h\to 0}\frac{f(h)-f(0)}h = \lim_{h\to 0}\frac{\frac{h^3}{\bar h}}h = \lim_{h\to 0} \frac{h^2}{\bar h}.$$ Note that $$\left|\frac{h^2}{\bar h}\right| = \left| h\cdot \frac{h}{\bar h}\right| = |h|\left|\frac h{\bar h}\right| = |h| \to 0,$$ and so $f'(0)=0$ and $f$ is complex differentiable at $0$.