# Show $f(z) = |z|^2$ is differentiable only at $z = 0$

So far I have rewritten the following expression:

$$f'(z) = \lim_{\Delta z\rightarrow0}\frac{f(z + \Delta{z})-f(z)}{\Delta{z}}$$

As such:

$$f'(z) =\lim_{\Delta z\rightarrow 0} (\bar{z} + \bar{\Delta{z}} + z\frac{\bar{\Delta{z}}}{\Delta{z}})$$

This limit only exists if each of the three terms in the summation have limits. Clearly the first two do, so I am currently evaluating the third term, that is, seeing if.. $$z(\lim_{\Delta{z}\rightarrow 0}\frac{\bar{\Delta{z}}}{\Delta{z}})$$

..exists for $z=0$. I don't think the limit above exists: if we allow $\Delta{z}$ to approach $0$ through the real numbers, then the numerator and denominator are the same and the limit evaluates to 1. Else, if we let $\Delta{z}$ approach $0$ from 'above' (through the imaginary numbers), we have the denominator is the negation of the numerator and so the limit is $-1$. Hence the limit is undefined.

However, would $0$ times a limit that does not exist be $0$? I'm just not sure how to prove this term does indeed evaluate to some value instead of being undefined; it's clear that for $z \neq 0$ that the expression is undefined, but for $z=0$ I'm confused if $0$ times an undefined limit is $0$?

• $\lim x_ny_n = (\lim x_n) (\lim y_n)$ is true when $\lim x_n$ and $\lim y_n$ exists. Similar argument can be done for complex limits. So taking out $z$ out of the limit there was a mistake. When $z=0$, the limit is clearly $0$, because it's the limit of $0$'s. Commented Sep 5, 2018 at 21:54

Think of a limit in $\mathbb C$ as a limit in $\mathbb R^2$. You can show a limit does not exist by taking two different paths.

Think of $\Delta z$ as $\Delta x +i \Delta y$ (where $\Delta x$ and $\Delta y$ are real), and take the limit for $(\Delta x, \Delta y)\to(0,0)$ along several curves. For instance: $$\lim_{(\Delta x,\Delta y)\to (0,0)} \frac{\overline {\Delta x+i \Delta y}}{\Delta x+i \Delta y},$$ becomes for $\Delta x=0$ $$\lim_{\Delta y\to 0} \frac{0-i \Delta y}{0+i \Delta y}=-1$$ and for $\Delta y=0$ $$\lim_{\Delta x\to 0} \frac{\Delta x-i \cdot 0}{\Delta x+i \cdot 0}=1.$$ This proves that the limit does not exist.

In any case, that doesn't let you conclude anything with your calculations. I suggest that you go back to the beginning and apply the same idea to the whole limit.

You should get that the limit does not exist if $z\neq 0$, that is if $z=x+iy$ with $(x,y)\neq (0,0)$. For the case $(x,y)=(0,0)$ it is not enough in general to see that both directions give the same limit, so you should use the definition.

However, it is possible that you have read or heard about Cauchy Riemann equations. The theorem states:

If $f(x+iy)=u(x,y)+i\cdot v(x,y)$, with $u, v\colon \mathbb R^2 \to \mathbb R$ two differentiable functions, then $f$ is differentiable at $x_0+iy_0$ IF AND ONLY IF $$\frac {\partial u}{\partial x}(x_0,y_0)=\frac {\partial v}{\partial y}(x_0,y_0)$$ and $$\frac {\partial u}{\partial y}(x_0,y_0)=-\frac {\partial x}{\partial y}(x_0,y_0).$$

Try using this theorem. Since for $z=x+iy$, $$|z|^2=|x+iy|^2=x^2+y^2=(x^2+y^2)+0i,$$ here we have $u(x,y)=x^2+y^2$ and $v(x,y)=0$.

• The problem is asking us to solve this without using the Cauchy-Riemann equations. But your explanation seems to make sense. Commented Sep 5, 2018 at 22:09
• So if I consider the whole limit, say when $\Delta{z}$ goes through the real values towards $0$ and assuming $z \neq 0$, I get $\Delta{z} = \Delta{x}$: $f'(z) = lim_{\Delta{x}\rightarrow 0}(\bar{z}+0+z\frac{\bar{\Delta{x}}}{\Delta{x}})=\bar{z}+z$ Then with $\Delta{z}$ approaching from above: $f'(z)=\bar{z} - z$ Which, if $z = x+iy$, means $\bar{z} - z = \bar{z} + z$ only when $z = -z$, or $x + iy = -x-iy$, i.e. $2x + 2iy = 0$. This only holds if $x,y$ are both $0$ ($z=0$), and in all other cases fails. Thus the limits are not equal when $z \neq 0$ and $f'(z)$ doesn't exist? Commented Sep 5, 2018 at 22:10
• Yes. And if $z=0$ you can write the limit for that special case and see that it reaches some result in $\mathbb C$. But this without considering alternative directions. Just $\Delta z \to 0$ in general. Commented Sep 5, 2018 at 22:15
• I understand now. Thanks a ton. Commented Sep 5, 2018 at 22:16