Review for Exam 1 Problem 1

Problem 1 on the Exam 1 review sheet asks us to show that the function $f(z)=|z|$ is nowhere differentiable using the definition of the derivative.

Let $x,y\in\mathbb{R}$ and let $z=x+iy$. If we approach $f(z)$ along the real axis then

\begin{align}
&\lim\limits_{h\rightarrow0}\frac{f((x+h)+iy)-f(x+iy)}{h}\\&=\lim\limits_{h\rightarrow0}\frac{|(x+h)+iy|-|x+iy|}{h}\\&=\lim\limits_{h\rightarrow0}\frac{\sqrt{(x+h)^2+y^2}-\sqrt{x^2+y^2}}{h}
\end{align}

Similarly, if we approach $f(z)$ along the imaginary axis then

\begin{align}
&\lim\limits_{h\rightarrow0}\frac{f(x+i(y+h))-f(x+iy)}{ih}\\&=\lim\limits_{h\rightarrow0}\frac{|x+i(y+h)|-|x+iy|}{ih}\\&=\lim\limits_{h\rightarrow0}\frac{\sqrt{x^2+(y+h)^2}-\sqrt{x^2+y^2}}{ih}
\end{align}

We can show that $f(z)$ is nowhere differentiable by showing that these two limits have different results. How can one show this?

Given $z_0 \in \mathbb C$, you might look at
$$\lim_{z\to z_0} \frac{|z-z_0|}{z-z_0}$$
along the line $\arg z=\arg z_0$ and along the circle $|z|=|z_0|$. I think the limit for one will be zero and for the other it will be one.

Thank you! Using your definition of the limit, I let $z=x+iy$ and approached the limit from two different directions. Approaching the limit from the real axis gave $1$ and approaching the limit from the imaginary axis gave $-i$.

@dgallimo I don't think I believe that. If $z_0=i$, I think that
$$\lim_{h\to 0,h\in\mathbb R}\frac{|i+h|-|i|}{h} = 0.$$
In general, I think that limit along the real axis of the difference quotient depends upon $z_0$.