I want to prove that $f(z) = \text{Im }z$ is not differentiable anywhere. I know how to prove it easily with the Cauchy-Riemann equations, however I'm also interested in proving it by just using the definition of differentiability.
I know that the definition of differentiability for complex functions is almost the same as for real functions, but I'm still having problem proving it since we're only considering the imaginary part.
Thanks in advance
An intuitive way to see that $f(z) = \text{Im }z$ is not differentiable is to note that a function $f$ that is differentiable at $z$ must, in the limit, map a small disk around $z$ to a small disk around $f(z)$ - see Needham's Visual Complex Analysis. The disk may be rotated by an anlge $\theta$ and scaled by a uniform real factor $r \ge 0$ (in which case $f'(z) = re^{i\theta}$) but it cannot be scaled by different factors in different directions.
For $f(z) = \text{Im }z$ any disk around $z$ is mapped to a segment of the real line, which is clearly not a disk, so $f(z)$ is not differentiable.
The definition of derivative for a function $\Bbb C\to\Bbb C$ is basically exactly the same as for real numbers. Given a function $f$ (complex or real), and a point $p$ in the domain of $f$, the derivative of $f$ at $p$, commonly denoted $f'(p)$, is a (complex or real) number such that $$ f(p+x)\approx f(p)+f'(p)\cdot x $$ for a certain, rigourously defined meaning of $\approx$.
In this case, we actually need that meaning, so here it is: $f(p+x)\approx f(p)+f'(p)\cdot x$ means $f(p+x)-f(p)- f'(p)\cdot x\approx0$, and more specifically: $$ \lim_{x\to0}\frac{f(p+x)-f(p)-x\cdot f'(p)}{|x|}=0 $$ You can easily check by inserting any complex number for $p$, and your particular function $\operatorname{Im}$, and see that there is no complex number we can insert for $f'(p)$ to make the limit true for complex $x$.
\Im(z)renders $\Im(z)$. – Jan 27 '20 at 14:40