I am trying to show that $f(x+iy)=3x+4yi$ is nowhere differentiable by definition.
I have shown by use of the Cauchy-Riemann Equations that $f(x+iy)$ is nowhere differentiable, but I am having trouble showing the same result by definition.
I let $z=x+iy$, so $f(z)=4z-\operatorname{Re}(z)$, so by definition:
\begin{align}
\lim_{h\to 0} \frac{f(z+h)-f(z)}{h}&=\lim_{h\to 0} \frac{4(z+h)-\operatorname{Re}(z+h)-4z+\operatorname{Re}(z)}{h} \\
&=\lim_{h\to 0} \frac{4h-\operatorname{Re}(h)}{h}
\end{align}
Letting $h=a+ib \ \ a,b\in\mathbb{R}$, we have
$$\lim_{(a,b)\to (0,0)} \frac{3a+4ib}{a+ib}$$
Now if we take the line along the positive real axis ($\operatorname{Im}(h)=0$), we have $$\lim_{a\to 0} \frac{3a}{a}=3$$
If we also take the line along the positive imaginary axis, ($\operatorname{Re}(h)=0$), we have
$$\lim_{b\to 0} \frac{4ib}{ib}=4$$
As these two limits aren't equal and by the uniqueness of limits, the limit doesn't exist. Hence $f(z)$ is nowhere differentiable.
Is the logic correct? Is there a better way of doing differentiation by definition?
Best Answer
Your approach is correct. However Cauchy-Riemann Equations don't hold and you showed with definition that this function is differentiable nowhere, one may observe $$f(z)=4z-{\bf Re}\ z=\dfrac72z-\dfrac12\overline{z}$$ isn't analytic and this approach is sometimes applicable as well.