Let $ \mathbb{R}_{-}=\{t \in \mathbb{R}: t \leq 0\}$ and definied $\log : \mathbb{C} \backslash \mathbb{R}_{-} \rightarrow \mathbb{C}$ as branch of logarithm.
Find real and imaginary part of $f(z)=(z-1) \cdot \operatorname{Re}(\log z)$ and clarify where this function $f$ have complex derivative.
Solution:
We have that:
$$\Re(\log(z)) = \Re(\log(|z|) + i\arg(z)) = \log(|z|)$$
So the real part is:
Let $z = a + ib$
$$ \Re(f(z)) = (a – 1)\log\sqrt{a^{2}+b^{2}}$$
$$ \Im(f(z)) = \Re(\log\sqrt{a^{2}+b^{2}}) b $$
For where f have complex derivative:
I think only in $z=0$ we have problem, but I don't build formally condition. How to show this?
Best Answer
Starting with your functions: $$ \Re(f(z)) = (a - 1)\log\sqrt{a^{2}+b^{2}} = \frac12(a - 1)\log (a^{2}+b^{2})$$ $$ \Im(f(z)) = b\log\sqrt{a^2+b^2}=\frac12b\log (a^{2}+b^{2})$$ where $z=a+bi$. This is not define for $a=b=0$, so we'll suppose that they are not both zero.
A complex function is derivable if it respect the Cauchy-Riemann equations. With respect to your notation, this gives : $$\frac{\partial\Re}{\partial a}= \frac{\partial\Im}{\partial b}\tag{1}$$ $$\frac{\partial\Re}{\partial b}= -\frac{\partial\Im}{\partial a}\tag{2}$$
This will give us conditions on $a$ and $b$ for derivability. $$\frac{\partial\Re}{\partial a}=\frac12\log (a^{2}+b^{2})+(a-1)\frac{a}{a^2+b^2}$$ $$\frac{\partial\Re}{\partial b}=(a-1)\frac{b}{a^2+b^2}$$ $$\frac{\partial\Im}{\partial a}=b\frac{a}{a^2+b^2}$$ $$\frac{\partial\Im}{\partial b}=\frac12\log (a^{2}+b^{2})+b\frac{b}{a^2+b^2}$$ In equation $(1)$, it gives $$\frac12\log (a^{2}+b^{2})+(a-1)\frac{a}{a^2+b^2}=\frac12\log (a^{2}+b^{2})+b\frac{b}{a^2+b^2}$$ $$(a-1)\frac{a}{a^2+b^2}=b\frac{b}{a^2+b^2}$$ $$(a-1)a=b^2\tag{3}$$ In equation $(2)$, it gives $$(a-1)\frac{b}{a^2+b^2}=-b\frac{a}{a^2+b^2}$$ $$(a-1)b=-ab$$ $$(2a-1)b=0\tag{4}$$ Equation $(4)$ is true only if $b=0$ or $a=\frac12$. Those are our candidates that will be tested in equation $(3)$.
If $b=0$ $$(a-1)a=0^2 \implies a=0 \text{ or } a=1$$ Since $b=0$, $a$ can't be $0$, so only one possiblity $z=1+0i$.
If $b=\frac12$ $$(\frac12-1)\frac12=b^2 \implies -\frac14=b^2$$ It is impossible, since $b\in \Bbb R$.
Conclusion, this function is only derivable at $z = 1$.