Show that $f(z) =\log z$ is analytic everywhere in the complex plane
except at the origin. Find its derivative.
I tried solving it using Cauchy Riemann equation. But for that, $f(z)$ needs to be separated as $f(z)=u+iv$. I am not able to split $\log z =u+iv$. Please help.
$z=x+iy$
Best Answer
That's simply not true. You can't even define a continuous logarithm function in $\mathbb{C}\setminus\{0\}$. To define logarithm you need to choose a branch of argument. For example we can choose $arg(z)$ to be in $[-\pi,\pi)$, in that case we define logarithm in $\mathbb{C}\setminus(-\infty,0]$. Then logarithm is a function, let's call it $L(z)$. It is really analytic. Take a point $z_0$ in the domain. We can write $w=L(z),w_0=L(z_0)$. Then:
$\lim_{z\to z_0}\frac{L(z)-L(z_0)}{z-z_0}=\lim_{w\to w_0}\frac{w-w_0}{e^w-e^{w_0}}=\frac{1}{e^{w_0}}=\frac{1}{z_0}$
I just used the fact that the derivative of $e^w$ at the point $w_0$ is $e^{w_0}$.