Curvature in complex analysis

Question

Apparently, the curvature of the image contour of the unit circle $|z| = 1$ under the conformal mapping $w = f(z)$ is $$\frac{1}{|zf'(z)|} \text{Re}\bigg(1 + \frac{zf''(z)}{f'(z)}\bigg).$$ How does one show this? How do we even define curvature in complex analysis? I could not find a single book that explains what curvature is in complex analysis.

Answer 1

As Sky said, the relevant concept of curvature is not something specific to complex analysis. It's the curvature of plane curves, defined as the rate of change of the unit tangent vector. In complex terms, the argument if the tangent vector at $f(z)$ is $\operatorname{Im} \log (iz f'(z))$.

Let $g(z)=\log (iz f'(z))$. Since $$g'(z) = \frac1z +\frac{f''(z)}{f'(z)}$$ the rate of change of $g$ as $z$ moves along the unit circle with unit speed is $iz g'(z) $. (Here $iz$ is the unit tangent vector for the circle.) Taking imaginary part yields $$ \operatorname{Im}\left(i+ \frac{iz f''(z)}{f'(z)}\right)= \operatorname{Re}\left(1+ \frac{z f''(z)}{f'(z)}\right) $$ And this needs to be divided by $|f'(z)|$ because when $z$ travels around the unit circle with unit speed, $f(z)$ traverses the image with speed $|f'(z)|$. The final answer is $$ \frac{1}{|f'(z)|}\operatorname{Re}\left(1+ \frac{z f''(z)}{f'(z)}\right) $$ which is equivalent to yours since $|z|=1$.