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.
[Math] Curvature in complex analysis
complex numberscomplex-analysiscurvature
Related Solutions
$$ | z - 1 | = 1 $$ $$ z = e^{i\theta} + 1 $$
$$ f(z)= z^2 $$
working in the w plane now.
$$\rho e^{i\phi} = (1 + e^ {i \theta})^2 $$
$$\rho e^{i\phi} = 1 + 2e^{i \theta} + e^{2i \theta} $$
$$\rho e^{i\phi} = e^{i\theta} ( e^{-i\theta} + 2 + e ^{i\theta}) $$
using an identity for the complex cosine function
$$ \rho e^{i\phi} = e^{i\theta} ( 2 + 2 cos\theta) $$
$$ \rho= ( 2 + 2 cos\theta) , (\phi = \theta )$$
Which is a cardioid in the w plane
Another method:
Parametric equation for the circle:
$x(t) = cos(t) + 1$ and $y(t) = sin(t) $
$f(z) = z^2 $ which takes $(u,v)$ to $(u^2 - v^2, 2uv)$
we get, after playing around a bit:
$ (1 + 2 cos(t) + cos(2t) , 2sin(t) + sin(2t) ) $
The equation of a cardioid.
Only, it should be in the form $ r = 2(1 + cos(t)) $ , same as above. So, a little work perhaps, it's the same cardioid...
Continuously differentiable (in particular, holomorphic) applications map boundaries into boundaries, hence it is enough to understand the image of $|z|=\frac{\pi}{2}$, made by points of the form $\frac{\pi}{2}e^{i\theta} = \frac{\pi}{2}\cos\theta + i\frac{\pi}{2}\sin\theta $ for some $\theta\in[0,2\pi)$. By exponentiating this thing we get
$$ e^{\frac{\pi}{2}\cos\theta}\left[\cos\left(\frac{\pi}{2}\sin\theta\right)+i\sin\left(\frac{\pi}{2}\sin\theta\right)\right] $$ i.e. a regular curve contained in the annulus $e^{-\pi/2}\leq |z|\leq e^{\pi/2}$ and in the half-plane $\text{Re}(z)\geq 0$:
$\hspace1in$
It is a sort of cardioid.
Best Answer
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 $|z|=1$.