Prove that \(f(z) = \abs{z}^2\) has a derivative at \(z=0\) and nowhere else.
Solution
The easy check: \(f\) is differentiable iff \(\delbar_z f = 0\), but \[ \delbar_z \abs{z}^2 = \delbar_z z\bar{z} = z \neq 0 ,\] unless of course \(z=0\).
A more explicit check: check the limits. \[ {f(z) - f(0) \over z-0} = { \abs{z}^2 \over z } = {z\bar z \over z} = \bar{z} \converges{z\to 0}\too 0 ,\] so \(f\) is differentiable at \(w=0\). Now taking \(w = Re^{i\theta} \neq 0\), \[ {f(z) -f(w) \over z-w} = {\abs{z}^2 - \abs{w}^2 \over z - w} = {\qty{\abs{z} + \abs{w} } \qty{\abs{z} - \abs{w}} \over z-w } = {\abs z - \abs w \over z-w}\cdot \qty{\abs{z} + \abs{w}} .\] First let \(z\to w\) along \(\bd \DD_{R'}(0)\) where \(R' \da \abs{w}\), so that the numerator vanishes and the limit is zero. Then let \(z\to w\) along the curve \(\ts{tw\st t\in [0, 1]}\), then \(\abs{z} = t \abs{w}\), so the ratio becomes \[ {\abs z - \abs w \over z-w}\cdot \qty{\abs{z} + \abs{w}} &= {t\abs{w} - \abs w \over tw-w}\cdot \qty{t\abs{w} + \abs{w}} \\ &= {\abs{w}\qty{t-1 } \over w(t-1)} \cdot \abs{w}(t+1) \\ &= { \abs{w}^2(t+1) \over w} \\ &= \bar{w}(t+1) \\ &\converges{t\to 1}\to 2\bar{w} ,\] which is nonzero is \(w\neq 0\).