Question

Which one of the following functions is differentiable at $z=0$ but NOT differentiable at any other point in the complex plane $\mathbb{C}$?

• A. $f(z)=z|z|,\;\; z\in \mathbb{C}$
• B. $f(z)=\sin(z),\;\; z\in \mathbb{C}$
• C. $f(z)= \begin{cases} \dfrac{1}{\bar z}, & z\neq 0,\\ 0, & z=0, \end{cases} \;\; z\in \mathbb{C}$
• D. $f(z)=e^{-z^2},\;\; z\in \mathbb{C}$

Hint:

Functions involving $\bar z$ or $|z|$ typically break analyticity; sometimes they are differentiable only at isolated points (like $z=0$). Always check by direct limit at that point.

Answer 1

The Correct Option is A

Step 1: Recall the definition of complex differentiability.
A function $f:\mathbb{C}\to\mathbb{C}$ is differentiable at $z_0$ if the limit \[ f'(z_0)=\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0} \] exists and is independent of the path of approach. For differentiability in an open set, the Cauchy–Riemann equations must also hold with continuous partial derivatives.
Step 2: Check option (B) $\sin(z)$.
$\sin(z)$ is entire (analytic everywhere in $\mathbb{C}$). It is differentiable at all points, not just $0$. Hence (B) is false.
Step 3: Check option (C) $\dfrac{1{\bar z}$ with $f(0)=0$.}
For $z\neq 0$, $f(z)=1/\bar z$ is nowhere complex-differentiable (since dependence on $\bar z$ violates analyticity). At $z=0$, check derivative: \[ \lim_{z\to 0}\frac{f(z)-f(0)}{z-0}=\lim_{z\to 0}\frac{1/\bar z}{z}. \] Take $z=x$ (real axis): $\frac{1/x}{x}=1/x^2\to\infty$. Limit fails. So not differentiable anywhere. Hence (C) is false.
Step 4: Check option (D) $e^{-z^2$.}
This is an entire function (analytic everywhere). Differentiable at all points. Not matching. So (D) is false.

Step 5: Check option (A) $f(z)=z|z|$.
Write $z=x+iy$, $|z|=\sqrt{x^2+y^2}$. Then \[ f(z)=(x+iy)\sqrt{x^2+y^2}. \] At $z=0$, compute derivative: \[ f'(0)=\lim_{z\to 0}\frac{f(z)-f(0)}{z-0}=\lim_{z\to 0}\frac{z|z|}{z}=\lim_{z\to 0}|z|=0. \] So $f$ is differentiable at $z=0$ with derivative $0$. At other points $z\neq 0$, check Cauchy–Riemann equations. Writing in terms of $x,y$, we find $u(x,y)=x\sqrt{x^2+y^2}$, $v(x,y)=y\sqrt{x^2+y^2}$. Partial derivatives do not satisfy Cauchy–Riemann except at $(0,0)$. Thus $f$ is not differentiable anywhere else. Final Answer: \[ \boxed{\,f(z)=z|z|\,} \]