Find the points where $f:\Bbb C\to\Bbb C,\, z\mapsto z|z|$ is differentiable.

Can someone check if this exercise is correctly done? Thank you.

Let $g(x,y):=(x\sqrt{x^2+y^2},y\sqrt{x^2+y^2})$, then under the isomorphism $z\mapsto (\Re(z),\Im(z))$ we have that $f\mapsto g$, and

$$\partial_x g(x,y)=\left(\frac{2x^2+y^2}{\sqrt{x^2+y^2}},\frac{xy}{\sqrt{x^2+y^2}}\right)\\\partial_y g(x,y)=\left(\frac{xy}{\sqrt{x^2+y^2}},\frac{2y^2+x^2}{\sqrt{x^2+y^2}}\right)$$

Then $g$ is totally differentiable at $\Bbb R^2\setminus\{(0,0)\}$ because it partial derivatives are continuous in this region. For $(x,y)=(0,0)$ it directional derivatives are

$$D_vg(0,0)=\lim_{t\to 0}\left((tv_1|tv|)',(tv_2|tv|)'\right)=(0,0),\quad v:=(v_1,v_2)\in\Bbb R^2\setminus\{(0,0)\}$$

Thus $g$ is also totally differentiable at zero. However to be $f$ complex differentiable at $(x,y)\neq (0,0)$ it must be the case that

$$\frac{2x^2+y^2}{\sqrt{x^2+y^2}}=\frac{2y^2+x^2}{\sqrt{x^2+y^2}}\quad\text{and}\quad \frac{xy}{\sqrt{x^2+y^2}}=-\frac{xy}{\sqrt{x^2+y^2}}$$

that is, the Cauchy-Riemann equations must hold.

From the second equation we have that $x=0$ or $y=0$, and from the first that $x^2=y^2$, hence $f$ is not complex differentiable in $\Bbb C\setminus\{0\}$. However for $(x,y)=(0,0)$ the equations holds trivially, so the unique point where $f$ is complex differentiable is at zero.


It seems fine. Just a word of caution: when showing the directional derivatives exist in all directions from $(0, 0)$, you should point out that the directional derivative is a linear function of the direction. In this case, it always returns $0$, which is linear of course, but it should be stated.

Alternatively, you could calculate the derivative of $f$ at $0$ from first principles: $$f'(0) = \lim_{z\rightarrow 0}\frac{f(z) - f(0)}{z} = \lim_{z\rightarrow 0}\frac{z|z|}{z} = \lim_{z\rightarrow 0} |z| = 0$$

  • $\begingroup$ f is differentiable at a point z other than 0 iff f(z)/z is. This makes the proof simpler. $\endgroup$ – Kavi Rama Murthy Sep 4 '17 at 6:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.