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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.

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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: $$f'(0) = \lim_{z\rightarrow 0}\frac{f(z) - f(0)}{z} = \lim_{z\rightarrow 0}\frac{z|z|}{z} = \lim_{z\rightarrow 0} |z| = 0$$

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  • $\begingroup$ 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

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