$\overline{z}$ is the conjugate of $z$.

And $z=x+iy$.

I don't know if it can be differentiated in any point for every complex number $z$.

I've tried to proof if it is differentiable in an exact point. I have separated the real part and the imaginary part and I have tried if the partial derivates are equal or not. But they are equal in an infinite number of points.

So, I don't know what can I try now.

  • $\begingroup$ Is $f(z) = \overline{z}$ entire? Do you know the definition? How would you show whether it is or isn't? $\endgroup$ – Zubin Mukerjee Apr 21 '20 at 19:37
  • $\begingroup$ Do you know the Cauchy-Riemann equations? $\endgroup$ – Zubin Mukerjee Apr 21 '20 at 19:38
  • $\begingroup$ @ZubinMukerjee $f:\mathbb{C}\rightarrow\mathbb{C}$. I have tried with the Cauchy-Riemann equations, but I got that this is not differentiable in every point. Now I don't know how can I get the points where $f$ is differentiable. $\endgroup$ – Robert Apr 21 '20 at 19:42

$$f(z) = z^2 e^{\overline{z}} = (x+iy)^2 e^{x-iy} = (x^2 -y^2 + 2ixy) e^{x-iy} = e^x(x^2 -y^2 + 2ixy)(\cos y - i \sin y)\\ =e^x[(x^2 - y^2)\cos y + 2xy \sin y] + i e^x[2xy\cos y - (x^2-y^2)\sin y].$$

Then apply the Cauchy Riemann equations, with $u(x,y) =e^x[(x^2 - y^2)\cos y + 2xy \sin y]$ and $v(x,y) = e^x[2xy\cos y - (x^2-y^2)\sin y]$.

This gives you a test of whether $f$ is differentiable.

  • $\begingroup$ I got there, and after applyint the Cauchy Riemann equations I got 2 equations: $x^2cos(y)-y^2cos(y)+2xysin(y)=0$ and $xsin(y)+ycos(x)=0$ $\endgroup$ – Robert Apr 21 '20 at 19:50
  • $\begingroup$ Then you have differentiability exactly when those equations are satisfied. They aren't always: e.g. if $x = 0$, $y = \pi$, then the second one gives $0 - 4 \pi =0 $ which is obviously false. Thus this gives you a precise characterization of the points at which you have differentiability. $\endgroup$ – Physical Mathematics Apr 21 '20 at 19:50

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