# Can I differentiate $f(z)= z^2 e^\overline{z}$ with z in complex numbers?

$$\overline{z}$$ 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.

• Is $f(z) = \overline{z}$ entire? Do you know the definition? How would you show whether it is or isn't? – Zubin Mukerjee Apr 21 '20 at 19:37
• Do you know the Cauchy-Riemann equations? – Zubin Mukerjee Apr 21 '20 at 19:38
• @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. – Robert Apr 21 '20 at 19:42

## 1 Answer

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

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.

• I got there, and after applyint the Cauchy Riemann equations I got 2 equations: $x^2cos(y)-y^2cos(y)+2xysin(y)=0$ and $xsin(y)+ycos(x)=0$ – Robert Apr 21 '20 at 19:50
• 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. – Physical Mathematics Apr 21 '20 at 19:50