Answer to Question #234108 in Complex Analysis for smi

Question

Show that f(z)=z²(e^z-1) is differentiable at z_o=0. Check whether f is analytic at O.

Answer 1

The derivative of the function $f(z)$ is: $f'(z)=2z(e^z-1)+z^2e^z=e^z(z^2+2z)-2z$.

$f(0)=0$. A necessary conditions for the function to be analytic are: $\frac{\partial f(z,\bar{z}) }{\partial{\bar{z}}}=0$. The latter holds. Rewrite the function $f(z)$ as: $f(z)=(x+iy)^2(e^{x+iy}-1)=(x^2-y^2+2xiy)(e^xe^{iy}-1)=(x^2-y^2)e^xe^{iy}-(x^2-y^2)+2xiye^xe^{iy}-2xyi=(x^2-y^2)e^x(cos(y)+i\,sin(y))+2xyie^x(cos(y)+i\,sin(y))-2xiy$

In case we split real and imaginary part as: $f(z)=u(x,y)+iv(x,y)$ it is obvious that partial derivatives $\frac{\partial u}{\partial x}$, $\frac{\partial u}{\partial y}$, $\frac{\partial v}{\partial x}$, $\frac{\partial v}{\partial y}$ are continuous. The latter are sufficient conditions for the function to be analytic. Thus, the function is analytic. In particular, at point 0.