iI have to show that the derivative of $f (z) = z (e ^ z)$ is $f '(z) = (e ^ z) + z (e ^ z)$
Then the solution would be something like:
$f(z) = (x+yi)[(e^x) \cos(y) + i(e^x) \sin(y)]$ $f(z) = x(e^x)\cos(y) + ix(e^x)\sin(y) + iy(e^x)\cos(y) - y(e^x)\sin(y)$
We organize
$f(z) = [x(e^x)\cos(y) - y(e^x)\sin(y)] + i [x(e^x)\sin(y) + y(e^x)cos(y)]$
We derive
$du/dx = (e^x)\cos(y) + x(e^x)\cos(y) - y(e^x)\sin(y)$ $dv/dy = (e^x)\cos(y) + x(e^x)\cos(y) - y(e^x)\sin(y)$
The first rule is met
$dv/dx = (e^x)\sin(y) + x(e^x)\sin(y) + y(e^x)\cos(y)$ $dv/dy = -(e^x)\sin(y) - x(e^x)\sin(y) - y(e^x)\cos(y)$
The second rule is met.
Now we have to check the answers
$f'(z) = du/dx + dv/dx$ $f'(z) = [(e^x)\cos(y) + x(e^x)\cos(y) - y(e^x)\sin(y)] + [(e^x)\sin(y) + x(e^x)\sin(y) + y(e^x)\cos(y)]$
$f'(z) = (e^x)[\cos(y) + x\cos(y) - y\sin(y)] + (e^x)[(\sin(y) + x \sin(y) + y \cos(y)]$
From this point I really do not know what I should do to demonstrate the derivative...
The answer I have to get is "$f'(z) = (e^z) + z(e^z)$"
Thanks for your help
The derivative should be $$ f'(z)=\frac{\partial u}{\partial x}+\color{red}{i}\frac{\partial v}{\partial x} $$ Thus $$ f'(z)=e^x\cos y + xe^x\cos y - ye^x\sin y+ i(e^x\sin y + xe^x\sin y + ye^x\cos y) $$ and so $$ f'(z)=\underbrace{e^x(\cos y+i\sin y)}_{A(z)}+ \underbrace{e^x(x\cos y-y\sin y+i(x\sin y+y\cos y))}_{B(z)} $$ The first summand is $A(z)=e^xe^{iy}=e^z$, the second is $$ B(z)=e^x(x+iy)(\cos y+i\sin y)=ze^xe^{iy}=ze^z $$ Just compute $$ (x+iy)(\cos y+i\sin y)=x\cos y-y\sin y+i(x\sin y+y\cos y)) $$ Hence $f'(z)=A(z)+B(z)=e^z+ze^z$.
But this is wasting time. The product rule easily extends to analytic complex functions, with the same proof as in the real case.