# Show $f(z)=e^x(x\cos y-y\sin y)+ie^x(y\cos y+x\sin y)$ is analytic

Show $$f(z)=e^x(x\cos y-y\sin y)+ie^x(y\cos y+x\sin y)$$ is analytic. Then find its derivative.

So I have $$u = e^x(x\cos y-y\sin y)$$ and $$v = e^x(y\cos y+x\sin y)$$.

Then I have $${\partial(u)\over\partial(x)} = e^x(x\cos y-y\sin y) + e^x(\cos y)$$, and $${\partial(v)\over\partial(x)} = e^x(y\cos y+x\sin y) + e^x(\sin y)$$.

I'm having trouble with finding $${\partial(u)\over\partial(y)}$$ and $${\partial(v)\over\partial(y)}$$. Also with finding the derivative of a complex function as well. I've seen examples with general form in $$z$$ but not with functions like these...

The given function is just $$f(z)=ze^{z}$$. So $$f'(z)=e^{z} (1+z)$$.
A much simpler solution is provided by Kavi. If you wish to proceed with Cauchy-Riemann, note that $$u=e^x(x\cos y-y\sin y)\quad\text{and}\quad v=e^x(y\cos y+x\sin y)$$ so \begin{align}u_x&=e^x(x\cos y-y\sin y)+e^x\cos y=e^x((x+1)\cos y-y\sin y)\\u_y&=e^x(-x\sin y-(\sin y+y\cos y))=-e^x(y\cos y+(x+1)\sin y)\\v_x&=e^x(y\cos y+x\sin y)+e^x\sin y=e^x(y\cos y+(x+1)\sin y)\\v_y&=e^x((\cos y-y\sin y)+x\cos y)=e^x((x+1)\cos y-y\sin y).\end{align} As all first-order partial derivatives exist at every point in $$\Bbb C$$ with $$u_x=v_y$$ and $$u_y=-v_x$$, the function $$f$$ is analytic on the whole of $$\Bbb C$$.