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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...

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The given function is just $f(z)=ze^{z}$. So $f'(z)=e^{z} (1+z)$.

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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 $\Bbb C$.

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