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Let $z=x+iy$ where $x,y\in\mathbb{R}$. The exponential function is $$e^z=e^x(\cos{y}+i\sin{y}).$$

Using the power series of $e^x$, $\cos{y}$ and $\sin{y}$, find a power series representation for $e^{z}$.


Recall that $$e^{x} = \sum_{k=0}^\infty \frac{x^k}{k!},\quad \cos{y} = \sum_{k=0}^\infty \frac{(iy)^{2k}}{(2k)!},\quad\sin{y} = \sum_{k=0}^\infty \frac{i^{2k}y^{2k+1}}{(2k+1)!}$$

Then

\begin{align} e^{z} &= \left[\sum_{k=0}^\infty \frac{x^k}{k!}\right]\left[\sum_{k=0}^\infty\left(\frac{(iy)^{2k}}{(2k)!}+i\frac{i^{2k}y^{2k+1}}{(2k+1)!}\right)\right]\\ &= \left[\sum_{k=0}^\infty \frac{x^k}{k!}\right]\left[\sum_{k=0}^\infty\left(\frac{(iy)^{2k}(2k+1)}{(2k+1)!}+i\frac{(iy)^{2k}y}{(2k+1)!}\right)\right]\\ &= \left[\sum_{k=0}^\infty \frac{x^k}{k!}\right]\left[\sum_{k=0}^\infty\frac{(iy)^{2k}}{(2k+1)!}\left(2k+1+iy\right)\right] \end{align}

Now I can't really see where to go.

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  • $\begingroup$ Shouldn't you have two sums in the first step? One running over the $k$'s of the exponential series and one running over the ones of $\sin$ and $\cos$? $\endgroup$ – noctusraid Feb 14 '16 at 12:44
  • $\begingroup$ @noctusraid I have modified my question to take this into account. $\endgroup$ – user2850514 Feb 14 '16 at 12:51
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Recall that $$ e^{z}=\sum_{n=0}^{+\infty}\frac{z^n}{n!},\;\;\;z\in\Bbb C $$ is one of the possible definition of the complex exponential.

If you want real exponential simply take $z$ real.

The formula $e^z=e^x(\cos y+i\sin y)$ is a consequence, and maybe you are confusing what comes first: taking the definition of complex exponential given above and reminding the Taylor expansion of $\sin$ and $\cos$, you can easily prove the Euler Identity: $$ e^{it}=\cos t+i\sin t,\;\;\; t\in\Bbb R $$ from which you immediately get $e^z=e^x(\cos y+i\sin y)$.

Maybe you are confused because you've tried to do the inverse path.

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