want to prove exponential identity with complex numbers I want to prove that $e^{x+y}=e^x*e^y$, where x and y are complex numbers. I only want to use that evey complex number is on the form $z=a+bi$, where $i^2=-1$.
And if z is a complex number we have $e^z=\Sigma_{n=0}^\infty\frac{z^n}{n!}$.
However I seem to get stuck, here is my attempt:
$x=a+bi$
$y=c+di$
$e^{x+y}=\Sigma_{n=0}^\infty\frac{(a+bi*c+di)^n}{n!}$. I am not sure if we can use the binomial theorem on the part inside the summation, because we have $i$ there?, can we?
If we can, I get
$=\Sigma_{n=0}^\infty[\Sigma_{k=0}^n\frac{n!}{k!(n-k)!}(a+bi)^k*(c+di)^{n-k}*\frac{1}{n!}]$. My problem is that I am not able to sepearte the n and the k, if I was able to show that this was equal to:
$\Sigma_{n=0}^\infty\frac{(a+bi)^n}{n!}*\Sigma_{k=0}^\infty\frac{(c+di)^k}{k!}$ then I would have provedwhat I wanted, is there a way to get here?
 A: Notice that the exponential series is absolutely convergent on $\Bbb C$ so we can use the Cauchy product of the series:
$$e^xe^y=\sum_{n=0}^\infty\frac{x^n}{n!}\sum_{n=0}^\infty\frac{y^n}{n!}=\sum_{n=0}^\infty\sum_{k=0}^n\frac{x^k}{k!}\frac{y^{n-k}}{(n-k)!}=\sum_{n=0}^\infty \frac1{n!}\sum_{k=0}^n{n\choose k}x^ky^{n-k}\\=\sum_{n=0}^\infty \frac{(x+y)^n}{n!}=e^{x+y}$$
A: $$e^{u}e^{v}=\sum_{r=0}^{\infty}\frac{u^{r}}{r!}\sum_{s=0}^{\infty}\frac{v^{s}}{s!}=\sum_{n=0}^{\infty}\sum_{r+s=n}\frac{u^{r}v^{s}}{r!s!}=\sum_{n=0}^{\infty}\frac{\left(u+v\right)^{n}}{n!}=e^{u+v}$$

edit
In general $$\sum_{r=0}^{\infty}a_{r}\times\sum_{s=0}^{\infty}b_{s}=a_{0}b_{0}+\left(a_{0}b_{1}+a_{1}b_{0}\right)+\left(a_{0}b_{2}+a_{1}b_{1}+a_{2}b_{0}\right)+\cdots=\sum_{n=0}^{\infty}c_{n}$$
where $$c_{n}=a_{0}b_{n}+\cdots+a_{n}b_{0}=\sum_{k=0}^{n}a_{k}b_{n-k}$$
Another way to notate $c_{n}$ is $$c_{n}=\sum_{r+s=n}a_{r}b_{s}$$
where $r$ and $s$ stand for nonnegative integers. 
A: There's no point in writing the real and imaginary parts of the complex numbers $x$ and $y$.
According to your formula for $e^{x+y}$, what is the coefficient of $x^ry^s$ for given $r,s\ge0$?
A: For z and w being complex,  $$
$$ $e^{z}*e^{w}$=$e^{(x+iy)}*e^{(a+ib)}$=$(e^{x}*e^{iy})*(e^{a}*e^{ib})$=$(e^{x}*e^{a})*(e^{iy}*e^{ib})=e^{(x+a)}*e^{i(b+y)}=e^{(x+a)+i(y+b)}=e^{z+w}$. I dont understand why I have to make my life complicated with the Taylor series Expansion for $e^{z}.$
