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  1. How is it the complex exponential converges for any value of $z$ in the complex plane? $$e^{z} = 1 + \frac{z}{1!} + \frac{z^2}{2!} \cdots\cdots$$

  2. How is it the "sum of exponents" rule holds for complex exponential, that is $e^{w}e^{z} = e^{w+z} $?

...using only the definition of $e^z$?

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Ratio test, binomial theorem. –  anon Jul 11 '12 at 20:01
Hint (for the second part): If you multiply the two power series together and collect up terms, and use the binomial theorem, you should be able to prove this. It can be justified by absolute convergence. –  Old John Jul 11 '12 at 20:02
i know 2nd expression can be proved by binomial theorem, but failed to do so. and for the first one, if only the postulates of "complex arithmetic" is given, how it converges? –  Aftnix Jul 11 '12 at 20:03
Latex output is not correct in my browser, sometimes its ok, sometimes it not. –  Aftnix Jul 11 '12 at 20:05
@Aftnix If you are familiar with the proof of the ratio test in the real case, you should see that it carries over easily for complex series. –  user17794 Jul 11 '12 at 20:17
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3 Answers

up vote 2 down vote accepted

$$e^z=\sum_{n=0}^\infty\frac{z^n}{n!}$$ Putting $$a_n:=\frac{1}{n!}\Longrightarrow R:=\frac{1}{\lim_{n\to\infty}\sqrt[n]{|a_n|}}=\lim_{n\to\infty}\sqrt[n]{n!}=\infty$$

So the series has infinite convergence radius and is thus absolutely convergent for any $\,z\in\Bbb C\,$ , and from here it follows that $$e^{w+z}=\sum_{n=0}^\infty\frac{(w+z)^n}{n!}=\sum_{n=0}^\infty\sum_{k=0}^n\binom{n}{k}\frac{w^kz^{n-k}}{n!}=\sum_{n=0}^\infty\sum_{k=0}^n\frac{w^kz^{n-k}}{k!(n-k)!}=$$ $$=\sum_{k=0}^\infty\frac{w^k}{k!}\sum_{n=0}^\infty\frac{z^n}{n!}=:e^we^z$$

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Thanks, thats what i was looking for. –  Aftnix Jul 12 '12 at 8:51
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  1. One can use the ratio test or notice that for every $z \in \mathbb{C}$ one has $|z| \in \mathbb{R}$ and $$ \sum_{n=0}^\infty\frac{|z|^n}{n!}=e^{|z|}, $$ i.e. the series $\sum_{n=0}^\infty z^n/n!$ converges absolutely, and therefore it converges for every $z \in \mathbb{C}$.

2. $$ e^we^z=(\sum_{n=0}^\infty\frac{w^n}{n!})(\sum_{n=0}^\infty\frac{z^n}{n!})=\sum_{n=0}^{\infty}A_n(w,z), $$ where with the use of Cauchy product one has \begin{eqnarray} A_n(w,z)&=&\sum_{k=0}^n\frac{w^k}{k!}\frac{z^{n-k}}{(n-k)!}=\sum_{k=0}^n\frac{1}{n!}{n\choose k}w^kz^{n-k}=\frac{1}{n!}\sum_{k=0}^n{n\choose k}w^kz^{n-k}=\frac{(w+z)^n}{n!}. \end{eqnarray}

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The second part can be proved also using the identity theorem: Two analytic functions are the same if they coincide on a set with a limit point. Look at $e^{w+z}$ and $e^w e^z$ with $w$ a fixed real number. The two functions (as functions of $z$) coincide on the real line. Therefore, they are equal for all $z \in \mathbb{C}$. Since $w$ was arbitrary they also coincide as functions of $w$ on the real line. Again by the identity theorem $e^{w+z}=e^w e^z$ for all $w \in \mathbb{C}$.

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