Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Euler famously used the Taylor's Series of $\exp$:

$$\exp (x)=\sum_{n=0}^\infty \frac{x^n}{n!}$$

and made the substitution $x=i\theta$ to find

$$\exp(i\theta) = \cos (\theta)+i\sin (\theta)$$

How do we know that Taylor's series even hold for complex numbers? How can we justify the substitution of a complex number into the series?

share|improve this question
3  
You should see Needham's book. –  J. M. Jul 13 '12 at 15:23
    
I think it would help if you asked yourself: what does it mean that "Taylor's series hold ..."? A series is just a series; it doesn't "hold", but it may or may not converge. –  Marc van Leeuwen Jul 13 '12 at 17:11

2 Answers 2

up vote 3 down vote accepted

The series $f(z)=\sum_{n=0}^{\infty} \frac{z^n}{n!}$ is convergent on $C$, and thus it defines an Analytic function.

Now there are few different ways to convince yourself that this has to be $e^z$.

For once, it is the only Analytic continuation of $e^x$ to the complex plane...

Or, alternately, you can prove that $f(z_1+z_2)=f(z_1)f(z_2)$ and $f(1)=e$. Also, you can show that it is the only differentiable function satisfying these two relations.

If you prefer differential equations, $f'(z)=f(z)$ and $f(1)=e$ uniquely determine a solution, and bot $e^z$ and $f(z)$ are solutions....

share|improve this answer

You define a complex function by the formula $f(z)=\sum_{n=0}^{\infty} \frac{z^n}{n!} $. You prove that it converges everywhere and defines a holomorphic function of $z$. Then you prove that for $z=x \in \mathbb{R}$ it agrees with the usuall exponential.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.