2
$\begingroup$

The proof for this $$ e^{i\theta} = \cos(\theta) + i\sin(\theta) $$ using the MacLaurin series is all right for a high school level, but I dont understand why the series that has been derived for the reals should hold for complex numbers too. Could someone give a sufficient reason why it is correct to use complex numbers in the series expansion for $ e^{x} $ ? I'm not completely convinced why you should be allowed to write $ e ^{ix}= 1 + \frac{ix}{1!} + \frac{{(ix)}^2}{2!} ....$ .

$\endgroup$
  • 2
    $\begingroup$ The series are the definitions of those functions. $\endgroup$ – velut luna Jul 4 '16 at 8:42
  • $\begingroup$ Well, that's my question. It is derived for the reals right? $\endgroup$ – user2277550 Jul 4 '16 at 8:45
  • 1
    $\begingroup$ Consider the radii of (absolute) convergence, and discover they are infinite $\endgroup$ – Henry Jul 4 '16 at 8:48
  • $\begingroup$ I'm sorry but does that affect my question? $\endgroup$ – user2277550 Jul 4 '16 at 8:50
  • 2
    $\begingroup$ The reason why $e^x=1+x+x^2/2!+\dots$ holds for all complex numbers is because $e^x$ is defined to be $1+x+x^2/2!+\dots$. $\endgroup$ – Wojowu Jul 4 '16 at 9:03
3
$\begingroup$

That's a good question. There are infinitely many possible ways to extend the definition of the exponential, cos or sin functions from the real numbers to the complex numbers. However, the uniqueness theorem for complex analytic functions tells you that if $f \colon \mathbb{R} \rightarrow \mathbb{R}$ is a function that can be extended to an analytic function $\tilde{f} \colon \mathbb{C} \rightarrow \mathbb{C}$ then the extension $\tilde{f}$ is unique. Thus, if we want to extend a function $f \colon \mathbb{R} \rightarrow \mathbb{R}$ in a way that the resulting extension is analytic, we have (at most) one way to do it. For the exponential, cos and sin functions, the extensions can be obtained by defining the functions using a power series expansion with the same coefficients as in the real case (only the argument and the result can be complex) so for example

$$ e^z := \sum_{n=0}^{\infty} \frac{z^n}{n!}. $$

Once those extensions are chosen, the proof of Euler's identity becomes a calculation using power series rules which are valid for complex power series just as they were valid for real power series.

$\endgroup$
  • $\begingroup$ Okay so the validity of it, rests on the uniqueness theorem is it? In that case, is it what Euler used to prove this? $\endgroup$ – user2277550 Jul 4 '16 at 9:41
  • 1
    $\begingroup$ I don't understand what you are asking. You can define $e^z$ for complex $z$ by the power series expansion and show that the power series converges for all $z \in \mathbb{C}$. Clearly this provides an extension of $e^x$ from real $x$ to complex $x$. This is completely valid and doesn't rest on anything and doing the same for $\sin,\cos$ you can prove Euler's identity. You can also ask - why have we defined $e^z$ for complex $z$ by the power series expansion? The uniqueness theorem tells you that no matter how you define $e^z$, if you want it to be analytic there is only one way to do it $\endgroup$ – levap Jul 4 '16 at 10:08
  • 1
    $\begingroup$ and so it might as well be the power series definition. This provides a motivation for the definition, not a justification. The definition is valid as it is. $\endgroup$ – levap Jul 4 '16 at 10:08
2
$\begingroup$

They are not $\textit{derived}$ from real Maclaurin series, they are $\textit{generalized}$ so that for example for complex $z$, $e^z$ is $\textit{defined}$ by $$e^z=1+z+\frac{z^2}{2!}+\cdots$$.

In this way, they are shown to have nice and useful properties (e.g., the Euler's formula you mentioned), and are consistent with the ancient geometric definitions of the functions.

For example, you don't need to worry whether $$\sin\frac{\pi}{6}=\frac{1}{2}$$ because from our work in real calculus, we've already proved that $$\frac{\pi}{6}-\frac{(\pi/6)^3}{3!}+\cdots=\frac{1}{2}.$$

$\endgroup$
  • $\begingroup$ Such a definition does not seem correct. What do you think of @levap 's answer? $\endgroup$ – user2277550 Jul 4 '16 at 9:22
  • $\begingroup$ By studying the radii of convergence, the definitions are valid. So it can only either be useful or not, but a definition cannot be incorrect. $\endgroup$ – velut luna Jul 4 '16 at 9:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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