# Why can we plug complex numbers into maclaurin series?

When finding a Maclaurin series for a function f(x). We evaluate f '(0), f ''(0), etc, to find our coefficients for each term.
I have done this for the standard functions, $$f (\mathbb{R}):->\mathbb{R}$$. However, to prove $$e^{ix} = cosx + i sinx$$, one can substitute ix into the Maclaurin series for $$e^x$$. So I am trying to understand how did we extend the domain of our function, and how can I possibly do the same with other functions, e.g. $$sinx$$ so that we can have $$x \in \mathbb{C}$$.

• Normally you'd prove $e^z$ is complex differentiable given its Taylor series. It's common to define the complex cosine and sine using $e^z$. The general process is called Analytic continuation. en.wikipedia.org/wiki/Analytic_continuation Sep 20 at 22:16
• How are you defining $e^z$, $\cos z$, and $\sin z$ for complex numbers?
– Dan
Sep 20 at 22:18
• @Dan, for real numbers x, I defined $e^x$ as the limit as n approaches $infinity$ of $(1+1/n)^n$ to the power of x, and sin and cos as the position against time for a point along the unit circle. Using those definitions, I should not be able to raise a number to the power of i, or evaluate sin(ix). Sep 20 at 22:24
• Most theorems to do with the convergences of series carry over from $\mathbb{R}$ into $\mathbb{C}-$ and, more generally, into any metric space $X$. In particular, if a complex series $\sum a_n$ converges absolutely (i.e. if the series of lengths of the $a_n$ converges) then $\sum a_n$ converges. Sep 20 at 22:30
• If you instead use the definition $$e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n,$$ the definition easily extends to complex numbers, since we are only raising to integer powers. (You'd still need to show that the limit exists of course.) Sep 21 at 0:19

What is typically done is some variation of the following:

1. We define limits of complex functions similarly to their real counterparts but using the modulus in place of the absolute value - so $$\lim_{z \rightarrow z_0} f(z) = L$$ if $$|f(z) - L| \rightarrow 0$$ as $$|z - z_0| \rightarrow 0$$, where the $$|\cdot|$$ represent complex modulus operations. We do something similar for the complex derivative.

2. We define $$e^z = \sum_{n = 0}^\infty \frac{z^n}{n!}$$, and we note that this gives us an entire function (i.e. it is analytic for all $$z \in \mathbb{C}$$) that aligns with the version of the function defined for real numbers. Again, all of the results from Taylor's theorem have complex analogues, which is why this series is absolutely convergent.

3. We then define $$\cos z = \frac{e^{iz} + e^{iz}}{2}$$ and $$\sin z = \frac{e^{iz} - e^{-iz}}{2i}$$.

4. We can then note that because $$e^z$$ is entire, we can directly derive the Taylor series for $$\cos z$$ and $$\sin z$$ and we get the versions that line up with the real versions, and also it's very easy to see that $$e^{iz} = \cos z + i \sin z$$.

The part where we see that the functions we've defined happen to line up with known functions on the real line is known as analytic continuation, although it's usually presented a little differently (we usually start with a function defined on the real numbers, find its Taylor series, and then look at a region of complex numbers where that series converges and say that the series defines an analytic continuation of our function in that region).

In short, because any function of the form $$f(z) = \sum_{n=0}^\infty a_n z^n$$ is analytic.

IOW, if you have any real function that's infinitely differentiable so that you can express it as a Maclaurin series, then you can just plug complex numbers into the same series and get a well-behaved infinitely differentiable complex function. We can then simply define the complex extension of the function to be that series.