# Abstract Nonsense

## The Exponential and Trigonometric Functions

Point of Post: In this post we define the exponential and trigonometric functions and note that they are holomorphic.

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Motivation

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Last time we proved that every function on an open subset of $\mathbb{C}$ that is locally representable by power series is necessarily holmorphic (in fact, infinitely differentiable!). That said, we didn’t actually give any honest to god examples of such functions. Thus, in this post we will finally lay down our first few non-trivial examples of  holomorphic functions. They will come in terms of what is, in a very precise sense we will make clear later on, the only extension of some of our favorite real valued functions: the exponential and trigonometric functions.

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The Exponential and Trigonometric Functions

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Let us define the exponential function $\text{exp}:\mathbb{C}\to\mathbb{C}$ by the rule

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$\displaystyle \text{exp}(z)=\sum_{n=0}^{\infty}\frac{z^n}{n!}$

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We shall often denote $\text{exp}(z)$ as $e^z$. Note that since

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$\displaystyle \lim \frac{1}{\sqrt[n]{n!}}=0$

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that the radius of convergence of $\text{exp}$ is $\infty$ and so it really is a well-defined function on $\mathbb{C}$.

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I leave it to the reader to verify that $\exp$ is a homomorphism $(\mathbb{C},+)\to(\mathbb{C}^\times,\cdot)$, or more explicitly that $\text{exp}(z+z')=\exp(z)\exp(z')$ and so in particular $\exp(z)\ne 0$ for all $z$ since $\text{exp}(z)\exp(-z)=\exp(0)=1$. Moreover, I leave it for you to check that $\text{exp}(\overline{z})=\overline{\text{exp}(z)}$ (just use the fact that the conjugation map is continuous) and that for integers $n\in\mathbb{Z}$ one has that $\text{exp}(nz)=\text{exp}(z)^n$ for all $z\in\mathbb{Z}$.

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Since $\text{exp}$ is representable by power series everywhere on $\mathbb{C}$ we know that $\exp\in\mathcal{O}(\mathbb{C})$–such globally holomorphic functions are called entire. We note moreover that, by definition,

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$\displaystyle \exp'(z)=\sum_{n=0}^{\infty}\frac{n}{n!}z^{n-1}=\sum_{n=1}^{\infty}\frac{1}{(n-1)!}z^{n-1}=\sum_{n=0}^{\infty}\frac{z^n}{n!}=\exp(z)$

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so that $\displaystyle \frac{\partial}{\partial z}\exp=\exp$. In fact, we have the following theorem:

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Theorem: The exponential function is the unique entire function $f$ such that $\displaystyle \frac{\partial f}{\partial z}=f$ and $f(0)=1$.

Proof: It clearly suffices to show uniqueness. Indeed, if $f$ were such a function then we’d have that $\displaystyle \frac{f}{\exp}$ is entire (note we can divide since $\exp$ is never zero). That said, using the division rule we see that

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$\displaystyle \frac{\partial}{\partial z}\frac{f}{\exp}=\frac{\exp f'-f\exp'}{\exp^2}=0$

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since $\exp f'=\exp f$ and $\exp'f=\exp f$. Thus, since $\mathbb{C}$ is connected we may conclude from previous proof that $\displaystyle \frac{f}{\exp}$ is constant. Noting that $\displaystyle \frac{f(0)}{\exp(0)}=\frac{1}{1}=1$ we may conclude that $f=\exp$ as desired. $\blacksquare$

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What we would now like to do is explain how the exponential and trigonometric functions are related. We begin with the observation that if $y$ is real then

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$\displaystyle \frac{\exp(iy)+\exp(-iy)}{2}=\sum_{n=0}^{\infty}\frac{(iy)^n+(-1)^n(iy)^n}{2n!}=\sum_{n=0}^{\infty}\frac{(-1)^ny^{2n}}{(2n)!}=\cos(y)$

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and similarly

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$\displaystyle \frac{\exp(iy)-\exp(-iy)}{2i}=\sin(y)$

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And thus, in particular,

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$\displaystyle \cos(y)+i\sin(y)=\frac{\exp(iy)+\exp(-iy)}{2}+\frac{\exp(iy)-\exp(-iy)}{2}=\exp(iy)$

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References:

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[1] Greene, Robert Everist, and Steven George Krantz. Function Theory of One Complex Variable. Providence: American Mathematical Society, 2006. Print.

[2] Conway, John B. Functions of One Complex Variable I. New York: Springer-Verlag, 1978. Print.

[3] Rudin, W. Real and Complex Analysis. New York,NY: McGraw-Hill, 1988. Print.

[4]  Ahlfors, Lars V. Complex Analysis; an Introduction to the Theory of Analytic Functions of One Complex Variable. New York: McGraw-Hill, 1966. Print.