# Equivalence of Definitions of Complex Exponential Function

## Theorem

The following definitions of the concept of the complex exponential function are equivalent:

### As a Sum of a Series

The exponential function can be defined as a (complex) power series:

 $\displaystyle \exp z$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}$ $\displaystyle$ $=$ $\displaystyle 1 + \frac z {1!} + \frac {z^2} {2!} + \frac {z^3} {3!} + \cdots + \frac {z^n} {n!} + \cdots$

### By Real Functions

The exponential function can be defined by the real exponential, sine and cosine functions:

$\exp z := e^x \paren {\cos y + i \sin y}$

where $z = x + i y$ with $x, y \in \R$.

Here, $e^x$ denotes the real exponential function, which must be defined first.

### As a Limit of a Sequence

The exponential function can be defined as a limit of a sequence:

$\displaystyle \exp z := \lim_{n \to \infty} \left({1 + \dfrac z n}\right)^n$

### As the Solution of a Differential Equation

The exponential function can be defined as the unique particular solution $y = \map f z$ to the first order ODE:

$\dfrac {\d y} {\d z} = y$

satisfying the initial condition $\map f 0 = 1$.

That is, the defining property of $\exp$ is that it is its own derivative.

## Proof

From Radius of Convergence of Power Series over Factorial: Complex Case, it follows that the power series $\displaystyle \sum_{n \mathop = 0}^\infty \dfrac {z^n} {n!}$ is absolutely convergent over the entirety of $\C$.

Hence, the definition of $\exp z$ as a sum of a series is valid.

It remains to demonstrate the logical equivalence of all the definitions.

### Sum of Series equivalent to Solution of Differential Equation

#### Sum of Series implies Solution of Differential Equation

Let $\exp z$ be the complex function defined as the sum of the power series:

$\exp z := \displaystyle \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}$

Let $y = \displaystyle \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}$.

Then:

 $\displaystyle \dfrac {\d y} {\d z}$ $=$ $\displaystyle \dfrac \d {\d z} \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 1}^\infty \frac {z^{n - 1} } {\paren {n - 1}!}$ Derivative of Complex Power Series $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}$ Translation of Index Variable of Summation $\displaystyle$ $=$ $\displaystyle y$

We show that $\displaystyle \sum_{n \mathop = 0}^\infty \dfrac {z^n} {n!}$ satisfies the initial condition:

$\exp \paren 0 = 1$.

Setting $z = 0$ we find:

 $\displaystyle y \paren 0$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty \frac {0^n} {n!}$ $\displaystyle$ $=$ $\displaystyle \frac {0^0} {0!} + \sum_{n \mathop = 1}^\infty \frac {0^n} {n!}$ $\displaystyle$ $=$ $\displaystyle \frac {0^0} {0!}$ as $0^n = 0$ for all $n > 0$ $\displaystyle$ $=$ $\displaystyle 1$ Definition of $0^0$

That is:

$\exp z$ is the particular solution of the differential equation:

$\dfrac {\d y} {\d z} = y$

satisfying the initial condition $y \left({0}\right) = 1$.

$\Box$

#### Solution of Differential Equation implies Sum of Series

Let $\exp z$ be the complex function defined as the particular solution of the differential equation:

$\dfrac {\d y} {\d z} = y$

satisfying the initial condition $y \paren 0 = 1$.

Let $f: \C \to \C$ be a solution to the differential equation $\dfrac {\d f} {\d z} = f$ with $f \paren 0 = 1$.

Then Holomorphic Function is Analytic shows that $f$ can be expressed as a power series:

$\displaystyle \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$

about any $\xi \in \C$.

When $\xi = 0$, we have for all $n \in \N_{\ge 1}$:

 $\displaystyle a_n$ $=$ $\displaystyle \dfrac {f^{\paren n} \paren 0} {n!}$ Power Series is Taylor Series $\displaystyle$ $=$ $\displaystyle \dfrac {f^{\paren {n - 1} } \paren 0} {n!}$ as $f^{\paren n} \paren 0 = f \paren 0 = f^{\paren {n - 1} } \paren 0$ $\displaystyle$ $=$ $\displaystyle \dfrac 1 n a_{n - 1}$ Power Series is Taylor Series

As $a_0 = \dfrac{f^{\paren 0} \paren 0} {0!} = 1$ by the initial condition, it follows inductively that:

$a_n = \dfrac 1 {n!}$

Hence:

$\displaystyle f \paren z = \sum_{n \mathop = 0}^\infty \dfrac 1 {n!} z^n$

$\Box$

### Sum of Series equivalent to Definition by Real Functions

We have the result:

Sum of Series equivalent to Solution of Differential Equation

which gives that the definition of $\exp z$ as the sum of the power series

$\exp z := \displaystyle \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}$

is equivalent to the definition of $\exp z$ as the solution of the differential equation:

$\dfrac {\d y} {\d z} = y$

satisfying the initial condition $y \paren 0 = 1$.

Let:

$e: \R \to \R$ denote the real exponential function
$\sin: \R \to \R$ denote the real sine function
$\cos: \R \to \R$ denote the real cosine function.

Then:

 $\displaystyle \exp z$ $=$ $\displaystyle \map \exp {x + i y}$ where $x, y \in \R$ $\displaystyle$ $=$ $\displaystyle \map \exp x \, \map \exp {i y}$ Exponential of Sum: Complex Numbers $\displaystyle$ $=$ $\displaystyle e^x \, \map \exp {i y}$ Definition (as Sum of Series) agrees with Definition of Real Exponential for all $x \in \R$ $\displaystyle$ $=$ $\displaystyle e^x \paren {\cos y + i \sin y}$ Euler's Formula, which can be proven using Definition as Sum of Series

$\Box$

### Sum of Series equivalent to Limit of Sequence

Let:

$\displaystyle s_n = \sum_{k \mathop = 0}^n \dfrac {z^k} {k!}$
$a_n = \left({1 + \dfrac z n}\right)^n$

Then we can express $a_n$ as follows:

 $\displaystyle a_n$ $=$ $\displaystyle \sum_{k \mathop = 0}^n \binom n k \left({\dfrac z n}\right)^k 1^{n-k}$ Binomial Theorem: Integral Index $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 0}^n \dfrac {z^k} {k!} \dfrac {n \paren {n - 1} \cdots \paren {n - k + 1} }{n^k}$ $\displaystyle$ $=$ $\displaystyle 1 + \sum_{k \mathop = 1}^n \dfrac {z^k} {k!} \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n}$ by algebraic manipulations

The limit of the difference between the $k$th terms of $a_n$ and $s_n$ is:

 $\displaystyle \lim_{n \mathop \to +\infty} \cmod {\dfrac {z^k} {k!} - \dfrac {z^k} {k!} \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n} }$ $=$ $\displaystyle \cmod {\lim_{n \mathop \to +\infty} \paren {\dfrac {z^k} {k!} - \dfrac {z^k} {k!} \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n} } }$ Modulus of Limit $\displaystyle$ $=$ $\displaystyle \cmod {\dfrac {z^k} {k!} \paren {1 - \prod_{j \mathop = 1}^{k - 1} \paren {1 - \lim_{n \mathop \to +\infty} \dfrac j n} } }$ Combination Theorem for Limits of Functions $\displaystyle$ $=$ $\displaystyle \cmod {\dfrac {z^k} {k!} \paren {1 - 1} }$ Sequence of Powers of Reciprocals is Null Sequence $\displaystyle$ $=$ $\displaystyle 0$

To show that $s_n$ and $a_n$ have the same limit, let $\epsilon \in \R_{>0}$.

From Tail of Convergent Series tends to Zero, it follows that we can find $M \in \N$ such that for all $m \ge M$:

$\displaystyle \sum_{k \mathop = m}^n \cmod {\dfrac {z^k} {k!} } < \dfrac \epsilon 2$

For all $k \in \left\{ {0, 1, \ldots, M - 1}\right\}$, we can find $N_k \in \N$ such that for all $n \ge N_k$:

$\displaystyle \cmod{ \dfrac {z^k} {k!} - \dfrac {z^k} {k!} \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n} } < \dfrac \epsilon {2 M}$

Then for all $n \ge \max \paren {M, N_0, N_1, \ldots, N_{M - 1} }$, we have:

 $\displaystyle \cmod {s_n - a_n}$ $=$ $\displaystyle \cmod {\sum_{k \mathop = 1}^{M-1} \dfrac{z^k}{k!} \left({ 1 - \prod_{j \mathop = 1}^{k-1} \left({1 - \dfrac j n}\right) }\right) + \sum_{k \mathop = M}^n \dfrac{z^k}{k!} \left({ 1 - \prod_{j \mathop = 1}^{k-1} \left({1 - \dfrac j n}\right) }\right) }$ $\displaystyle$ $\le$ $\displaystyle \sum_{k \mathop = 1}^{M-1} \cmod {\dfrac{z^k}{k!} \left({ 1 - \prod_{j \mathop = 1}^{k-1} \left({1 - \dfrac j n}\right) }\right) } + \sum_{k \mathop = M}^n \cmod {\dfrac{z^k}{k!} } \cmod {1 - \prod_{j \mathop = 1}^{k-1} \left({1 - \dfrac j n}\right) }$ Triangle Inequality for Complex Numbers $\displaystyle$ $\le$ $\displaystyle \sum_{k \mathop = 1}^{M-1} \cmod {\dfrac{z^k}{k!} \left({ 1 - \prod_{j \mathop = 1}^{k-1} \left({1 - \dfrac j n}\right) }\right) } + \sum_{k \mathop = M}^n \cmod {\dfrac{z^k}{k!} }$ $\displaystyle$ $<$ $\displaystyle \sum_{k \mathop = 1}^{M-1} \dfrac \epsilon {2 M} + \dfrac \epsilon 2$ $\displaystyle$ $=$ $\displaystyle \epsilon$

As an Absolutely Convergent Series is Convergent, $\left\langle{s_n}\right\rangle$ converges.

Then:

 $\displaystyle 0$ $=$ $\displaystyle \lim_{n \mathop \to +\infty} \cmod {s_n - a_n}$ $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to +\infty} s_n - \lim_{n \mathop \to +\infty} a_n$ Combination Theorem for Limits of Functions $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 0}^\infty \dfrac {z^k} {k!} - \lim_{n \mathop \to +\infty} \left({1 + \dfrac z n}\right)^n$

The result follows.

$\blacksquare$