Proof of The Complex Exponential: the Functional Equation, Conjugates, Real Arguments and the Modulus

Routine facts about numbers, maps and finite sums are used without comment (Routine Facts, Continued: Typed Arithmetic, Maps and Predicativity); the results on conjugates and moduli cited below are read for the present complex numbers, conjugate and modulus by The Earlier and the New Versions of the Definitions of the Natural Numbers, the Complex Numbers, the Conjugate and the Modulus Agree, and the Two Versions of Real Correspond (ii), The Earlier and the New Versions of the Definitions of the Natural Numbers, the Complex Numbers, the Conjugate and the Modulus Agree, and the Two Versions of Real Correspond (iii) and The Earlier and the New Versions of the Definitions of the Natural Numbers, the Complex Numbers, the Conjugate and the Modulus Agree, and the Two Versions of Real Correspond (iv), and the results on real series and the real exponential in the present definitions of sequences and series by The Earlier and the New Versions of the Definitions of Sequences, Limits, Cauchy Sequences, Series and Iterated Operations Agree (i) and The Earlier and the New Versions of the Definitions of Sequences, Limits, Cauchy Sequences, Series and Iterated Operations Agree (v). Elements of $\mathbb{N}_{0}$, such as factorials and binomial coefficients, are read in $\mathbb{C}$ through the embedding $\iota$ (The Complex Numbers, with the Real Numbers Identified with a Subset of the Complex Numbers (iv), The Complex Numbers, with the Real Numbers Identified with a Subset of the Complex Numbers (vi)).

For $u\in\mathbb{C}$ and $k\in\mathbb{N}$ write

$$a_{k}(u)=\frac{u^{k-1}}{(k-1)!},\qquad\text{so that}\qquad \exp u=\sum_{k=1}^{\infty}a_{k}(u)$$

by the definition of the exponential (The Complex Exponential Function (i)).

The value at zero. We have $0^{0}=1$ and $0!=1$, so the first term is $a_{1}(0)=1$; and $0^{j}=0$ for $j\ge1$ (Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (ii)), so $a_{k}(0)=0$ for every $k\ge2$. Consequently every partial sum of $\sum_{k}a_{k}(0)$ equals $1$. A constant sequence converges to its value (Limits of Complex Sequences: Uniqueness, Real and Imaginary Parts, Arithmetic, Real Sequences, Conjugates, Moduli, the Cauchy Criterion and the Modulus Metric (iii)), so the series converges with sum $1$ (Series of Complex Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence (ii)). This proves (i).

The Cauchy product of two exponential series. Let $z,w\in\mathbb{C}$, and for $n\in\mathbb{N}$ let

$$c_{n}=\sum_{k=1}^{n}a_{k}(z)\,a_{n+1-k}(w)$$

be the $n$-th term of the Cauchy product of the series of $\exp z$ and $\exp w$. Put $m=n-1$ and substitute $j=k-1$; as $k$ runs over $1,\dots,n$, $j$ runs over $0,\dots,m$ (Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals (viii)), and $n+1-k=m-j+1$. The factorial formula $j!\,(m-j)!\binom{m}{j}=m!$ (Binomial Coefficients: Existence and Uniqueness by Pascal's Recursion, Vanishing above the Diagonal, Diagonal and First Values, the Factorial Formula and Symmetry (v)), for the binomial coefficients of The Binomial Coefficient (i), holds in $\mathbb{N}_{0}$ and therefore in $\mathbb{C}$; since $j!\,(m-j)!\neq0$ and $m!\neq0$ in $\mathbb{C}$ (Complex Power Series: the Sequence of Terms and the Sum as a Function on the Set of Convergence (ii)), it gives $1/(j!\,(m-j)!)=\binom{m}{j}/m!$. Hence, using the binomial theorem (The Binomial Theorem in a Commutative Ring (i)) in the last step,

$$c_{n}=\sum_{j=0}^{m}\frac{z^{j}w^{m-j}}{j!\,(m-j)!}=\frac{1}{m!}\sum_{j=0}^{m}\binom{m}{j}z^{j}w^{m-j}=\frac{(z+w)^{n-1}}{(n-1)!}=a_{n}(z+w).$$

The functional equation. The series $\sum_{k}a_{k}(z)$ and $\sum_{k}a_{k}(w)$ converge absolutely (The Complex Exponential Series Converges Absolutely, and Its Sum Is a Function on ℂ (ii)). By the theorem on Cauchy products (Series of Complex Numbers: Absolute Convergence, Comparison, Linearity, Index Shifts, Real and Imaginary Parts, Conjugates, Real Series and the Cauchy Product (viii)), the series $\sum_{n}c_{n}$ therefore converges, with sum $\exp z\cdot\exp w$. By the previous paragraph its terms are $a_{n}(z+w)$, so its sum is also $\exp(z+w)$. This proves (ii).

Reciprocals. Taking $w=-z$ in (ii) and using (i),

$$\exp z\cdot\exp(-z)=\exp0=1.$$

So $\exp z\neq0$, and $\exp(-z)=1/\exp z$. This proves (iii).

Conjugates. Conjugation preserves sums and products and fixes exactly the elements of $\mathbb{R}$ (Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (i)). By induction on the exponent it carries $z^{k-1}$ to $\overline{z}^{\,k-1}$, and it fixes the real number $1/(k-1)!$; hence $\overline{a_{k}(z)}=a_{k}(\overline{z})$ for every $k$. Since conjugation commutes with the sum of a convergent series (Series of Complex Numbers: Absolute Convergence, Comparison, Linearity, Index Shifts, Real and Imaginary Parts, Conjugates, Real Series and the Cauchy Product (vi)),

$$\overline{\exp z}=\sum_{k=1}^{\infty}\overline{a_{k}(z)}=\sum_{k=1}^{\infty}a_{k}(\overline{z})=\exp\overline{z}.$$

This proves (iv).

Real arguments. Let $x$ be a real number. Each term $a_{k}(x)$ is $\iota\big(x^{k-1}/(k-1)!\big)$, the $k$-th term of the real exponential series at $x$ read in $\mathbb{C}$. The real series $\sum_{k}x^{k-1}/(k-1)!$ converges (The Exponential Series Converges Absolutely, and Its Sum Is a Map on ℝ (ii)), so the complex series converges to $\iota$ of the real sum (Series of Complex Numbers: Absolute Convergence, Comparison, Linearity, Index Shifts, Real and Imaginary Parts, Conjugates, Real Series and the Cauchy Product (vii)); and the real sum is $\exp_{\mathbb{R}}(x)$ (The Real Exponential Function (i), The Real Exponential Function (ii)). Thus $\exp x=\exp_{\mathbb{R}}(x)$, a real number. This proves (v).

The modulus. We use, in turn, $u\overline{u}=|u|^{2}$ (Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (iii)), (iv), (ii), $z+\overline{z}=2\operatorname{Re}z$ (Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (ii)), (v), and the functional equation of the real exponential (Basic Properties of the Exponential Function (ii)):

$$|\exp z|^{2}=\exp z\cdot\overline{\exp z}=\exp z\cdot\exp\overline{z}=\exp(z+\overline{z})=\exp(2\operatorname{Re}z)=\exp_{\mathbb{R}}(2\operatorname{Re}z)=\exp_{\mathbb{R}}(\operatorname{Re}z)^{2}.$$

Both $|\exp z|\ge0$ (Conjugation and Modulus of Complex Numbers: Field Automorphism, Real and Imaginary Parts, Multiplicativity, Inverses, Component Bounds, Triangle Inequalities and Finite Sums (iii)) and $\exp_{\mathbb{R}}(\operatorname{Re}z)>0$ (Basic Properties of the Exponential Function (iii)), and two nonnegative reals with equal squares are equal; so $|\exp z|=\exp_{\mathbb{R}}(\operatorname{Re}z)$. For a real number $y$ we have $\operatorname{Re}(iy)=0$ and $\exp_{\mathbb{R}}(0)=1$ (Basic Properties of the Exponential Function (i)), hence $|\exp(iy)|=1$. This proves (vi).