Complex numbers

Multiplying two numbers in polar form multiplies their moduli and adds their arguments: $r_1(\cos\theta_1+i\sin\theta_1)\cdot r_2(\cos\theta_2+i\sin\theta_2) = r_1r_2\big(\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)\big)$, by the angle-sum identities for cosine and sine. Multiplying $n$ copies of the same number adds its argument to itself $n$ times.

We prove $(\cos\theta+i\sin\theta)^n = \cos n\theta+i\sin n\theta$ by induction on $n \ge 0$; the general modulus $r = 1$ case and negative $n = -m$ are then immediate.

**Base case ($n = 0$).** $(\cos\theta+i\sin\theta)^0 = 1 = \cos 0 + i\sin 0$. ✓

Inductive step. Assume $(\cos\theta+i\sin\theta)^n = \cos n\theta + i\sin n\theta$. Multiply both sides by $(\cos\theta+i\sin\theta)$: $(\cos\theta+i\sin\theta)^{n+1} = (\cos n\theta + i\sin n\theta)(\cos\theta + i\sin\theta)$. Expanding and applying the angle-addition identities $\cos(A+B) = \cos A\cos B - \sin A\sin B$ and $\sin(A+B) = \sin A\cos B + \cos A\sin B$, the real part becomes $\cos n\theta\cos\theta - \sin n\theta\sin\theta = \cos(n+1)\theta$ and the imaginary part becomes $\sin n\theta\cos\theta + \cos n\theta\sin\theta = \sin(n+1)\theta$. So the formula holds for $n+1$, completing the induction.

Extension to general modulus and negative exponent. For modulus $r = 1$: $(r(\cos\theta+i\sin\theta))^n = r^n(\cos n\theta + i\sin n\theta)$. For $n = -m$ with $m > 0$: since $|\cos m\theta + i\sin m\theta| = 1$, the reciprocal is $\overline{\cos m\theta + i\sin m\theta} = \cos(-m\theta) + i\sin(-m\theta)$, giving $(\cos\theta+i\sin\theta)^{-m} = \cos(-m\theta) + i\sin(-m\theta)$.

The Taylor series of $e^x$, $\cos x$ and $\sin x$ still make sense when $x$ is replaced by $i\theta$: $e^{i\theta} = \sum_{n=0}^{\infty}\dfrac{(i\theta)^n}{n!}$. Splitting the sum into even and odd $n$, and simplifying each power of $i$ using $i^2=-1$, gives exactly $\cos\theta = \sum \dfrac{(-1)^k\theta^{2k}}{(2k)!}$ as the real part and $\sin\theta = \sum \dfrac{(-1)^k\theta^{2k+1}}{(2k+1)!}$ as the imaginary part. Making this fully rigorous — manipulating an infinite series of complex terms like a finite sum — belongs to the theory of functions of a complex variable.

Step 1 — Recall the Taylor series. For all real $x$: $e^x = \sum_{n=0}^{\infty}\dfrac{x^n}{n!}$, $\cos x = \sum_{k=0}^{\infty}\dfrac{(-1)^k x^{2k}}{(2k)!}$, $\sin x = \sum_{k=0}^{\infty}\dfrac{(-1)^k x^{2k+1}}{(2k+1)!}$. All three series converge absolutely for every real (and complex) argument.

**Step 2 — Substitute $x = i\theta$ into the exponential series.** Absolute convergence justifies term-by-term substitution: $e^{i\theta} = \sum_{n=0}^{\infty}\dfrac{(i\theta)^n}{n!} = 1 + i\theta + \dfrac{(i\theta)^2}{2!} + \dfrac{(i\theta)^3}{3!} + \cdots$.

**Step 3 — Simplify using $i^2 = -1$ and separate real and imaginary parts.** The powers of $i$ cycle with period 4: $i^0=1,\,i^1=i,\,i^2=-1,\,i^3=-i,\,i^4=1,\ldots$ Separating even-indexed terms (real) and odd-indexed terms (imaginary): $e^{i\theta} = \left(1 - \dfrac{\theta^2}{2!} + \dfrac{\theta^4}{4!} - \cdots\right) + i\left(\theta - \dfrac{\theta^3}{3!} + \dfrac{\theta^5}{5!} - \cdots\right) = \cos\theta + i\sin\theta$. Making this fully rigorous requires absolute convergence of the complex exponential series, which holds because $\sum |a_n|$ converges.

Over $\mathbb{R}$, a polynomial such as $x^2+1$ can fail to have a root because its graph never crosses the $x$-axis. Complex numbers remove this obstruction: $\mathbb{C}$ contains a root for every non-constant polynomial, with no exceptions left. Once one root $z_1$ is found, dividing $p(z)$ by $(z-z_1)$ gives a polynomial of degree $n-1$, to which the same fact applies again.

There is no purely algebraic proof: every known proof borrows a tool from analysis or topology. The shortest uses Liouville's theorem from the theory of functions of a complex variable — see "Functions of a complex variable" for the full argument.