[proofplan] We prove the identity for every integer $n$ by splitting into three cases and handling each separately. For $n \geq 1$, we induct on $n$, using the [Multiplication of Complex Numbers in Polar Form](/theorems/911) to pass from $n = k$ to $n = k+1$. For $n = 0$, the identity is a direct evaluation of the zeroth power. For $n < 0$, we write $n = -m$ with $m \geq 1$, apply the case $m \geq 1$ already established, and invert the resulting unit-modulus complex number using the formula $\frac{1}{a+bi} = \frac{a-bi}{a^2+b^2}$ together with the Pythagorean identity $\cos^2\theta + \sin^2\theta = 1$. The parity identities $\cos(-\varphi) = \cos\varphi$ and $\sin(-\varphi) = -\sin\varphi$ then put the result into the required form. [/proofplan] [step:Establish the identity for $n \geq 1$ by induction on $n$] Fix $\theta \in \mathbb{R}$ and let $P(n)$ denote the statement $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$. **Base case ($n = 1$).** Both sides equal $\cos\theta + i\sin\theta$, so $P(1)$ holds. **Inductive step.** Assume $P(k)$ for some integer $k \geq 1$, that is, \begin{align*} (\cos\theta + i\sin\theta)^k = \cos(k\theta) + i\sin(k\theta). \end{align*} Multiplying both sides by $\cos\theta + i\sin\theta$: \begin{align*} (\cos\theta + i\sin\theta)^{k+1} = \bigl(\cos(k\theta) + i\sin(k\theta)\bigr)\bigl(\cos\theta + i\sin\theta\bigr). \end{align*} The right-hand side is the product of two complex numbers in polar form, each of modulus $1$, with arguments $k\theta$ and $\theta$. By the [Multiplication of Complex Numbers in Polar Form](/theorems/911), the product has modulus $1 \cdot 1 = 1$ and argument $k\theta + \theta = (k+1)\theta$: \begin{align*} \bigl(\cos(k\theta) + i\sin(k\theta)\bigr)\bigl(\cos\theta + i\sin\theta\bigr) = \cos((k+1)\theta) + i\sin((k+1)\theta). \end{align*} This establishes $P(k+1)$. By induction, $P(n)$ holds for every integer $n \geq 1$. [guided] We prove $P(n) : (\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$ for every integer $n \geq 1$ by induction on $n$, with $\theta \in \mathbb{R}$ fixed throughout. **Why induction?** The exponent $n$ is a positive integer, and the identity we want to prove increases the exponent by one each step. Induction is the natural tool: we check the smallest case, then show that the identity propagates from $n = k$ to $n = k+1$. **Base case ($n = 1$).** We must verify $P(1)$. The left-hand side is $(\cos\theta + i\sin\theta)^1 = \cos\theta + i\sin\theta$. The right-hand side is $\cos(1 \cdot \theta) + i\sin(1 \cdot \theta) = \cos\theta + i\sin\theta$. The two agree, so $P(1)$ holds. **Inductive step.** Fix an integer $k \geq 1$ and assume $P(k)$: \begin{align*} (\cos\theta + i\sin\theta)^k = \cos(k\theta) + i\sin(k\theta). \end{align*} We aim to deduce $P(k+1)$. The key move is to factor the $(k+1)$-th power as the product of a $k$-th power and a single factor: \begin{align*} (\cos\theta + i\sin\theta)^{k+1} = (\cos\theta + i\sin\theta)^k \cdot (\cos\theta + i\sin\theta). \end{align*} Applying the inductive hypothesis $P(k)$ to the first factor gives \begin{align*} (\cos\theta + i\sin\theta)^{k+1} = \bigl(\cos(k\theta) + i\sin(k\theta)\bigr)\bigl(\cos\theta + i\sin\theta\bigr). \end{align*} How do we multiply these two expressions? Each is a complex number in polar form $r(\cos\varphi + i\sin\varphi)$: the first has modulus $r_1 = 1$ and argument $\varphi_1 = k\theta$; the second has modulus $r_2 = 1$ and argument $\varphi_2 = \theta$. We apply the [Multiplication of Complex Numbers in Polar Form](/theorems/911), which states that the product of two complex numbers in polar form has modulus equal to the product of the moduli and argument equal to the sum of the arguments. The hypotheses of theorem 911 are satisfied since both factors are written in polar form and $\theta, k\theta \in \mathbb{R}$. We conclude \begin{align*} \bigl(\cos(k\theta) + i\sin(k\theta)\bigr)\bigl(\cos\theta + i\sin\theta\bigr) = 1 \cdot 1 \cdot \bigl(\cos(k\theta + \theta) + i\sin(k\theta + \theta)\bigr) = \cos((k+1)\theta) + i\sin((k+1)\theta). \end{align*} This is exactly $P(k+1)$. By the principle of mathematical induction, $P(n)$ holds for every integer $n \geq 1$. [/guided] [/step] [step:Verify the identity for $n = 0$ by direct evaluation] For any complex number $z$ we have $z^0 = 1$ by convention, so in particular \begin{align*} (\cos\theta + i\sin\theta)^0 = 1. \end{align*} On the other hand, $\cos(0 \cdot \theta) + i\sin(0 \cdot \theta) = \cos 0 + i\sin 0 = 1 + i \cdot 0 = 1$. Both sides equal $1$, so the identity holds for $n = 0$. [/step] [step:Reduce the case $n < 0$ to the case $n \geq 1$ via reciprocation] Let $n$ be a negative integer and write $n = -m$ with $m \geq 1$. By the definition of a negative integer power of a nonzero complex number, \begin{align*} (\cos\theta + i\sin\theta)^n = (\cos\theta + i\sin\theta)^{-m} = \frac{1}{(\cos\theta + i\sin\theta)^m}. \end{align*} The denominator is nonzero because $|\cos\theta + i\sin\theta| = 1$, so the reciprocal is well-defined. Applying the case $m \geq 1$ established in the first step: \begin{align*} (\cos\theta + i\sin\theta)^m = \cos(m\theta) + i\sin(m\theta), \end{align*} hence \begin{align*} (\cos\theta + i\sin\theta)^n = \frac{1}{\cos(m\theta) + i\sin(m\theta)}. \end{align*} We now compute the reciprocal on the right. For any real $a, b$ with $a^2 + b^2 \neq 0$, \begin{align*} \frac{1}{a + bi} = \frac{a - bi}{(a+bi)(a-bi)} = \frac{a - bi}{a^2 + b^2}. \end{align*} Applying this with $a = \cos(m\theta)$ and $b = \sin(m\theta)$: \begin{align*} \frac{1}{\cos(m\theta) + i\sin(m\theta)} = \frac{\cos(m\theta) - i\sin(m\theta)}{\cos^2(m\theta) + \sin^2(m\theta)}. \end{align*} By the Pythagorean identity $\cos^2\varphi + \sin^2\varphi = 1$ applied with $\varphi = m\theta$, the denominator equals $1$, so \begin{align*} (\cos\theta + i\sin\theta)^n = \cos(m\theta) - i\sin(m\theta). \end{align*} It remains to rewrite this in terms of $n = -m$. Using the parity identities $\cos(-\varphi) = \cos\varphi$ (cosine is even) and $\sin(-\varphi) = -\sin\varphi$ (sine is odd) applied with $\varphi = m\theta$: \begin{align*} \cos(m\theta) - i\sin(m\theta) = \cos(-m\theta) + i\sin(-m\theta) = \cos(n\theta) + i\sin(n\theta). \end{align*} This establishes the identity for $n < 0$. [guided] Let $n$ be a negative integer. Our strategy is to reduce to the case we have already proved (namely $n \geq 1$) and then manipulate the result into the desired form. **Setting up the reduction.** Write $n = -m$ where $m \geq 1$ is a positive integer. By definition of a negative integer power, \begin{align*} (\cos\theta + i\sin\theta)^n = (\cos\theta + i\sin\theta)^{-m} = \frac{1}{(\cos\theta + i\sin\theta)^m}. \end{align*} We must first check that the denominator is nonzero, otherwise the reciprocal is undefined. The modulus of $\cos\theta + i\sin\theta$ is $\sqrt{\cos^2\theta + \sin^2\theta} = \sqrt{1} = 1 \neq 0$, so $\cos\theta + i\sin\theta \neq 0$, and therefore $(\cos\theta + i\sin\theta)^m \neq 0$. **Invoking the positive-exponent case.** Since $m \geq 1$, the first step of this proof gives \begin{align*} (\cos\theta + i\sin\theta)^m = \cos(m\theta) + i\sin(m\theta), \end{align*} so we now have \begin{align*} (\cos\theta + i\sin\theta)^n = \frac{1}{\cos(m\theta) + i\sin(m\theta)}. \end{align*} Our task reduces to computing this reciprocal. **Computing the reciprocal.** Given a nonzero complex number $a + bi$, we rationalise by multiplying numerator and denominator by the conjugate $a - bi$: \begin{align*} \frac{1}{a + bi} = \frac{1}{a+bi} \cdot \frac{a-bi}{a-bi} = \frac{a-bi}{(a+bi)(a-bi)} = \frac{a-bi}{a^2 + b^2}. \end{align*} This identity requires $a^2 + b^2 \neq 0$; in our setting we will verify this momentarily. Apply it with $a = \cos(m\theta)$ and $b = \sin(m\theta)$: \begin{align*} \frac{1}{\cos(m\theta) + i\sin(m\theta)} = \frac{\cos(m\theta) - i\sin(m\theta)}{\cos^2(m\theta) + \sin^2(m\theta)}. \end{align*} **Simplifying the denominator.** By the Pythagorean identity, $\cos^2\varphi + \sin^2\varphi = 1$ for every real $\varphi$. Applied with $\varphi = m\theta \in \mathbb{R}$, this yields $\cos^2(m\theta) + \sin^2(m\theta) = 1$, which simultaneously confirms $a^2 + b^2 = 1 \neq 0$ (so the reciprocal formula was valid) and collapses the denominator: \begin{align*} (\cos\theta + i\sin\theta)^n = \cos(m\theta) - i\sin(m\theta). \end{align*} **Recasting in terms of $n$.** We want the right-hand side in the form $\cos(n\theta) + i\sin(n\theta)$, where $n = -m$. Recall two basic parity facts about the trigonometric functions: - Cosine is an even function: $\cos(-\varphi) = \cos\varphi$ for every $\varphi \in \mathbb{R}$. - Sine is an odd function: $\sin(-\varphi) = -\sin\varphi$ for every $\varphi \in \mathbb{R}$. Applying these with $\varphi = m\theta$: \begin{align*} \cos(m\theta) &= \cos(-m\theta) = \cos(n\theta), \\ -\sin(m\theta) &= \sin(-m\theta) = \sin(n\theta). \end{align*} Substituting: \begin{align*} (\cos\theta + i\sin\theta)^n = \cos(m\theta) - i\sin(m\theta) = \cos(n\theta) + i\sin(n\theta). \end{align*} This establishes the identity for every negative integer $n$. [/guided] [/step] [step:Combine the three cases to conclude] The three preceding steps cover, respectively, the cases $n \geq 1$, $n = 0$, and $n \leq -1$. Together they exhaust all integers, so \begin{align*} (\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta) \end{align*} for every $\theta \in \mathbb{R}$ and every $n \in \mathbb{Z}$. $\blacksquare$ [/step]