De Moivre's Theorem Calculator

For a complex number in polar form $z = r(\cos\theta + i\sin\theta)$, raising it to the $n$th power gives $z^n = r^n(\cos(n\theta) + i\sin(n\theta))$. The modulus is raised to the power $n$ and the argument (angle) is multiplied by $n$. This works for all integers $n$, including negative values.

In rectangular form $(a + bi)$, computing $z^5$ requires expanding five multiplications and tracking all cross-terms. In polar form, you simply raise the modulus $r$ to the 5th power and multiply the angle by 5 -- two simple arithmetic operations instead of dozens.

Set $r = 1$ and expand $(\cos\theta + i\sin\theta)^n$ using the binomial theorem, then equate real and imaginary parts with $\cos(n\theta)$ and $\sin(n\theta)$. For $n = 2$ this gives the double-angle formulas: $\cos(2\theta) = \cos^2\theta - \sin^2\theta$ and $\sin(2\theta) = 2\sin\theta\cos\theta$.

Yes. For $z^{-n}$, the modulus becomes $r^{-n} = 1/r^n$ and the argument becomes $-n\theta$. This is equivalent to computing $1/z^n$, which is useful for dividing complex numbers in polar form.

Euler's formula states $e^{i\theta} = \cos\theta + i\sin\theta$. Substituting into De Moivre's theorem gives $(e^{i\theta})^n = e^{in\theta}$, which is simply the exponent rule for exponentials. De Moivre's theorem is the trigonometric form of this exponential identity.