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Imagine expanding (1+i3)10(1 + i\sqrt{3})^{10} using the binomial theorem β€” you would face eleven terms, alternating powers of ii, and endless arithmetic.
Every complex number z=x+iyz = x + iy can be written in modulus-argument form as z=r(cos⁑heta+isin⁑heta)z = r(\cos heta + i\sin heta), where r=x2+y2r = \sqrt{x^2 + y^2} is the distance from the origin and heta=atan2⁑(y,x) heta = \operatorname{atan2}(y, x) is the angle with the positive real axis.
What happens if you multiply a complex number zz by itself over and over again?

The Power of Repeated Multiplication

Multiplying z=r(cos⁑heta+isin⁑heta)z = r(\cos heta + i\sin heta) by itself gives z2=r2(cos⁑2heta+isin⁑2heta)z^2 = r^2(\cos 2 heta + i\sin 2 heta) because you multiply the lengths (rimesr=r2r imes r = r^2) and add the angles (heta+heta=2heta heta + heta = 2 heta).
In 1707, French mathematician Abraham de Moivre formalized this pattern into a single theorem while exploring trigonometry and complex expressions: for any integer nn, [r(cos⁑heta+isin⁑heta)]n=rn(cos⁑nheta+isin⁑nheta)[r(\cos heta + i\sin heta)]^n = r^n(\cos n heta + i\sin n heta).
How can we turn a messy algebraic power into a simple three-step calculation using this identity?

Computing Powers Step-by-Step

Let's evaluate (1+i3)6(1 + i\sqrt{3})^6. We first find the modulus r=12+(3)2=2r = \sqrt{1^2 + (\sqrt{3})^2} = 2 and argument heta = \arctan(\sqrt{3}/1) = rac{\pi}{3}, giving polar form 2(\cos rac{\pi}{3} + i\sin rac{\pi}{3}).

Common Mistakes to Avoid