# Ask Uncle Colin: A Complex Battle

Dear Uncle Colin,

I’m supposed to solve $(1+i)^N = 16$ for $N$, and I don’t know where to start!

-- Don’t Even Mention Other Imaginary Variations – Reality’s Enough

Hello, DEMOIVRE, there are a couple of ways to attack this.

The simplest way (I think) is to convert the problem into polar form: $(1+i) = \sqrt{2} e^{i \frac{\pi}{4}}$, which means $(1+i)^N = \sqrt{2^N} e^{i \frac{N\pi}{4}}$.

For the multiplier to work, $16 = \sqrt{2^N}$, so $2^N = 256$ and $N=8$.

For the exponent to work, $\frac{N\pi}{4} = 2k\pi$, for some integer $k$, to make sure the resulting complex number lies on the positive real axis. That gives $N = 8k$, so (as far as the angle is concerned), $N$ can be any multiple of 8.

The only number that satisfies both conditions is $N=8$.

Alternatively, you can work it out by trial and error: $(1+i)^2 = 2i$, so $(1+i)^4 = -4$ and $(1+i)^8 = 16$. That comes out nicely in this case, but you really can’t guarantee it in general.

-- Uncle Colin