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