Ask Uncle Colin: A Complex Battle

Ask Uncle Colin is a chance to ask your burning, possibly embarrassing, maths questions -- and to show off your skills at coming up with clever acronyms. Send your questions to and Uncle Colin will do what he can.

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


Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.


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