# 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 colin@flyingcoloursmaths.co.uk 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

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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##### Where do you teach?

I teach in my home in Abbotsbury Road, Weymouth.

It's a 15-minute walk from Weymouth station, and it's on bus routes 3, 8 and X53. On-road parking is available nearby.