# Ask Uncle Colin: Powers and polar form

Dear Uncle Colin,

I’ve been given $u = (2\sqrt{3} - 2\i)^6$ and been told to express it in polar form. I’ve got as far as $u=54 -2\i^6$, but don’t know where to take it from there!

- Not A Problem I’m Expecting to Resolve

Hello, NAPIER, and thanks for your message!

I fear you’ve fallen into one of the classical mathematical traps: you *cannot* generally say that $(a+b)^n = a^n + b^n$. Why not? If you could, you could write $2^5$ as $(1+1)^5$, ‘expand’ it as $1^5 + 1^5$ and conclude that $32=2$. Which it isn’t.

I know of three more-or-less reasonable ways to do this, one of which is so much easier than the others, I hesitate to call them reasonable.

### Method 1: expand the brackets

It’s simple! ‘Just’ work out $(2\sqrt{3} - 2\i)(2\sqrt{3} - 2\i)(2\sqrt{3} - 2\i)(2\sqrt{3} - 2\i)(2\sqrt{3} - 2\i)(2\sqrt{3} - 2\i)$. The first pair of brackets give you $(8 - 8\sqrt{3}\i)$, which makes the whole thing $(8 - 8\sqrt{3}\i)(8 - 8\sqrt{3}\i)(8 - 8\sqrt{3}\i)$.

Expanding the first pair of brackets here gives you $(-128 - 128\sqrt{3}\i)(8-8\sqrt{3}\i)$, which you can expand to get $-4096$, which is $4096 e^{\pi \i}$ in polar form.

But that’s a silly way to do it. We know a better way to expand many copies of the same bracket.

### Method 2: binomial expansion

Using the traditional binomial expansion routine with $a=2\sqrt{3}$, $b = -2\i$ and $n=6$ gives:

$ 1 \times 1728 \times 1 + \\ 6 \times 288\sqrt{3} \times (-2\i) + \\ 15 \times 144 \times(-4) + \\ 20 \times 24\sqrt{3} \times (8\i) + \\ 15 \times 12 \times 16 + \\ 6 \times 2\sqrt{3} \times -32\i + \\ 1 \times 1 \times -64 = \\ 1728 - 3456\sqrt{3}\i - 8640 + 3840\sqrt{3}\i + 2880 - 384 \sqrt{3}\i - 64 = - 4096$

… as before. Again, that’s $4096 e^{\pi \i}$.

### Method 3: the proper way

By far the simplest way is to take your complex number and turn it directly into polar form: $(2\sqrt{3}-2\i) = 4 e^{-\frac{\pi}{6}\i}$. Taking the sixth power of that is as simple as taking the sixth power of the modulus ($4^6 = 4096$) and multiplying the argument by 6 $\left(-\frac{\pi}{6} \times 6 = -\pi\right)$. (In argument terms, $-\pi$ and $\pi$ are the same, because angles.)

I hope that clears it up!

-- Uncle Colin