Ask Uncle Colin: Powers and polar form

Dear Uncle Colin,

I’ve been given $u=(2\sqrt{3}-2\mathrm{i}{)}^6$ and been told to express it in polar form. I’ve got as far as $u=54-2{\mathrm{i}}^6$, 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 $(2\sqrt{3}-2\mathrm{i})(2\sqrt{3}-2\mathrm{i})(2\sqrt{3}-2\mathrm{i})(2\sqrt{3}-2\mathrm{i})(2\sqrt{3}-2\mathrm{i})(2\sqrt{3}-2\mathrm{i})$. The first pair of brackets give you $(8-8\sqrt{3}\mathrm{i})$, which makes the whole thing $(8-8\sqrt{3}\mathrm{i})(8-8\sqrt{3}\mathrm{i})(8-8\sqrt{3}\mathrm{i})$.

Expanding the first pair of brackets here gives you $(-128-128\sqrt{3}\mathrm{i})(8-8\sqrt{3}\mathrm{i})$, which you can expand to get $-4096$, which is $4096e^{\pi \mathrm{i}}$ 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 $a=2\sqrt{3}$, $b=-2\mathrm{i}$ and $n=6$ gives:

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

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

Method 3: the proper way

By far the simplest way is to take your complex number and turn it directly into polar form: $(2\sqrt{3}-2\mathrm{i})=4e^{-\frac{\pi }{6}\mathrm{i}}$. 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 $(-\frac{\pi }{6}\times 6=-\pi )$. (In argument terms, $-\pi$ and $\pi$ are the same, because angles.)

I hope that clears it up!

-- Uncle Colin