Ask Uncle Colin: Powers and polar form

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'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

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.

Share

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Sign up for the Sum Comfort newsletter and get a free e-book of mathematical quotations.

No spam ever, obviously.

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.

On twitter