In the equation: $z = (\sqrt3+i)^n$ determine the values of $n$ for which $z$ is a real number.

Question

Question : In the equation: $z = (\sqrt3+i)^n$ determine the values of $n$ for which $z$ is a real number.

Working:

$z = (\sqrt3+i)^n $

$z^\frac1n = (\sqrt3+i)$

$z^\frac1n = (2cis(\frac\pi6 + 2k\pi))$

I'm not quite if my working is correct, its just similar to what we have done in class I don't really know how to approach the question after this.

Any help is appreciated, Thanks

Instead of worrying about such things as "$z^{\frac1n}$", why not compute $(\sqrt3+i)^n$? — Angina Seng, Feb 11 '18 at 06:01
Why not calculate $z^2$ and $z^3$ and ... stop right there, and look for a good hunch. — dxiv, Feb 11 '18 at 06:04
@YuriyS that was my answer, but I thought there would be more to it — Plus Twenty, Feb 11 '18 at 06:13
@Conal It means that $z^6 = -64 \in \mathbb{R}$, and therefore $z^{6n} = \left(-64\right)^n \in \mathbb{R}$ for all integer $n$. — dxiv, Feb 11 '18 at 06:20
@Conal, you've got great answers already, but I would also suggest you brush up on the vector interpretation of complex numbers, where multiplication corresponds to both stretching and rotation. Then all this will become clear to you — Yuriy S, Feb 11 '18 at 06:36
@YuriyS Will definitely look it up, this is just an extension question I was given in class we'd never looked at before. — Plus Twenty, Feb 11 '18 at 08:46

score 5 · Accepted Answer · answered Feb 11 '18 at 06:16

5

You have $$ z=2\left(\frac {\sqrt3}2+\frac i2\right)=2 (\cos\frac\pi6+i\sin\frac\pi6). $$ Thus $$z^n=2^n (\cos\frac {n\pi}6+i\sin\frac {n\pi}6). $$ For the imaginary part to be zero, we need $\sin n\pi/6=0$. The sine is zero precisely on the integer multiples of $\pi $. So $z^n $ will be real when $n $ is a multiple of $ 6$.

answered Feb 11 '18 at 06:16

Martin Argerami

205,756

Thanks for the explanation! – Plus Twenty Feb 11 '18 at 06:29
1

Glad I could help! :) – Martin Argerami Feb 11 '18 at 06:43

In the equation: $z = (\sqrt3+i)^n$ determine the values of $n$ for which $z$ is a real number.

1 Answers1