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

Answer 1

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$.