If $z = −3 − \sqrt{ 3}i$, then what is $\arg(z ^{66})$?

Question

Having a little trouble with this question. I've graphed $z$ and found the angle to be $\dfrac{7\pi}{6}$. However I'm not really sure how to proceed from this. Some help would be appreciated thanks :)

Answer 1

Hint: Express $z$ as $\lvert z\rvert e^{i\operatorname{arg} z}$ then use the laws of indices. You will get $\operatorname{arg}(z^{66})=77\pi$, which is equivalent to $\operatorname{arg}(z^{66})=\pi$.

Answer 2

$$z=-2\sqrt3\left(\dfrac{\sqrt3}2+\dfrac i2\right)=-2\sqrt3(\cos30^\circ+i\sin30^\circ)$$

Now use How to prove Euler's formula: $e^{it}=\cos t +i\sin t$?

OR De Moivre's Theorem and it's domain

Answer 3

Hint 1: What is the relationship between the arguments of two given complex numbers, and the argument of their product?

Hint 2: If you have one complex number, and multiply it with itself $66$ times, what is the relationship between the argument of the original number and the argument of the result?

Answer 4

Hint: By hand, compute $\mathrm{Arg}(1),\mathrm{Arg}(z),\mathrm{Arg}(z^2),$ and $\mathrm{Arg}(z^3)$.

Answer 5

$$ z=-3-\sqrt{3}i=|z|e^{i\theta}\\ \theta=\tan^{-1}\frac{-\sqrt{3}}{-3}=-\frac{5\pi}{6}\\ \arg z^{66}=66~\theta=-50~\pi\mapsto \pi $$