complex number with high powers

Question

The question:

$\frac{(-1-\sqrt{3}i)^{73}}{2^{73}}$

I really do not even know where to begin. Am I suppose to expand $(-1-\sqrt{3}i)$ 73 times?

Answer 1

Hint: $$\frac{-1-\sqrt{3} i}{2} = r e^{i \theta}$$ for some real number $\theta$ and nonnegative number $r$. Can you find what $\theta$ and $r$ are? Euler's formula may be helpful. Raising $r e^{i \theta}$ to the $73$rd power is a bit easier.

Answer 2

Hint:

$$\frac{(-1-\sqrt{3}i)^{73}}{2^{73}}=(-\frac{1}{2}-\frac{\sqrt{3}}{2}i)^{73}$$ and $$(-\frac{1}{2}-\frac{\sqrt{3}}{2}i)^{3}=1$$

Answer 3

Hint: $\quad\dfrac{-1-i\sqrt 3}2=\mathrm e^{\tfrac{4i\pi}3}.$

Answer 4

A systematic approach goes like this.

1) Represent any complex number $z\in\mathbb{C}$, your example being $z=\frac{-1-\sqrt{3} i}{2}$ in polar coordinates $$z = r e^{i \theta},$$ where $$r=\sqrt{\textrm{Re}{z}^2+\textrm{Im}{z}^2}$$ and $$\theta = \arg{z} = \arctan{\frac{\textrm{Re}{z}}{\textrm{Im}{z}}}$$ unless $\textrm{Im}{z}=0$. In your example, we find $r=\sqrt{\frac{1}{4}+\frac{3}{4}}=1$ and $\theta=-\frac{2}{3}2\pi$

2) Take the power $$z^n=r^ne^{i\theta n}$$ for $n\in\mathbb{Z}$. In your example, we obtain $z^{73}=1^{73}e^{-i73\frac{2}{3}2\pi}=e^{-i\frac{2}{3}2\pi}=z$, because $e^{-i72\frac{2}{3}2\pi}=e^{-i48\times 2\pi}=1$.

Answer 5

Ok, what I got was:

$(-\frac{1}{2}-\frac{\sqrt{3}}{2}i)^{73}$

$|z| = \sqrt{(-1/2)^2+\sqrt{3}/2} = \sqrt{1} = 1$

tan$^{-1}$ ($-.5/ \sqrt{3}/2) = \frac{-2}{3}2\pi$

$z^{73} = 1($cos$(\frac{-2}{3}2\pi)+i$sin$(\frac{-2}{3}2\pi)$)

$z = 1^{73}($cos$(\frac{-2}{3}2\pi*73)+i$sin$(\frac{-2}{3}2\pi8*73)$)

= $-.5 + \frac{\sqrt{3}}{2}$ which is the same answer had I not multiplied by 73