Question

Precal help if you could could you answer both. Me finishing this helps me graduate.

Answer 1

Exercise 1:

The easiest way to compute powers of complex numbers is to write them in the form

$z = \rho e^{i\theta} = \rho(\cos(\theta)+i\sin(\theta))$

In this form, you have

$z^n = \rho^n e^{in\theta} = \rho^n(\cos(n\theta)+i\sin(n\theta))$

The magnitude of the number is given by

$z=a+bi \implies \rho = \sqrt{a^2+b^2}$

So, we have

$z=-1+\sqrt{3}i \implies \rho = \sqrt{1+3}=2$

As for the angle, we have

$z=a+bi \implies \theta = \text{atan2}\dfrac{b}{a}$

So, we have

$z=-1+\sqrt{3}i \implies \theta = \text{atan2}\left(\dfrac{-\sqrt{3}}{-1}\right) = -\dfrac{2\pi}{3}$

Finally,

$z=-1-\sqrt{3}i = 2\left(\cos\left( -\dfrac{2\pi}{3}\right) + i\sin\left( -\dfrac{2\pi}{3}\right)\right) \implies z^6 = 2^6\left(\cos\left( -4\pi\right) + i\sin\left(-4\pi\right)\right) = 64$

Exercise 2:

You simply have to compute the trigonometric function:

$\cos\left(-\dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2},\quad \sin\left(-\dfrac{\pi}{4}\right) = -\dfrac{\sqrt{2}}{2}$

So, we have

$z = \sqrt{2}\left(\dfrac{\sqrt{2}}{2} - i\dfrac{\sqrt{2}}{2}\right) = 1-i$