Evaluate $(\overline{z}_3)^4$ given that $z_3 = -\frac{1}{2}+j\frac{\sqrt{3}}{2}$

Question

Given that:
$$z_3 = -\frac{1}{2}+j\frac{\sqrt{3}}{2}$$
evaluate the following:
$(\overline{z}_3)^4$
Solution:
$$(\overline{z}_3)^4 = [-\frac{1}{2}-j\frac{\sqrt{3}}{2}]^4$$ $$=[1\angle(-\frac{2\pi}{3})]^4$$ $$=1\angle(-\frac{8\pi}{3})$$ $$=1(\cos(-\frac{8\pi}{3})+j\sin(-\frac{8\pi}{3}))$$ $$=-\frac{1}{2}-j\frac{\sqrt{3}}{2}$$

To write $-\frac{1}{2}-j\frac{\sqrt{3}}{2}$ in polar form we first note that $|-\frac{1}{2}-j\frac{\sqrt{3}}{2}|=\sqrt{(\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}=1$. With $a=-\frac{1}{2}$ and $b=-\frac{\sqrt{3}}{2}$, we obtain the reference angle of $\tan^-1(\frac{b}{a})=\tan^-1(\sqrt{3})=\frac{\pi}{3}$ radians. However $-\frac{1}{2}-j\frac{\sqrt{3}}{2}$ is in the third quadrant. Hence to get the argument of $-\frac{1}{2}-j\frac{\sqrt{3}}{2}$, we need to ratate $\frac{\pi}{3}$ (clockwise) by $\pi$ radians to get $\theta=\frac{\pi}{3}-\pi=-\frac{2\pi}{3}$.

My questions:
1. Why do we have to note that $|-\frac{1}{2}-j\frac{\sqrt{3}}{2}|=\sqrt{(\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}=1$?
2. How do we know that $-\frac{1}{2}-j\frac{\sqrt{3}}{2}$ is in the third quadrant?

Answer 1

Given: $z = -\frac{1}{2}-\frac{\sqrt{3}}{2}i$

Find: $z^4$

In order to avoid multiplying (FOIL) numbers of the form $a+bi$ three times we make use de Moivre's of formula which is: $\left(\cos\theta+i\sin\theta\right)^n = \cos n\theta + i\sin n\theta$.

This means we need to transform the complex number $z$ into trig (polar) form: $r\left(\cos\theta+i\sin\theta\right)$.

As you noted, $r = \left|z\right| = 1$. In order to solve for $\theta$ we use the fact that $\tan\theta = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$. This means $\theta = \tan^{-1}\sqrt{3} = \frac{\pi}{3}$. However this is only the reference angle and not the full angle we need.

In order to find the quadrant where our complex number is located we can plot it using the x-axis for the real part and the y-axis for the complex part. So left $\frac{1}{2}$ and down $\frac{\sqrt{3}}{2}$. The point is in the third quadrant.

So the angle is $\theta = \pi+\frac{\pi}{3} = \frac{4\pi}{3}$.

Using de Moivre's formula:

$z = -\frac{1}{2}-\frac{\sqrt{3}}{2}i = 1\left(\cos\theta+i\sin\theta\right)\\ \Rightarrow z^4 = \left(1\left(\cos\theta+i\sin\theta\right)\right)^4\\ \Rightarrow z^4 = 1^4\left(\cos\theta+i\sin\theta\right)^4\\ \Rightarrow z^4 = 1^4\left(\cos 4\theta+i\sin 4\theta\right)\\ \Rightarrow z^4 = \cos 4\theta+i\sin 4\theta$

I included a couple more steps than needed in the last part to show how it all fits together.