Question 9.1.10: Determine the modulus, an argument, and a polar form of z1 =...

We have

\left|z_1\right| = \left|2 – 2i\right| = \sqrt{2^2 + (-2)^2} = 2\sqrt{2}

Any argument \theta of z_1 satisfies

2 = 2\sqrt{2} \cos \theta       and       − 2 = 2\sqrt{2} \sin \theta

So, \cos \theta = \frac{1}{\sqrt{2}} and \sin \theta = − \frac{1}{\sqrt{2}}, which gives \theta = − \frac{\pi }{4} + 2\pi k , k \in \mathbb{Z}. Hence, a polar form of z_1 is

z_1 = 2\sqrt{2} \left(\cos (- \frac{\pi }{4} ) + i \sin (- \frac{\pi }{4} ) \right)

For z_2, we have

\left|z_2\right| = \left|-1 + \sqrt{3} i \right| = \sqrt{(-1)^2 + (\sqrt{3})^2} = 2

Since −1 = 2 \cos \theta and \sqrt{3} = 2 \sin \theta, we get \theta = \frac{2\pi }{3} + 2\pi k, k \in \mathbb{Z}. Thus, a polar form of z_2 is

z_2 = 2 \left(\cos ( \frac{2 \pi }{3} ) + i \sin ( \frac{2 \pi }{3} ) \right)