Complex number polar form equation

I've been struggling with a complex numbers algebra question for a few days now, and the tutor says I still haven't got it right.

Express $z_4 =−\sqrt{3} + i$ in polar form. Hence solve the equation

$$z^2= z_4,$$

for $z$ a complex number. You may leave the answer in polar form.

So far I've got

$$\begin{align} z&=\sqrt{2}\ \text{cis} \sqrt{150}\\ z&=\sqrt{2}\ \text{cis}\left(\dfrac{5\pi}{6}\right)+2k\pi\\ \text{and}\\ z&=\sqrt{2}\ \text{cis}\left(\dfrac{17\pi}{6}\right)+2k\pi \end{align}$$

I'm pretty sure these are all just different forms of the same equation though...

Can anyone help?

Many thanks,

John

Answer

If a complex number $z=x+iy$, then the polar form of $z$ is

$$r=|z|=\sqrt{x^2+y^2}$$
and
$$\theta=\arg(z)=\tan^{-1}\left(\frac{y}{x}\right).$$
For more detail explanation about $\arg(z)$, you may refer to this. Therefore
$$z=re^{i\theta}.$$
Now, to express $z=-\sqrt{3}+i$ in polar form, we have
$$r=|z|=\sqrt{(-\sqrt{3})^2+1^2}=2$$

and
$$\theta=\arg(z)=\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)=-\frac{\pi}{6}+k\pi,\quad\text{where }k=1,3,5,7,\cdots$$
Note that, the period of the tangent is $\pi$ rad. Hence, $$\Large z_4=\Large2e^{\left(-\frac{\pi}{6}+k\pi\right)i}.$$
Thus,
$$\begin{align} \Large z^2&=\Large z_4\\ \Large z&=\Large \left(z_4\right)^{\frac{1}{2}}\\ &=\Large\left[2e^{\left(-\frac{\pi}{6}+k\pi\right)i}\right]^{\frac{1}{2}}\\ &=\Large\pm\sqrt{2}\ e^{\left(-\frac{\pi}{12}+\frac{k\pi}{2}\right)i}\\ \Large z&=\Large\pm\sqrt{2}\ \text{cis}\left(-\frac{\pi}{12}+\frac{k\pi}{2}\right). \end{align}$$

$$\\$$

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$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$