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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

Tunk-Fey
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user133422
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2 Answers2

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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, $$ \begin{align} \Large z^2&=\Large z_4\\ \Large z&=\Large \left(z_4\right)^{\frac{1}{2}}\\ &=\Large\left[2e^{\left(-\frac{\pi}{6}+k\pi\right)i}\right]^{\frac{1}{2}}\\ &=\Large\pm\sqrt{2}\ e^{\left(-\frac{\pi}{12}+\frac{k\pi}{2}\right)i}\\ \Large z&=\Large\pm\sqrt{2}\ \text{cis}\left(-\frac{\pi}{12}+\frac{k\pi}{2}\right). \end{align} $$

$$\\$$

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

Tunk-Fey
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Remember that a complex number $a+ib$ really can be thought of as a point in the (complex) plane, so $a$ is your "$x$-coordinate" and $b$ is your "$y$-coordinate".

With this in mind, it's easier just to notice that $z_4=-\sqrt3+i=2(-\frac{\sqrt3}{2}+i\frac{1}{2})$.

If $-\frac{\sqrt3}{2}+i\frac{1}{2}$ doesn't look familiar, look at this http://commons.wikimedia.org/wiki/File:Unit_circle_angles_color.svg; it's just the point $(-\frac{\sqrt3}{2},\frac{1}{2})$ written differently.

Any complex number on the unit circle can be written as $e^{i\theta}$ where $\theta\in[\pi,-\pi)$. Looking at the picture of the unit circle you know right away that $z_4=-\sqrt3+i=2(-\frac{\sqrt3}{2}+i\frac{1}{2})=2e^{i\frac{5\pi}{6}}$

Now we know that if you take a number in polar coordinates, and you rotate it a full circle, i.e. $2\pi$, you end up at the same number. Since we want all the solutions to the equation $z^2=z_4$, we need all the $z_4$'s, i.e. $z_4=2e^{i(\frac{5\pi}{6}+2\pi k)}$ where $k=1,2,\dots$

Now we can solve $z^2=2e^{i(\frac{5\pi}{6}+2\pi k)}$, take square root of both sides:

$$z=2^{\frac{1}{2}}e^{\frac{1}{2}i(\frac{5\pi}{6}+2\pi k)}=\sqrt2e^{i(\frac{5\pi}{12}+\pi k)}$$

Just so we're sure we check that $(\sqrt2e^{i(\frac{5\pi}{12}+\pi k)})^2=z_4$.

Edit: $\sqrt{2}e^{i(\frac{5\pi}{12}+\pi k)}=\sqrt{2}(\cos(\frac{5\pi}{12}+\pi k)+i\sin(\frac{5\pi}{12}+\pi k))$ by Euler's formula. But you definitely want to use the form $re^{i\theta}$ when computing with complex numbers.

patience
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