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Use De Moivre's theorem to solve the equation $z^2=-3-3i$. (Give your answers in polar form)

Can you please explain why there are two answers? I cannot seem to understand why.

By the way, the answers are $$z=18^{1/4}\mathrm{cis}(-3\pi/8)$$ $$z=18^{1/4}\mathrm{cis}(5\pi/8)$$

Thanks for the help.

NasuSama
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cla1n
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    If $w$ is solution of $z^2=a$, then so is $-w$. –  Jan 20 '14 at 01:01
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    Fundamental Theorem of Algebra. The polynomial has degree $2$, it must have two roots! – mathematics2x2life Jan 20 '14 at 01:05
  • Ok... But how does it apply to this question? I'll try to think through this. – cla1n Jan 20 '14 at 01:05
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    @Steve0101 Every polynomial of degree $n$ with complex coefficients has $n$ roots. Notice $z^2=-3-3i$ iff $z^2+(3+3i)=0$ is a complex polynomial of degree $2$. It has to have $2$ answer by FTA. As for the notation, $\text{cis}(\theta)=e^{i\theta}$, but the latter is far more common. – mathematics2x2life Jan 20 '14 at 01:28
  • Now I understand that there are two different answers, but how are the angles obtained. That is my main question. – cla1n Jan 20 '14 at 01:45

2 Answers2

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Notice that $-3 - 3i$ is equivalent to $3\sqrt{2}e^{-\frac{3\pi}{4}i}$. To see how, write $-3-3i$ as $3(-1 - i)$. Look at

$$-1 - i = re^{i\theta}$$

where $\theta$ is the angle and $r$ is the radius. Then

$$\begin{aligned} -1 &= r\cos(\theta)\\ -1 &= r\sin(\theta) \end{aligned}$$

With some algebra, we have

$$\begin{aligned} (r\cos(\theta))^2 + (r\sin(\theta)^2) &= (-1)^2 + (-1)^2\\ r^2(\cos^2(\theta) + \sin^2(\theta)) &= 2\\ r^2 &= 2\\ r &= \sqrt{2}\\ \Rightarrow -1 &= \sqrt{2}\cos(\theta)\\ \Rightarrow -1 &= \sqrt{2}\sin(\theta)\\ \Longrightarrow \theta &= -\dfrac{3\pi}{4} \end{aligned}$$

So we have $3\sqrt{2}e^{-\frac{3\pi}{4}i}$, which by Euler's identity is equivalent to

$$3\sqrt{2}\left(\cos\left(-\dfrac{3\pi}{4}\right) + i\sin\left(-\dfrac{3\pi}{4}\right)\right)$$

which gives us

$$\begin{aligned} z^2 &= 3\sqrt{2}\left(\cos\left(-\dfrac{3\pi}{4} \right) - i\sin\left(\dfrac{3\pi}{4}\right)\right)\\ z &= \left(3\sqrt{2}\left(\cos\left(-\dfrac{3\pi}{4} \right) - i\sin\left(\dfrac{3\pi}{4}\right)\right) \right)^{\frac{1}{2}}\\ \end{aligned}$$

with the following solutions found by De Moivre's Theorem:

$$\begin{aligned} z_1 &= (3\sqrt{2})^{\frac{1}{2}}\left(\cos\left(\dfrac{-\frac{3\pi}{4}}{2} \right) + i\sin\left(-\dfrac{\frac{3\pi}{4}}{2} \right) \right)\\ &= (3\sqrt{2})^{\frac{1}{2}}\left(\cos\left(-\dfrac{3\pi}{8} \right) + i\sin\left( -\dfrac{3\pi}{8}\right)\right)\\ &= (3\sqrt{2})^{\frac{1}{2}}e^{-\frac{3\pi}{8}i}\\ &= (\sqrt{18})^{\frac{1}{2}}e^{-\frac{3\pi}{8}i}\\ &= 18^{\frac{1}{4}}e^{-\frac{3\pi}{8}i}\\ z_2 &= (3\sqrt{2})^{\frac{1}{2}}\left(\cos\left(\dfrac{-\frac{3\pi}{4} + 2\pi}{2} \right) + i\sin\left(-\dfrac{\frac{3\pi}{4} + 2\pi}{2} \right) \right)\\ &= 18^{\frac{1}{4}}e^{\frac{5\pi}{8}i} \end{aligned}$$

By mathematics2x2life's comment, those solutions are equivalent to the following:

$$\begin{aligned} z_1 &= 18^{\frac{1}{4}}\mathrm{cis}\left(-\dfrac{3\pi}{8} \right)\\ z_2 &= 18^{\frac{1}{4}}\mathrm{cis}\left(\dfrac{5\pi}{8} \right) \end{aligned}$$

References

NasuSama
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$$\pm\sqrt[4]{18}\left(\frac{\sqrt{2-\sqrt{2}}}2-\frac{\sqrt{2+\sqrt{2}}}{2}i\right)$$

Senex Ægypti Parvi
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