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I understand the basics of converting to polar form but I have just come across a question that I haven't seen before.

Usually the complex number is expressed as $z=a+bi$, but this time the complex number I was given is $z^3=-4+4{\sqrt3}i$. Do I need to somehow remove the 3 power? Do I just simply use my usually formulas and ignore the power? Thanks.

EDIT: I thought I should add additional information. I need to convert to polar form, as I will then use Moivre's rule to calculate the roots of the complex number.

I found the polar form for the RHS, $z=8(\cos2.09+i\sin1.05)$. Do I need to cube root this to find the polar form of the original complex number?

Moivre expression is in the picture attached Moivre Expression

where $n=3$ and $k=0,1,2$ I am to find the roots $z0, z1, z2$

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    $\begingroup$ You could try first expressing the right hand side in polar form and then taking the cube root of it. $\endgroup$ – J P Oct 12 '19 at 21:09
  • $\begingroup$ Depends on what you want to do. Do you want to figure out what $z$ is? Then the question becomes $z = \root 3 \of {-4 + 4 \sqrt{-3}}$. $\endgroup$ – Bob Happ Oct 12 '19 at 21:17
  • $\begingroup$ The angle is unique in the polar form, you cannot have $z=r(\cos \theta + i \sin \varphi)$ with $\theta \neq \varphi + 2k\pi$ for any $k \in \mathbb{Z}$ $\endgroup$ – user289143 Oct 12 '19 at 21:32
  • $\begingroup$ Ohh I see. Yeah I made a silly error. Thanks for spotting it. $\endgroup$ – Twigleg Oct 12 '19 at 21:39
  • $\begingroup$ "Do I need to somehow remove the 3 power? " Eventually. but you can do it when it is easiest. There now reason to take the third root in rectangular form and then convert the third root to polar when you can just as legitimately convert the third power to polar first and then take the third root. Finding the third roots of complex numbers in polar form is ridiculously easy. Finding the third roots of complex numbers in rectangular form is ridiculously hard. $\endgroup$ – fleablood Oct 12 '19 at 22:40
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Now $z^3=8(\cos \frac{2\pi}{3}+i \sin \frac{2 \pi}{3})$: using DeMoivre's formula we get $$ z_0=\sqrt[3]{8}\left(\cos \left(\frac{1}{3}\cdot\frac{2\pi}{3}+\frac{2 \cdot 0 \cdot \pi}{3}\right)+i \sin \left(\frac{2 \pi}{9}+\frac{2 \cdot 0 \cdot \pi}{3}\right)\right)=2\left(\cos \frac{2\pi}{9}+i \sin \frac{2 \pi}{9}\right)\\ z_1=\sqrt[3]{8}\left(\cos \left(\frac{1}{3}\cdot\frac{2\pi}{3}+\frac{2 \cdot 1 \cdot \pi}{3}\right)+i \sin \left(\frac{2 \pi}{9}+\frac{2 \cdot 1 \cdot \pi}{3}\right)\right)=2\left(\cos \frac{8\pi}{9}+i \sin \frac{8 \pi}{9}\right)\\ z_2=\sqrt[3]{8}\left(\cos \left(\frac{1}{3}\cdot\frac{2\pi}{3}+\frac{2 \cdot 2 \cdot \pi}{3}\right)+i \sin \left(\frac{2 \pi}{9}+\frac{2 \cdot 2 \cdot \pi}{3}\right)\right)=2\left(\cos \frac{14\pi}{9}+i \sin \frac{14 \pi}{9}\right) $$

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  • $\begingroup$ Ok, so the $z^3$ doesn't matter that it is to the power of 3? I can still simply use the polar form I found? $\endgroup$ – Twigleg Oct 12 '19 at 21:33
  • $\begingroup$ @Twigleg the power is important, because it's telling you which $n$-th root take. Since $n=3$ the argument of the roots will be the the cubic root of the argument of $z^3$ and the angles will be the angle of $z^3$ divided by $3 + \frac{2k \pi}{3}$ with $k=0,1,2$ $\endgroup$ – user289143 Oct 12 '19 at 21:39
  • $\begingroup$ So if I was told to find the real and imaginary parts of $z_0$ , $z_1$ and $z_2$ to two decimal places, I would just need to put these straight into a calculator and get the answers? $\endgroup$ – Twigleg Oct 12 '19 at 21:50
  • $\begingroup$ Yeah, passing from polar form to cartesian form is merely calculating explicitly $\cos \theta$ and $\sin \theta$ $\endgroup$ – user289143 Oct 12 '19 at 21:51
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You need to write $z^3=r(\cos\theta+i\sin\theta)$. To find $r$, you take the absolute value, and you found $r=8$. Then $$z^3=-4+4\sqrt 3i=8(\cos\theta+i\sin\theta)$$ This means $$\cos\theta=-\frac 12\\\sin\theta=\frac{\sqrt 3}2$$ From here you can find $\theta=\frac 23\pi$. Then, applying Moivre, you get $$z=\sqrt[3] 8\left(\cos\left(\frac{2\pi}9+\frac{2k\pi}3\right)+i\sin\left(\frac{2\pi}9+\frac{2k\pi}3\right)\right)$$ Here $k=0,1,2$, and you can also use $\sqrt[3] 8=2$.

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The modulus of $z^3$ is $\sqrt{16(1+3)}=8$, so $$z^3=8\Bigl(-\frac12+\frac{\sqrt 3}2i\Bigr)=8\mathrm e^{\tfrac{2i\pi} 3}$$ Therefore, if $\;z=r\mathrm e^{i\theta}$, we have to solve \begin{cases} r^3=8\enspace (r\in\mathbf R^+)\iff r=2,\\[1ex] 3\theta\equiv \frac{2\pi}3\mod 2\pi\iff\theta\equiv \frac {2\pi}9\mod\frac{2\pi}3. \end{cases} This results in $3$ solutions since the complex exponential has period $2i\pi$, $$z_1=2\mathrm e^{\tfrac{2i\pi}9}, \quad z_2=2\mathrm e^{\tfrac{8i\pi}9}, \quad z_3=2\mathrm e^{\tfrac{14i\pi}9}\;(=2\mathrm e^{-\tfrac{4i\pi}9}).$$

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