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I would like to know how to do this question.

(a) Determine all of the solutions of the equation $$8z^3+1=0.$$

Express your solutions in the form $z = re^{i\theta}$ where $-\pi<\theta \leq \pi$.

(b) Plot all of the solutions from part (a) in the complex plane.

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closed as off-topic by mrf, user88595, hardmath, user91500, Grigory M Jun 7 at 14:06

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2 Answers 2

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Clearly $$ 8z^3+1=0\Longleftrightarrow 8z^3=-1 $$ but you know that $$ -1=e^{i\pi}=e^{i\pi+2k\pi i},\;\; k\in\mathbb Z. $$ Hence $$ 8z^3=-1\Longleftrightarrow z^3=\frac18e^{i\pi+2k\pi i}\Longleftrightarrow z=\frac12e^{i\frac{\pi}3(1+2k)},\;\; k\in\mathbb Z.$$ But as $k\in\mathbb Z$ you have only three different values for $z$ which are the tree different solutions of your equation.

These solution will form a triangle whose vertices are the roots of your initial polynomial: \begin{align*} z_0=&\frac12e^{i\frac{\pi}3}=\left(\frac14,\frac{\sqrt3}4\right)\\ z_1=&\frac12e^{i\pi}=-\frac12\\ z_2=&\frac12e^{-i\frac{\pi}{3}}=\left(\frac14,-\frac{\sqrt3}4\right) \end{align*} I wrote the cartesian coordinates for the vertices of the triangle, using the identification of $\mathbb C$ with the real plane $\mathbb R^2$.

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But for giving the complete solution (and thus leaving nothing for the OP to do), how is this different from what I wrote more than half an hour ago? –  DonAntonio Jun 7 at 13:52

Hints:

$$z^3=-\frac18=\frac18e^{\pi i+2k\pi i}=\frac18e^{\pi i(1+2k)}$$

Now apply de Moivre's Formula and use $\;k=0,1,2\;$ ....

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