# How to solve this equation? $8z^3+1=0.$ [closed]

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.

-

## closed as off-topic by mrf, user88595, hardmath, Deutsch Mathematiker, Grigory MJun 7 '14 at 14:06

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – mrf, user88595, hardmath, Deutsch Mathematiker, Grigory M
If this question can be reworded to fit the rules in the help center, please edit the question.

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 $\mathbb R^2$.

-
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 '14 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 $\;k=0,1,2\;$ ....

-