How to find all values of $z$ such that $z^3=-8i$

Question

If I am asked to find all values of $z$ such that $z^3=-8i$, what is the best method to go about that?

I have the following formula: $$z^{\frac{1}{n}}=r^\frac{1}{n}\left[\cos\left(\frac{\theta}{n}+\frac{2\pi k}{n}\right)+i\sin\left(\frac{\theta}{n}+\frac{2\pi k}{n}\right)\right]$$

for $k=0,\pm1, \pm2,...$

Applying this formula, I find the cubed root of $8$, which is $2$. And then when I apply it to the formula, I get the following:

$$z = 2\left[\cos\left(\frac{\pi}{3}+\frac{2\pi k}{3}\right)+i\sin\left(\frac{\pi}{3}+\frac{2\pi k}{3}\right)\right]$$ for $k=0,\pm1, \pm2,...$

I am confused, because the given solution is as follows: $$z = 2\left[\cos\left(\frac{\pi}{2}+\frac{2\pi k}{3}\right)+i\sin\left(\frac{\pi}{2}+\frac{2\pi k}{3}\right)\right]$$ for $k=0,\pm1, \pm2,...$

Where did I go wrong? How would my approach changed if I was asked to find all values for $-8$, or $8i$?

-i is laready at an angle in the complex plane. It can be written as $\cos(-pi/2)+i*\sin(-\pi/2)$. This is where you start (seeing that $\theta=-\pi/2$, or, if you prefer, $\theta=3\pi/2$ which is the same). — orion, Mar 26 '17 at 13:03
Sorry for my dumb question, but how do you know the angle is $(-\pi / 2)$? — Gary, Mar 26 '17 at 13:05
Related: http://math.stackexchange.com/q/1242129/42969, http://math.stackexchange.com/q/603714/42969. — Martin R, Mar 26 '17 at 13:21

Warren Hill · Answer 1 · 2017-03-26T13:40:21.280

There are several ways to solve this but one of the simplest is to write $ z $ of the form $ r \angle \theta $.

In this case $ z = 8 \angle \frac{-\pi}{2} $ We want cube roots and the cube root of 8 is 2. We get the first one by dividing $ \theta$ by 3

$ 2 \angle \frac{-\pi}{6}$ is thus one solution. There are three unique solutions and we get a solution by adding or subtracting $ \frac{2 \cdot \pi}{3} $ we can get a solution every time we do this but to avoid repeated solutions we are only interested in the ones where $ -\pi \lt \theta \le \pi$.

I'll leave this as an exercise, If required you can now convert these solutions back into $ a + i b$ form.

score 3 · Accepted Answer · answered Mar 26 '17 at 13:07

3

Short way write it as $$z^{ 3 }=-8i\\ z^{ 3 }=8{ e }^{ i\frac { 3\pi }{ 2 } }\\ z=\sqrt [ 3 ]{ 8 } { e }^{ i\left( \frac { \frac { 3\pi }{ 2 } +2\pi k }{ 3 } \right) }=2{ e }^{ i\left( \frac { 3\pi +4\pi k }{ 6 } \right) },k=0,1,2\quad $$

answered Mar 26 '17 at 13:07

haqnatural

21,578

This is how we solve it usually but introductory courses work with sine and cosine. I think it has some pedagogical value and I don't think the teacher/professor will be excited about a solution with more advanced tools that the student might have no theoretical basis yet. – Džuris Mar 26 '17 at 15:41
I guess the complex number with the Euler form is included in the elementary school programme. – haqnatural Mar 26 '17 at 15:45
What do you mean by elementary school and how do you which country should you talk about? I enciuntered complex numbers in secondary school and only in algebraic form. And few years ago they got replaced by statistics topics and are not taught at all in common Latvian schools. – Džuris Mar 26 '17 at 16:04

JMP · Answer 3 · 2017-03-26T14:01:37.017

3

You are claiming $\theta=\pi$, whereas it doesn't - actually $\theta=\dfrac{3\pi}{2}$, which gives the solution you list.

For $-8$ use $\theta=\pi$, and for $8i$ use $\theta=\dfrac{\pi}{2}$.

edited Mar 26 '17 at 14:01

answered Mar 26 '17 at 13:54

JMP

21,771

score 1 · Answer 4 · answered Mar 26 '17 at 13:46

The problem with you solution is the following: you didn't wrote properly the modulus and the exponential part of $w:=-8i$.

Clearly $$ r=|w|=8\\ -i=\exp{\left(\frac{3}{2}\pi i\right)}=\exp{\left(\frac{3}{2}\pi i+2ki\pi\right)}\;,\;\;k\in\Bbb Z $$ so that $$ z^3=r\exp[{i\pi\left(3/2+2k\right)}]\\ \Longrightarrow z_k =\sqrt[3]r\exp\left[{\frac{i\pi}3\left(\frac32+2k\right)}\right] =2\exp\left[{\frac{i\pi}2+\frac{2ki\pi}3}\right] $$ and as $k\in\Bbb Z$ you will notice that $z_k$ assumes only three different values, which are the solution of your initial equation.

How to find all values of $z$ such that $z^3=-8i$

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