Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to find $$(-64\mathrm{i})^{\frac{1}{3}}$$ This is a complex variables question. I need help by show step by step. Thanks a lot.

share|cite|improve this question
This is a bizarre selection of tags. – mrf Jan 30 '13 at 21:05
Could one not use the formula r^1/n \times$ the trigometric expansion? Setting k=0,1,2 – MathsPro Feb 1 '15 at 18:17

Let $y=(-64i)^{\frac13}\implies y^3=-64i=64i^3=(4i)^3$


$(y-4i)\{y^2+y\cdot 4i+(4i)^2\}=0$

If $y-4i=0,y=4i$

else $y^2+y\cdot 4i-16=0\implies y=\frac{-4i\pm\sqrt{(-4i)^2-4\cdot1(-16)}}2=\pm2\sqrt3-2i$

So, $y=4i,\pm2\sqrt3-2i$

share|cite|improve this answer
@Sunny88, why? do the other not satisfy the $y^3=-64i?$ what is the square root of $i?$ – lab bhattacharjee Dec 15 '12 at 17:18
@Sunny88, what to specify when I'm interested in all the roots? Also may I request you to specify some reference to your statement. – lab bhattacharjee Dec 15 '12 at 17:42
I tried to find reference and realized that there are many different conventions. I guess I was wrong to say that you are not correct. – Sunny88 Dec 15 '12 at 19:05

If you transform $-64i$ to polar form, you get $r=\sqrt{0+(-64)^2}=64$ and $\theta=-\pi/2$. Then you have $$(-64i)^{1/3} = r^{1/3}\cdot (\cos(\theta*\frac{1}{3})+i\sin(\theta*\frac{1}{3})) = 64^{1/3}\cdot (\cos((-\pi/2)*\frac{1}{3})+i\sin((-\pi/2)*\frac{1}{3})$$ $$= 4\cdot (\cos(-\pi/6)+i\sin(-\pi/6))$$ Given that $$\cos(-\pi/6)=\frac{\sqrt{3}}{2}$$ and $$\sin(-\pi/6) = -\frac{1}{2}$$ We have: $$4\cdot (\cos(-\pi/6)+i\sin(-\pi/6)) = 4\cdot (\frac{\sqrt{3}}{2}-\frac{1}{2}i) = 2\sqrt{3}-2i$$ The other roots can be found by adding $2\pi$ and $4\pi$ to $\theta$. So, $$4\cdot (\cos((\theta+2\pi)\cdot \frac{1}{3})+i\sin((\theta+2\pi)\cdot \frac{1}{3})) =4i$$ and $$4\cdot (\cos((\theta+4\pi)\cdot \frac{1}{3})+i\sin((\theta+4\pi)\cdot \frac{1}{3})) = -2\sqrt{3}-2i$$

share|cite|improve this answer
The other two?${}{}{}$ – André Nicolas Dec 15 '12 at 16:44
You are using a convention that is not the same as the one I am used to. – André Nicolas Dec 15 '12 at 16:59
+1 for pointing out the definition of the principal cubic root. – s1lence Dec 15 '12 at 17:01
@AndréNicolas Sorry, you are right, I searched on the internet and it seems that my convention is not the popular one. – Sunny88 Dec 15 '12 at 19:02

For any $n\in\mathbb{Z}$, $$\left(-64i\right)^{\frac{1}{3}}=\left(64\exp\left[\left(\frac{3\pi}{2}+2\pi n\right)i\right]\right)^{\frac{1}{3}}=4\exp\left[\left(\frac{\pi}{2}+\frac{2\pi n}{3}\right)i\right]=4\exp\left[\frac{3\pi+4\pi n}{6}i\right]=4\exp \left[\frac{\left(3+4n\right)\pi}{6}i\right]$$

The cube roots in polar form are: $$4\exp\left[\frac{\pi}{2}i\right] \quad\text{or}\quad 4\exp\left[\frac{7\pi}{6}i\right] \quad\text{or}\quad 4\exp\left[\frac{11\pi}{6}i\right]$$

and in Cartesian form: $$4i \quad\text{or}\quad -2\sqrt{3}-2i \quad\text{or}\quad 2\sqrt{3}-2i$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.