# How to find $(-64\mathrm{i}) ^{1/3}$?

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.

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This is a bizarre selection of tags. – mrf Jan 30 '13 at 21:05
Could one not use the formula r^1/n \times$the trigometric expansion? Setting k=0,1,2 – MathsPro Feb 1 '15 at 18:17 ## 3 Answers Let$y=(-64i)^{\frac13}\implies y^3=-64i=64i^3=(4i)^3$So,$y^3-(4i)^3=0(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$- @Sunny88, why? do the other not satisfy the$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 $$4\cdot (\cos((\theta+4\pi)\cdot \frac{1}{3})+i\sin((\theta+4\pi)\cdot \frac{1}{3})) = -2\sqrt{3}-2i$$ - 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: $$4i \quad\text{or}\quad -2\sqrt{3}-2i \quad\text{or}\quad 2\sqrt{3}-2i$$

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