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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

3 Answers 3

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

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@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 $$4\cdot (\cos((\theta+4\pi)\cdot \frac{1}{3})+i\sin((\theta+4\pi)\cdot \frac{1}{3})) = -2\sqrt{3}-2i$$

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The other two?${}{}{}$ –  André Nicolas Dec 15 '12 at 16:44
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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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