# simplify $(-2 + 2\sqrt3i)^{\frac{3}{2}}$?

How can I simplify $(-2 + 2\sqrt3i)^{\frac{3}{2}}$ to rectangular form $z = a+bi$?

(Note: Wolfram Alpha says the answer is $z=-8$. My professor says the answer is $z=\pm8$.)

I've tried to figure this out for a couple hours now, but I'm getting nowhere. Any help is much appreciated!

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It helps to visualize the number $z' = -2 + 2\sqrt{3} i$ as a vector in the complex plane. Can you find the length $R$ and angle $\theta$ of this vector with the positive real axis? Then you know that $z' = R e^{\theta i}$, and taking the $3/2$-power suddenly becomes easy. – TMM Jun 2 '12 at 19:38
Using $R = \sqrt{x^2 + y^2}$ and $\theta = \arctan(\frac{y}{x})$, $R = 3i$ and $\theta = \frac{\pi}{3}$. So $z^{'}$ would then be $3i\cdot e^{\frac{\pi}{3}i}$? – mr_schlomo Jun 2 '12 at 19:53
Using $x = -2$ and $y = 2 \sqrt{3}$ you get $R = \sqrt{(-2)^2 + (2 \sqrt{3})^2} = \sqrt{4 + 12} = 4$. For $\theta$, you should get $2\pi/3$. So $z' = 4 e^{2 \pi i/3}$, and $(z')^{3/2} = \ldots$. Then translate back to the form $x + iy$ to get your answer. – TMM Jun 2 '12 at 20:02
So it becomes $z = \pm8 + cis(\pi) = \pm8 + \cos(\pi) + \sin(\pi)i = \pm8$! Now it makes sense. – mr_schlomo Jun 2 '12 at 20:10

$$(-2 + 2\sqrt3i) = 4 \exp\left(\frac{2\pi}{3}i\right) = 4 \cos \left(\frac{2\pi}{3}\right) + 4 \sin \left(\frac{2\pi}{3}\right) i$$

and I would say $$\left(4 \exp\left(\frac{2\pi}{3}i\right)\right)^{\frac{3}{2}} = 4^{\frac{3}{2}} \exp\left(\frac{3}{2} \times \frac{2\pi}{3}i\right) =8 \exp(\pi i) = -8.$$

I think using $(-2 + 2\sqrt3i) = 4 \exp\left(\frac{8\pi}{3}i\right)$ or $4 \exp\left(-\frac{4\pi}{3}i\right)$ here would be unconventional.

To get an answer of $\pm 8$ you would need to believe $\sqrt{-2 + 2\sqrt3i} = -1-\sqrt3 i$ as well as $1+\sqrt3 i$ and while the square of each of them gives the intended value, I would take what I regard as the principal root giving a single answer, and so does Wolfram Alpha.

It is like saying $\sqrt{4} = 2$ alone even though $(-2)^2=4$ too.

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Agreed. If it said "Let $z$ be a solution of $z^2 = (-2 + 2\sqrt{3}i)^3$" it would be a different story, but square roots should be considered single-valued functions. – TMM Jun 2 '12 at 19:32

$\:\sqrt{-2+2\sqrt{-3}}\:$ can be denested by a radical denesting formula that I discovered as a teenager.

Simple Denesting Rule $\rm\ \ \ \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \ \color{brown}{divide\ out}\ \sqrt{trace}$

Recall $\rm\: w = a + b\sqrt{n}\:$ has norm $\rm =\: w\:\cdot\: w' = (a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2 - n\: b^2$

and, furthermore, $\rm\:w\:$ has trace $\rm\: =\: w+w' = (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\: 2\:a$

Here ${-}2+2\sqrt{-3}\:$ has norm $= 16.\:$ $\rm\ \: \color{blue}{subtracting\ out}\ \sqrt{norm}\ = -4\$ yields $\ 2+2\sqrt{-3}\:$

and this has $\rm\ \sqrt{trace}\: =\: 2,\ \ hence\ \ \ \color{brown}{dividing\ it\ out}\$ of this yields the sqrt: $\:1+\sqrt{-3}.$

Checking we have $\ \smash[t]{\displaystyle \left(1+\sqrt{-3}\right)^2 =\ 1-3 + 2\sqrt{-3}\ =\ -2 + 2 \sqrt{-3}}$

Therefore $\quad\ \begin{eqnarray}\rm\:(-2 + 2\sqrt{-3})^{3/2} &=&\ (-2+2\sqrt{-3})\ (-2+2\sqrt{-3})^{1/2} \\ &=&\ -2\,(1-\sqrt{-3})\ (1+\sqrt{-3}) \\ &=&\ -8\rm\ \ \ (up\ to\ sign) \end{eqnarray}$

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Try something like this, perhaps:

$$(-2+2\sqrt{3}i)^{3/2}=\exp \left(\frac{3}{2}\operatorname {Log} (-2+2\sqrt{3}i)\right)$$

We know the principal value of $\operatorname {Log z}$ is given by $\operatorname {Log z}=\log |z|+i\arg \theta=$ for $n \in \mathbb{Z}$:

$$\exp \left(\frac{3}{2}\operatorname {Log} (-2+2\sqrt{3}i)\right)=\exp \left(\frac{3}{2}(\log 4+i \frac{2\pi}{3})\right)=8$$

Keep in mind that because we only used the single values $\operatorname {Log z}$, we only get the positive answer.

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So this can be written as $$4^{3/2} \cdot (-\frac{1}{2} + \frac{\sqrt{3}}{2} i)^{3/2} = 8 \cdot (\cos\frac{2\pi}{3} + i\cdot \sin\frac{2\pi}{3})^{3/2}$$

and $4^{3/2} =8$. Use De moivre now. And you can also pull out $-4$ and get going. Hence your $z = \pm{8}$.

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Ah ha! So it just becomes $z = 8\cdot cis(\pi)$ which then simplifies to $z = 8\sqrt{-1}(-1) = \pm8$! – mr_schlomo Jun 2 '12 at 19:15
@mr_schlomo: In the same way you can remove $-4$ and get a negative value – user9413 Jun 2 '12 at 19:17

$$-2+2\sqrt{3}i = 4 \exp(2 \pi/3 i) = 4 \exp(2 \pi/3 i + 2 k \pi i) \text{ where }k\in \mathbb{Z}$$ Hence, $$\left(-2+2\sqrt{3}i \right)^{3/2} = \left(4 \exp(2 \pi/3 i + 2k \pi i) \right)^{3/2} = 8 \exp(\pi i + 3k \pi i) = \pm 8$$

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