# Thread: [SOLVED] Solving for roots -&gt; Z^3 = -i

1. ## [SOLVED] Solving for roots -&gt; Z^3 = -i

Hi all,

I'm trying to work out the roots for the equation above. I've worked out one of them as being z = i, as z^3= i^3 = -i.

Wolfram Alpha gives
z = 1/2 (sqrt(3)-i) and z = 1/2 (-sqrt(3)-i)

as the other answers, but I'm unsure how they arrived at this?

Thanks.

2. Hello, Ashgasm!

Tricky . . .

Find all roots of: . $z^3 \:=\:-i$

We have: . $z^3 + i \:=\:0$

Since $(\text{-}i)^3 \:=\:i$, we have: . $z^3 + (\text{-}1)^3 \;=\;0$ . . . a sum of cubes

Factor: . $\bigg[z + (\text{-}i)\bigg]\,\bigg[z^2 -(\text{-}1)(z) + (\text{-}i)^2\bigg] \;=\;0$

. . . . . . . . . $(z - i)\,(z^2 + iz - 1)\;=\;0$

We have two equations to solve:

$z - i \:=\:0 \quad\Rightarrow\quad \boxed{z \:=\:i}$

$z^2 + iz - 1 \:=\:0$
Quadratic Formula: . $z \:=\:\frac{-i \pm\sqrt{i^2 - 4(1)(\text{-}1)}}{2(1)} \;=\;\boxed{\frac{-i \pm\sqrt{3}}{2}}$

The three roots are: . $i,\;\;\frac{\sqrt{3}-i}{2},\;\;\frac{-\sqrt{3}-i}{2}$

3. Originally Posted by Ashgasm
Hi all,

I'm trying to work out the roots for the equation above. I've worked out one of them as being z = i, as z^3= i^3 = -i.

Wolfram Alpha gives
z = 1/2 (sqrt(3)-i) and z = 1/2 (-sqrt(3)-i)

as the other answers, but I'm unsure how they arrived at this?

Thanks.
Since you know one of the roots, use the fact that the arguments of the roots differ by 2pi/3.

-i = cis(-pi/2). Therefore, in polar form, the other two roots are cis(-pi/2 + 2pi/3) and cis(-pi/2 + 4pi/3). Now convert each of these into cartesian form.

4. Big thanks to both of you. That certainly clears it up for me.

Rep++.