# Calculate the cubic root of $\frac{-1-i}{-1+i}$

I need to calculate the cubic root of this complex number $$\frac{-1-i}{-1+i}$$ for later plotting, so:

I have to occupy the following formula $$z_k = \sqrt[n]{|w|} (\cos(\frac{\alpha+(360°)k}{n})+i\sin(\frac{\alpha+(360°)k}{n}))$$ where $$w = \frac{-1-i}{-1+i}$$ and $$n = 3$$.

But first i need to calculate $$|w|$$ and $$\alpha$$ where $$\alpha = \tan^{-1}(\frac{Im(w)}{Re(w)})$$

My question is how I have to represent w. I mean, do I have to calculate $$\frac{-1-i}{-1+i}$$ first and then do the other calculations?

My attempt :

$$w=i=\cos(90^{\rm{o}})+i\sin(90^{\rm{o}})$$

The three cubic roots of $$\;i\;$$ are

$$z_k=\sqrt[3]{|i|}\left[\cos\left(\dfrac{90^{\rm{o}}+360^{\rm{o}}k}3\right)+i\sin\left(\dfrac{90^{\rm{o}}+360^{\rm{o}}k}3\right)\right]$$

where $$\;k\in\big\{0,1,2\big\}\;.$$

Hence ,

$$z_0=\cos(30^{\rm{o}})+i\sin(30^{\rm{o}})=\dfrac{\sqrt3}2+\dfrac12 i$$

$$z_1=\cos(150^{\rm{o}})+i\sin(150^{\rm{o}})=-\dfrac{\sqrt3}2+\dfrac12i$$

$$z_2=\cos(270^{\rm{o}})+i\sin(270^{\rm{o}})=-i$$

Consequently , the three cubic roots of $$\;i\;$$ are

$$z_0=\dfrac{\sqrt3}2+\dfrac12i$$

$$z_1=-\dfrac{\sqrt3}2+\dfrac12i$$

$$z_2=-i$$

• Hint: First simplify $w$ by using $\frac{a+ib}{c+id}=\frac{(a+ib)(c-id)}{(c+id)(c-id)}$ Commented Jul 23, 2021 at 20:27
• I would calculate $\frac{-1-i}{-1+i}$ first. Commented Jul 23, 2021 at 20:28
• Why the cubic root for a complex number? Commented Jul 23, 2021 at 20:31
• That may be a problem with the language :) @Bernard Commented Jul 23, 2021 at 20:42

First reduce the number to a simpler form: $$w=\frac{-1-i}{-1+i}=\frac{1+i}{1-i}=\frac{(1+i)^2}{1+1}=i=e^{i\pi/2}$$ Now it should be easier.
• $w=i=\cos(90^o)+i\sin(90^o)$ The three cubic roots of $\;i\;$ are $z_k=\sqrt[3]{|i|}\left[\cos\left(\dfrac{90^{\rm{o}}+360^{\rm{o}}k}3\right)+i\sin\left(\dfrac{90^{\rm{o}}+360^{\rm{o}}k}3\right)\right]$ where $\;k\in\big\{0,1,2\big\}\;.$ Hence , $z_0=\cos(30^{\rm{o}})+i\sin(30^{\rm{o}})=\dfrac{\sqrt3}2+\dfrac12 i$ $z_1=\cos(150^{\rm{o}})+i\sin(150^{\rm{o}})=-\dfrac{\sqrt3}2+\dfrac12i$ $z_2=\cos(270^{\rm{o}})+i\sin(270^{\rm{o}})=-i$ Consequently , the three cubic roots of $\;i\;$ are $z_0=\dfrac{\sqrt3}2+\dfrac12i$ $z_1=-\dfrac{\sqrt3}2+\dfrac12i$ $z_2=-i$ Commented Jul 23, 2021 at 22:29