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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

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

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

$z_2=-i$

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

1 Answer 1

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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.

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    $\begingroup$ You should give a full answer to the original poster. $\endgroup$
    – Angelo
    Commented Jul 23, 2021 at 21:51
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    $\begingroup$ $w=i=\cos(90^o)+i\sin(90^o)$ 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 $z_0=\dfrac{\sqrt3}2+\dfrac12i$ $z_1=-\dfrac{\sqrt3}2+\dfrac12i$ $z_2=-i$ $\endgroup$
    – Angelo
    Commented Jul 23, 2021 at 22:29
  • $\begingroup$ @Angelo Maybe, but I chose not to, since the OP seems to know the twchnique $\endgroup$
    – egreg
    Commented Jul 24, 2021 at 6:37

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