[SOLVED] Solving for roots -> Z^3 = -i

Mar 2008
28
0
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.
 

Soroban

MHF Hall of Honor
May 2006
12,028
6,341
Lexington, MA (USA)
Hello, Ashgasm!

Tricky . . .


Find all roots of: .\(\displaystyle z^3 \:=\:-i\)

We have: .\(\displaystyle z^3 + i \:=\:0\)


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


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

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



We have two equations to solve:

\(\displaystyle z - i \:=\:0 \quad\Rightarrow\quad \boxed{z \:=\:i}\)


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



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


 

mr fantastic

MHF Hall of Fame
Dec 2007
16,948
6,768
Zeitgeist
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.
 
Mar 2008
28
0
Big thanks to both of you. That certainly clears it up for me.

Rep++. :)