Finding Solutions to z^4 = 1-i√3

[Diode]

Hello all,

I've got a bit of a problem trying to find and plot solutions to this equation:

z^{4}=1-i\sqrt{3}

I'm ok to plot things on an Argand diagram and I know there will be 4 to find, but my sources only explain explicitly how to find the nth roots of unity. Any help would be greatly appreciated =]

 

[queenofbabes]

express the right hand side in the polar form: |r| e^(x + 2*n*pi)

then you can take roots on both sides explicity and solve from there.

 

[Diode]

Would I be right in doing this this?

z^{4}=2e^{in\frac{\pi }{3}}

z=2e^{\frac{in\frac{\pi }{3}}{4}}

with different roots being multiples of n, up to 5?

 

[queenofbabes]

If you draw it out on an Argand idagram, the argument is - tan^{-1} \sqrt{3} = -\frac{\pi}{3}

so z^4 = 2e^{(-\frac{\pi}{3} + 2n\pi)}

 

[Elucidus]

Would I be right in doing this this?

z^{4}=2e^{in\frac{\pi }{3}}

z=2e^{\frac{in\frac{\pi }{3}}{4}}

with different roots being multiples of n, up to 5?

1-i\sqrt{3} = 2 e^{i(5\pi/3)}

z^n=re^{i\theta} \Rightarrow z = \sqrt[n]{r}\; e^{i(\theta +2k\pi)/n} \text{ for }k=0,1,2, \dots, n-1

--Elucidus