Square roots of complex number

Question: Find the square roots of 8root3i-8

z= 16cos(pi/3+isinpi/3)
z^1/2 = 4cos(pi/6+ isinpi/6)
z^1/2 = +- 4( root3/2 + 1/2i)
z^1/2 = +-2root+2i?
Is that right.

Btw is this using de moivre's theorem because it isn't mentioned till a later chapter in the book.

cheers

Reply 1

Monsor

2

Original post by Super199

Question: Find the square roots of 8root3i-8

z= 16cos(pi/3+isinpi/3)
z^1/2 = 4cos(pi/6+ isinpi/6)
z^1/2 = +- 4( root3/2 + 1/2i)
z^1/2 = +-2root+2i?
Is that right.

Btw is this using de moivre's theorem because it isn't mentioned till a later chapter in the book.

cheers

Bro you're meant to use Charles Darwin's theory about evolution

Posted from TSR Mobile

Reply 2

B_9710

18

It is De Movire's theorem. Sometimes you may find it quicker and more convenient to express complex numbers in exponential form,

e^{i\theta}

. But yeah everything you've done is fine.

Reply 3

Super199

OP

20

Original post by Monsor

Bro you're meant to use Charles Darwin's theory about evolution

Posted from TSR Mobile

snm

Reply 4

Zacken

22

I don't think that's quite right (at a glance can't check for sure) and it is using DeMoivre's. You might want to haveanother go using (a+ib)^2 = -8 + 8rt(3)i and gettig two simultaneous equations by comparing real and imaginary.

Reply 5

Super199

OP

20

Original post by Zacken

I don't think that's quite right (at a glance can't check for sure) and it is using DeMoivre's. You might want to haveanother go using (a+ib)^2 = -8 + 8rt(3)i and gettig two simultaneous equations by comparing real and imaginary.

Yh I would have lol but I don't think you can in FP3

Reply 6

Zacken

22

Original post by Super199

Yh I would have lol but I don't think you can in FP3

You certainly can.

Reply 7

Super199

OP

20

Original post by Zacken

You certainly can.

But obvs for higher powers its better to use de moivre's theorem?

Reply 8

B_9710

18

Sorry, I didn't actually check your answer for some reason but as Zacken thinks there is a mistake, you can bet there must be.

Reply 9

Zacken

22

Original post by Super199

But obvs for higher powers its better to use de moivre's theorem?

Any power higher than 2 is almost impossible using the simultaneous method and DeMoivre is a necessity, yes. Anywho, I'm not sure your answer is corrct. So check it using the other method I suggested.