Emergency Need Help (1 Viewer)

[FroZenWaffleS]

How do you find the square root of:

-2(squareroot)3 - 2i

And to make sure, how do you find the cube roots of:

27(1/2 + (squareroot)3 i/ 2)

 

[Riviet]

To find the square root of any complex number, let (a+ib)²=your complex number
then expand the LHS and equate real and imaginary parts to get two equations and solve them simultaneously. After you find a and b, substitute into a+ib to get your 2 roots.

For cube roots or higher powers, let z=rcis@ be a root,
then z⁵=r⁵cis5@=your complex number
equate moduli and argument to find it. Sorry that it's very brief. But it's for your urgent reply.

 

[FroZenWaffleS]

does that always work?

could you perhaps elaborate at bit on what you did with the cube roots?

 

[klaw]

does that always work?

could you perhaps elaborate at bit on what you did with the cube roots?

Demoivre's theorem.
(rCis@)^n=r^ncis(n@)

 

[Riviet]

Sorry, i got confused with something else... try this instead. Works for all complex numbers.
To find the n roots of a+ib we write:
a+ib=r(cos@+isin@)
.: the nth roots are given by:
z=r^1/ncis[(@+2kpi)/n], where k=1,2,3,...,(n-1) e.g for cube roots it's for k=0,1,2 since there are three roots.
Substitute the value of k and n into z to obtain your complex roots.
That should help. I can answer any quick questions if you don't understand any of the above.

 

[FroZenWaffleS]

Why dont someone do the question for me so i fully understand how to do it?

It will be very appreciated.

EDIT: actually dont worry about, i get what you mean now

Thanks for the help

 

[Riviet]

My pleasure.