maths help: Complex Equations:

[crooked i]

Find the square roots of 3-4j

Can anyone help with this?

btw, j = square root(-1)

 

[crooked i]

HOW did you manage that!

 

[crooked i]

(a^2-b^2)+a*b*j - so 3 = a^2-b^2 and -4 = a*b

EDIT: Think of it as factorising a quadratic - the roots will be of the form a+bj, so you just need to expand (a+bj)(a+bj)

thats what i thought, but why do we do that. thats what confuses me, we are sqrting yet somehow i have to factorise.

i mean yeah i could do that, and get my marks, but itll eat me till i know WHY thats what we do.

 

[crooked i]

so wait, the question isnt square root, its the roots?