Hello,zabije cie!
Are you allowed to use DeMoivre's Theorem?
Find the square roots of: .Z .= .-½ - (½√3)i
The polar form is: .Z .= .cos(4π/3) + i·sin(4π/3)
Then: .Z^½ .= .±[cos(2π/3) + i·sin(2π/3)] .= .±[-½ + (½√3)i]
could somone help me do this math problem, i've got all the information i think i need, but i cannot get the right answer. its for a math make-up, would really appriciate your help. thanks.
Info:
A=rcosΘ
B=rsinΘ
Z=A+Bi = (rcosΘ+isinΘ)
tanΘ=B/A
r=sqrt((A^2)+(B^2))
Problem: Find all the square roots of the complex number Z= -1/2 - (sqrt3)/2. You can use polar coordinates but convert them back to retangular form for the answer.
Hello again, zabije cie!
DeMoivre's Theorem says: .(cosθ + i·sinθ)^n .= .cos(nθ) + i·sin(nθ)
In baby-talk: the exponent "jumps in and joins the angle".
We have: θ = 4π/3 and n = ½
[cos(4π/3) + i·sin(4π/3)]^½ .= .cos(½·4π/3) + i·sin(½·4π/3) .= .cos(2π/3) + i·sin(2π/3)
And I do have two square roots . . . did you see the ± ?