Solving Complex Number Equation: Z^3 = 2+2i

[string_656]

hey again,

im have a problem with 1 of the questions I am doing.

Z^3 = 2+2i, and it asks to solve for Z.

does it want me to actually get a number for Z? or does it simply want me to write...

z = (2+2i)^(1/3)

but ^^^^ this seems way to easy

 

[Repainted]

Well... Using De Moivre's theorem, there are actually 3 solutions for that equation.

Because 2 + 2i can be expressed as 2\sqrt{2}e^i(\pi/4) = 2\sqrt{2}e^{i(\pi/4+2k\pi)}.

so Z³ = 2\sqrt{2}e^{i(\pi/4+2k\pi)}

=> Z = (2\sqrt{2}e^{i(\pi/4+2k\pi)})^(1/3) for any 3 consecutive values of k.

 

[lurflurf]

Poorly written instructions perhaps. You could write Z in a+ib form.

 

[string_656]

thanks repainted, lurflurf how could i rearange it so that its in a + ib form?

 

[rock.freak667]

thanks repainted, lurflurf how could i rearange it so that its in a + ib form?

Use Euler's formula of e^iθ=cosθ+isinθ