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string_656
Sep11-09, 01:42 AM
hey again,

im have a problem with 1 of the questions im 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
Sep11-09, 02:04 AM
Well... Using De Moivre's theorem, there are actually 3 solutions for that equation.

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

so Z3 = 2\sqrt{2}ei(\pi/4+2k\pi)

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

lurflurf
Sep11-09, 02:12 AM
Poorly written instructions perhaps. You could write Z in a+ib form.

string_656
Sep11-09, 02:17 AM
thanks repainted, lurflurf how could i rearange it so that its in a + ib form?

rock.freak667
Sep11-09, 09:55 AM
thanks repainted, lurflurf how could i rearange it so that its in a + ib form?

Use Euler's formula of eiθ=cosθ+isinθ