View Full Version : complex numbers (2)
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
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