Calculations involve complex no. HELP!!!

1. Expand (a + bi)^(1/3).

First, convent a + bi to polar form

a + bi = r((a/r) + (b/r)i) = r(cos θ + i sin θ)
Where r = √(a² + b²) , tan θ = b/a

Then, use the formula
ⁿ√[r(cos θ + i sin θ)] = ⁿ√r × [cos ((θ + 2kπ)/n) + i sin ((θ + 2kπ)/n)],
where k = 0, 1, 2, …, (n – 1).

In this case, n = 3
∴(a + bi)^(1/3)
= ³√[r(cos θ + i sin θ)]
= ³√r × [cos ((θ + 2kπ)/3) + i sin ((θ + 2kπ)/3)] , where k = 0, 1, 2
= ³√√(a² + b²) × [cos ((2kπ + tan^(– 1) (b/a))/3) + i sin (((2kπ + tan^(– 1) (b/a))/3)]
= (a² + b²)^(1/6) [cos ((2kπ + tan^(– 1) (b/a))/3) + i sin (((2kπ + tan^(– 1) (b/a))/3)] ,
where k = 0, 1, 2

2. Expand (a – bi)^(1/3).

First, convent a – bi to polar form

a – bi = r((a/r) – (b/r)i) = r(cos φ + i sin φ)
Where r = √(a² + b²) , tan φ = – b/a

Then, use the formula
ⁿ√[r(cos θ + i sin θ)] = ⁿ√r × [cos ((θ + 2kπ)/n) + i sin ((θ + 2kπ)/n)],
where k = 0, 1, 2, …, (n – 1).

In this case, n = 3
∴(a – bi)^(1/3)
= ³√[r(cos φ + i sin φ)]
= ³√r × [cos ((φ + 2kπ)/3) + i sin ((φ + 2kπ)/3)] , where k = 0, 1, 2
= ³√√(a² + b²) × [cos ((2kπ + tan^(– 1) (– b/a))/3) + i sin (((2kπ + tan^(– 1) (– b/a))/3)]
= (a² + b²)^(1/6) [cos ((2kπ – tan^(– 1) (b/a))/3) + i sin (((2kπ – tan^(– 1) (b/a))/3)] ,
where k = 0, 1, 2

3. Expand (a + bi)³.

(a + bi)³ = a³ + 3a²(bi) + 3a(bi)² + (bi)³
　　　　= a³ + 3a²bi – 3ab² – b³i
　　　　= (a³ – 3ab²) + (3a²b – b³)i


P.S. ⁿ√[r(cos θ + i sin θ)] = ⁿ√r × [cos ((θ + 2kπ)/n) + i sin ((θ + 2kπ)/n)] 的證明。

ⁿ√[r(cos θ + i sin θ)]
= ⁿ√r × ⁿ√(cos θ + i sin θ)
= ⁿ√r × ⁿ√[cos (θ + 2kπ) + i sin (θ + 2kπ)] , k belongs to any integer
= ⁿ√r × ⁿ√[cos (n[(θ + 2kπ)/n]) + i sin (n[(θ + 2kπ)/n])] , k, n belongs to any +ve integer
= ⁿ√r × ⁿ√([cos ((θ + 2kπ)/n) + i sin ((θ + 2kπ)/n)]ⁿ) (De Moivre’s Theorem)
= ⁿ√r × [cos ((θ + 2kπ)/n) + i sin ((θ + 2kπ)/n)]

In fact, it is suitable for any +ve integer k
However, when k ≧ n , the value is repeated.
So we simplify say that it is suitable for k = 0, 1, 2, …, (n – 1).

轉z=a+bi做polar form
即係z=r(cos(x)-isin(x))
r=sqrt(a+b)
x=tan^-1(b/a)

根據戴魔根定理
z^m=r^m(cos(x)-isin(x))^m
=r^m(cos(mx)-isin(mx))

你應用上面條式
唔需要expand, 但可以搵倒答案
唔明可以send email問我^^

2007-04-19 14:32:39 補充：
To be honest, converting a complex number from cartesian form to polar form is definitely different from expanding a complex number in cartesian form.That's the reason why nobody had answered your question before me.I don't understand why the other copied my answer.

2007-04-19 14:34:33 補充：
I appreciate the other person showed you the equation for solving a complex number in the form (a bi)^n. However, the questioner stated clearly that the solution is not required. So your answer may stray away from the question.Plx, I don't admire plagiarism.