Calculation about complex no (a+bi)^1/3......

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).

For solving (a+bi)^1/3,
Let (c+di)^3 = (a+bi)
　c^3 + 3c^2 (di) + 3c (di)^2 + (di)^3 = (a+bi)
　 　　c^3 + 3c^2 di - 3c d^2 - d^3 i = (a+bi)
　 　(c^3 - 3c d^2) + (3c^2 d - d^3)i = (a+bi)
By comparing the real and imaginary parts,
　{ c^3 - 3c d^2 = a
　{ 3c^2 d - d^3 = b

Similarly for solving (a-bi)^1/3.

Besides, (a+bi)^3 = a^3 + 3a^2 (bi) + 3a (bi)^2 + (bi)^3
　　　　　　　　= a^3 + 3a^2 bi - 3a b^2 - b^3 i
　　　　　　　　= (a^3 - 3a b^2) + (3a^2 b - b^3)i //