Further Complex Numbers 2

a) Modulus = √[(√6)2 + (√2)2] = 2√2

Argument = tan-1 (√2/√6) = π/6

b) z3/4 = √6 + √2 i = 2√2 [cos (π/6) + i sin (π/6)]

= 2√2 eiπ/6

z = (2√2 eiπ/6)4/3

= (23/2eiπ/6)4/3

= 4 ei2π/9

2012-08-09 08:51:46 補充：
If there are 3 roots then we should solve it as:

z^(3/4) = 2√2 e^(iπ/6)

z^3 = 64 e^(i2π/3)

z = 4 e^[i(2π/9 + 2kπ/3)] where k = 0, 1 or 2

Hence:

z = 4 e^(i2π/9), 4 e^(i8π/9) or 4 e^(i14π/9)