[XXXNICEGUYXXX]

The complex number z has modulus 4 and argument pi/3
Find in the form a+bi,

(a) z^2
(b) 1/z
(c) i^3z

Don't worry. this isn't on any exam. Heinemann p4 pg 34 3B question 13
thx

[ot209]

firstly,
4e^i(pi/3) = 4( cos(pi/3) + isin(pi/3) ) = 2 + i(2root3)

1. z^2 = ( 2 + i(2root3) )^2 = 4 + i(8root3) - 12 = -8 + i(8root3)

2. 1/z = 1/(2+i(2root3)) x ( 2 - i(2root3)/2 - i(2root3) )
...which comes to 1/8 - i(1/8 root3)

3. i^3z = i^(2 + i(2root3)) = i^2 x i^(i2root3)
= -1 x (i^i)^2root3 = -(e^(-pi/2))^2root3
...which has imaginary part equal to zero.