`(1/3+3i)^3=(1/3)^3+(3i)^3+3(1/3)(1/3+3i)`

(1+3i)(1-3i)

(-2-(1)/(3)i)^(3)

(-2-i(1)/(3))^(3)

Convert the following in the form of (a+ib) : (i) (1+i)^(4) (ii) (-3+(1)/(2)i)^(3) (iii) (1-i)(3+4i) (iv) (1+i)(1+ 2i)(1+ 3i) (v) (3+5i)/(6-i) (vi) ((2+3i)^(2))/(2+i) (vii) ((1+ i)(2+i))/((3+i)) (viii) (2-i)^(-3)

Convert the following in the form of (a+ib) : (i) (1+i)^(4) (ii) (-3+(1)/(2)i)^(3) (iii) (1-i)(3+4i) (iv) (1+i)(1+ 2i)(1+ 3i) (v) (3+5i)/(6-i) (vi) ((2+3i)^(2))/(2+i) (vii) ((1+ i)(2+i))/((3+i)) (viii) (2-i)^(-3)

For i^2= -1, (1 - 3i)^3= ?

((1-i)^(3))/(1-i^(3))=-2

If frac{(1+i)^3}{(1-i)^3}-frac{(1-i)^3}{(1+i)^3} = x+iy , then