Question

How to find the Im: z, when `z=((1+i)/(1-i*sqrt(3)))^(-2i)`?

Answer 1

`z=e^((7pi)/6)(cos(ln(1/2))-isin(ln(1/2)))`

Explanation:

In polar form `1+i` can be written as `sqrt2(cos(pi/4)+isin(pi/4))` or `sqrt2xxe^(ipi/4)`

and `1-isqrt3` can be written as `2(cos(-pi/3)+isin(-pi/3))` or `2xxe^(-ipi/3)`

Hence `z=((1+i)/(1-isqrt3))^(-2i)`

= `((sqrt2xxe^(ipi/4))/(2xxe^(-ipi/3)))^(-2i)`

= `(1/sqrt2xxe^(i(pi/4+pi/3)))^(-2i)`

= `(1/sqrt2)^(-2i)xxe^(-2i^2(7pi)/12)`

= `(e^(ln(1/sqrt2)))^(-2i)xxe^(2*(7pi)/12)`

= `e^((7pi)/6)e^(i(-2ln(1/sqrt2))`

= `e^((7pi)/6)(cos(-2ln(1/sqrt2))+isin(-2ln(1/sqrt2)))`

= `e^((7pi)/6)(cos(2ln(1/sqrt2))-isin(2ln(1/sqrt2)))`

= `e^((7pi)/6)(cos(ln(1/2))-isin(ln(1/2)))`