(1+i)/(1-isqrt3)=(1+i)(1+isqrt3)/4=1/4(1-sqrt3+i(1+sqrt3))
Now we will reduce this complex number to the exponential form by making
x+iy=sqrt(x^2+y^2)e^(iphi) where phi=arctan (y/x)
so making x = 1-sqrt3 and y=1+sqrt3 we have
sqrt(x^2+y^2)=2sqrt2 and phi=arctan((1 + sqrt[3])/(1 - sqrt[3]))
now proceeding
((1+i)/(1-isqrt3))^(-2i)=(2sqrt2)^(-2i)(e^(iphi))^(-2i)=(e^lambda)^(-2i)(e^(iphi))^(-2i)=e^(-2ilambda)e^(2phi)
Here lambda is such that e^lambda=2 sqrt2 or lambda=log(2 sqrt2)
Finally, using de Moivre's identity
e^(ix)=cosx+isinx we arrive at
((1+i)/(1-isqrt3))^(-2i)=(cos(2log(2sqrt2))+isin(2log(2sqrt(2))))e^(2phi)
and the imaginary component is
sin(2log(2sqrt(2)))e^(2phi) with phi=arctan((1 + sqrt[3])/(1 - sqrt[3]))