#(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]))#
