.
#a^2+b^2=1# (equation #1#)
#(sqrt(1+x)-isqrt(1-x))/(sqrt(1+x)+isqrt(1-x))=((sqrt(1+x)-isqrt(1-x))(sqrt(1+x)+isqrt(1-x)))/((sqrt(1+x)+isqrt(1-x))(sqrt(1+x)+isqrt(1-x)))=(1+x-i^2(1-x))/(1+x+i^2(1-x)+2isqrt(1-x^2))=#
#(1+x-(-1)(1-x))/(1+x+(-1)(1-x)+2sqrt(i^2(1-x^2)))=#
#(1+x+1-x)/(1+x-1+x+2sqrt(x^2-1))=2/(2x+2sqrt(x^2-1))=1/(x+sqrt(x^2-1))=#
#(x-sqrt(x^2-1))/((x+sqrt(x^2-1))(x-sqrt(x^2-1)))=(x-sqrt(x^2-1))/(x^2-x^2+1)=#
#x-sqrt(x^2-1)=x-sqrt((-1)(1-x^2))=x-sqrt(i^2(1-x^2))=#
#x-isqrt(1-x^2)#
If we set #x=a# then from equation #1# we get:
#x^2+b^2=1#
#b^2=1-x^2#
#b=sqrt(1-x^2)#
We can plug this in to get:
#(sqrt(1+x)-isqrt(1-x))/(sqrt(1+x)+isqrt(1-x))=x-isqrt(1-x^2)=a-ib#
