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Q.
If $a^{2}+b^{2}=1$, then $\frac{1+b+i a}{1+b-i a}=$
Complex Numbers and Quadratic Equations
Solution:
Given that $a^{2}+b^{2}=1$, therefore
$\frac{1+b+i a}{1+b-i a}=\frac{(1+b+i a)(1+b+i a)}{(1+b-i a)(1+b+i a)}$
$=\frac{(1+b)^{2}-a^{2}+2 i a(1+b)}{1+b^{2}+2 b+a^{2}}$
$=\frac{\left(1-a^{2}\right)+2 b+b^{2}+2 i a(1+b)}{2(1+b)}$
$=\frac{2 b^{2}+2 b+2 i a(1+b)}{2(1+b)}=b+i a$