Question

If $x^2+y^2=1$ then value of $\frac{1+x+iy}{1+x-iy}$ is

• A. $x-iy$
• B. $2x$
• C. $-2iy$
• D. $x+iy$

Answer 1

The correct option is D $x+iy$

Since $x^2+y^2=1$

So let $x=\cos \theta$ and $y=\sin \theta$

$\therefore \frac{1+x+iy}{1+x-iy}=\frac{(1+\cos \theta )+i\sin \theta }{(1+\cos \theta )-i\sin \theta }$

$=\frac{2{\cos }^2\frac{\theta }{2}+2i\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{\cos }^2\frac{\theta }{2}-2i\sin \frac{\theta }{2}\cos \frac{\theta }{2}}$

$=\frac{\cos \frac{\theta }{2}+i\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}-i\sin \frac{\theta }{2}}$

$=\frac{\cos \frac{\theta }{2}+i\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}-i\sin \frac{\theta }{2}}\times$ $\frac{\cos \frac{\theta }{2}+i\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}+i\sin \frac{\theta }{2}}$

$=\frac{{(\cos \frac{\theta }{2}+i\sin \frac{\theta }{2})}^2}{{\cos }^2\frac{\theta }{2}-i^2{\sin }^2\frac{\theta }{2}}$

$=\frac{{\cos }^2\frac{\theta }{2}+i^2{\sin }^2\frac{\theta }{2}+2i\cos \frac{\theta }{2}\cos \frac{\theta }{2}}{{\cos }^2\frac{\theta }{2}+{\sin }^2\frac{\theta }{2}}$

$=\frac{{\cos }^2\frac{\theta }{2}-{\sin }^2\frac{\theta }{2}+2i\cos \frac{\theta }{2}\cos \frac{\theta }{2}}{1}$

$=\cos \theta +i\sin \theta =x+iy$