Question

If $\left|z\right|=1$ then the value of $\frac{1+z}{1+\overline{z}}$ is

• A. $z$
• B. $\overline{z}$
• C. $z+\overline{z}$
• D. None of these

Answer 1

Answer: (A)

$z$

Let $z=x+iy$

and $\left|z\right|=1$

$\sqrt{x^2+y^2}=1$

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

$=\frac{1+z}{1+\overline{z}}$

$=\frac{1+x+iy}{1+x-iy}$

$=\frac{(1+x)+iy}{(1+x)-iy}$ $\times \frac{(1+x)+iy}{(1+x)+iy}$

$=\frac{\left(\right(1+x)+iy{)}^2}{(1+x{)}^2-(iy{)}^2}$

$=\frac{(1+x{)}^2+(iy{)}^2+2(1+x)\left(iy\right)}{1+x^2+2x+y^2}$

$=\frac{(1+x{)}^2-y^2+2(1+x\left)\right(iy)}{1+x^2+y^2+2x}$

from eq. 1

$=\frac{(1+x{)}^2-(1-x^2)+2(1+x\left)\right(iy)}{1+1+2x}$

$=\frac{(1+x{)}^2-(1-x\left)\right(1+x)+2(1+x\left)\right(iy)}{2+2x}$

$=\frac{(1+x)\left[\right(1+x)-(1-x)+2(iy\left)\right]}{2(1+x)}$

$=\frac{\left[\right(1+x)-(1-x)+2(iy\left)\right]}{2}$

$=\frac{[1+x-1+x+2(iy\left)\right]}{2}$

$=\frac{(2x+2(iy\left)\right)}{2}$

$=x+iy$

$=z$