Question

If $|u|=|v|=1,uv\ne -1,$ and $z=\frac{u-v}{1+uv}$, then

• A. $|z|=1$
• B. $Re\left(z\right)=0$
• C. $Im\left(z\right)=0$
• D. $Re\left(z\right)=Im\left(z\right)$

Answer 1

The correct option is B $Re\left(z\right)=0$

Given that $|u|=|v|=1$
Let $u=\cos A+i\sin A$ and $u=\cos A+i\sin Av=\cos B+i\sin B$
$z=\frac{u-v}{1+uv}$
$\Rightarrow z=\frac{\cos A+i\sin A-\cos B-i\sin B}{1+(\cos A+i\sin A)(\cos B+i\sin B)}$
$\Rightarrow z=\frac{\cos A-\cos B+i(\sin A-\sin B)}{1+\cos (A+B)+i\sin (A+B)}$
$\Rightarrow z=\frac{-2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)+2\cos \left(\frac{A+B}{2}\right)i\sin \left(\frac{A-B}{2}\right)}{2{\cos }^2\left(\frac{A+B}{2}\right)+2i\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A+B}{2}\right)}$
$\Rightarrow z=\frac{i\sin \left(\frac{A-B}{2}\right)}{\cos \left(\frac{A+B}{2}\right)}\left[\frac{i\sin \left(\frac{A+B}{2}\right)+\cos \left(\frac{A+B}{2}\right)}{\cos \left(\frac{A+B}{2}\right)+i\sin \left(\frac{A+B}{2}\right)}\right]$
$\Rightarrow z=\frac{i\sin \left(\frac{A-B}{2}\right)}{\cos \left(\frac{A+B}{2}\right)}$
Therefore, $Re\left(z\right)=0$
Ans: B