Question

$\frac{(1-i)}{(1+i)}$ is equal to

• A. $\cos \frac{\mathrm{\pi }}{2}+i\sin \frac{\mathrm{\pi }}{2}$
• B. $\cos \frac{\mathrm{\pi }}{2}-i\sin \frac{\mathrm{\pi }}{2}$
• C. $\sin \frac{\mathrm{\pi }}{2}+i\cos \frac{\mathrm{\pi }}{2}$
• D. $Noneofthese$

Answer 1

The correct option is B

$\cos \frac{\mathrm{\pi }}{2}-i\sin \frac{\mathrm{\pi }}{2}$

Explanation for the correct option:

Multiplying numerator and denominator by the conjugate of denominator

Let $z=\frac{(1-i)}{(1+i)}$

$\Rightarrow z=\left(\frac{1-i}{1+i}\right)\cdot \left(\frac{1-i}{1-i}\right)=\frac{1-2i+i^2}{1-i^2}$

Now $i^2=-1$

$\Rightarrow$ $z=\frac{1-2i+i^2}{1-i^2}=\frac{1-2i-1}{1-(-1)}$

$=\frac{-2i}{2}$

$=-\mathit{i}$

$\Rightarrow z=\cos \frac{\mathrm{\pi }}{2}-i\sin \frac{\mathrm{\pi }}{2}$ $\left[\mathbf{\because }\mathbf{-}\mathit{i}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{-}\mathit{i}\mathit{s}\mathit{i}\mathit{n}\frac{\mathbf{\pi }}{\mathbf{2}}\right]$

Hence, Option ‘B’ is Correct.