Q87 / 155 · WBJEE 2017 · Mathematics
The expression $\frac{(1+i)^{n}}{(1-i)^{n-2}}$ equals
- A$-i^{n+1}$
- B$i^{n+1}$
- Correct answerC$-2 i^{n+1}$
- D1
Solution
$(\mathrm{c}) \frac{(1+i)^{n}}{(1-i)^{n-2}}=\frac{(1+i)^{n}}{(1-i)^{n}(1-i)^{-2}}$
$$
\begin{aligned}
&=\left(\frac{1+i}{1-i}\right)^{n}(1-i)^{2} \\
=&\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^{n}\left(1+i^{2}-2 i\right) \\
=&\left(\frac{1+2 i+i^{2}}{1-i^{2}}\right)^{n}\left(1+i^{2}-2 i\right) \\
=&\left(\frac{1+2 i-1}{1-(-1)}\right)^{n}(1-1-2 i) \\
=&\left(\frac{2 i}{2}\right)^{n}(-2 i) \\
=& i^{n}(-2 i)=-2^{n+1}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.