Question

$\frac{{(1+i)}^{2n+1}}{{(1-i)}^{2n-1}}nϵN$

• A. $2,n\pi$.
• B. ${(-1)}^n,$0 or $\pi$ depending upon n
• C. $2,0$
• D. $1,0$

Answer 1

The correct option is A $2,n\pi$.

$1+i=\sqrt{2}\left(\frac{1+i}{\sqrt{2}}\right)$
$=\sqrt{2}{e}^{i\frac{\pi }{4}}$.
Similarly $1-i$ $=\sqrt{2}{e}^{i\frac{-\pi }{4}}$.
Substituting in the expression we get
$\frac{e^{i\frac{(2n+1)\pi }{4}}}{e^{-i\frac{(2n-1)\pi }{4}}}.{\sqrt{2}}^{2n+1-(2n-1)}$
$=e^{i(2n+1+2n-1)\frac{\pi }{4}}.2$
$=2e^{i.4n\frac{\pi }{4}}$
$=2.e^{in\pi }$
$=2cos\left(n\pi \right)$
Hence, $2.(-1{)}^n$