Question

What is $\frac{(1+i{)}^{4n+5}}{(1-i{)}^{4n+3}}$ equal to, where n is a natural number and $i=\sqrt{-1}$?

• A. $2$
• B. $2i$
• C. $-2i$
• D. $i$

Answer 1

The correct option is A $2$

Multiplying with ${(1+i)}^{4n+3}$ in the both numerator and denominator, we get

$\Rightarrow \frac{{(1+i)}^{4n+5}}{{(1-i)}^{4n+3}}\times \frac{{(1+i)}^{4n+3}}{{(1+i)}^{4n+3}}$

$\Rightarrow \frac{{(1+i)}^{(4n+3)+(4n+5)}}{{\left(\right(1-i\left)\right(1+i\left)\right)}^{4n+3}}$

$\Rightarrow \frac{{(1+i)}^{(8n+8)}}{(1-i^2{)}^{4n+3}}$

$\Rightarrow \frac{{{\left(\right(1+i)}^2)}^{(4n+4)}}{(1+1{)}^{4n+3}}$

$\Rightarrow \frac{{(1^2+i^2+2i)}^{(4n+4)}}{2^{4n+3}}$

$\Rightarrow \frac{{(1-1+2i)}^{(4n+4)}}{2^{4n+3}}$

$\Rightarrow \frac{2^{4n+4}\times i^{4n+4}}{2^{4n+3}}$

$\Rightarrow 2\times {\left(i^4\right)}^{n+1}$ $(\therefore i^4=1)$

$\Rightarrow 2\times 1=2$