Question

If $\frac{(1-i{)}^{2n}}{(1+i{)}^{2n+1}}$ is real number, then

• A. $n=1$
• B. $n=2$
• C. $novalueofn$
• D. $n=3$

Answer 1

The correct option is C $novalueofn$

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

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

Thus the given number is real for no value of n.