Question

Find the conjugates of the following complex number : $\frac{(1+i{)}^{2n+1}}{(1-i{)}^{2n-1}},nϵN$

Answer 1

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

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

$=2i\left(i^{2n-1}\right)$

$=2i^{2n}$

$=2(-1{)}^n$

Thus, the conjugate $=2(-1{)}^n$