First slide

Algebra of complex numbers

Question

If $n \in N,$ then $\frac{(1 + i)^{n}}{(1 - i)^{n - 2}}$ is equal to-

Moderate

A

$i^{n + 1}$
B

$-2 i^{n + 1}$
C

$i^{n + 2}$
D

$- 2 i^{n + 2}$

Solution

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

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