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$\frac{(1+i)^{2016}}{(1-i)^{2014}}$ is equal to
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Solution:
1544 Upvotes
Verified Answer
The correct answer is:
$-2i$
We have,
$\begin{aligned}
& \frac{(1+i)^{2016}}{(1-i)^{2014}}=\frac{\left[(1+i)^2\right]^{1008}}{\left[(1-i)^2\right]^{1007}}=\frac{\left(1+2 i+i^2\right)^{1008}}{\left(1-2 i+i^2\right)^{1007}} \\
& =\frac{(2 i)^{1008}}{(-2 i)^{1007}}=\frac{(2 i)^{1008}}{(-1)^{1007}(2 i)^{1007}} \\
& =-(2 i)^{1008-1007}=-2 i
\end{aligned}$
$\begin{aligned}
& \frac{(1+i)^{2016}}{(1-i)^{2014}}=\frac{\left[(1+i)^2\right]^{1008}}{\left[(1-i)^2\right]^{1007}}=\frac{\left(1+2 i+i^2\right)^{1008}}{\left(1-2 i+i^2\right)^{1007}} \\
& =\frac{(2 i)^{1008}}{(-2 i)^{1007}}=\frac{(2 i)^{1008}}{(-1)^{1007}(2 i)^{1007}} \\
& =-(2 i)^{1008-1007}=-2 i
\end{aligned}$
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