Question:
Prove that $(1-\mathrm{i})^{\mathrm{n}}\left(1-\frac{1}{\mathrm{i}}\right)^{\mathrm{n}}=2^{\mathrm{n}}$ for all values of $\mathrm{n} \mathrm{N}$
Solution:
L.H.S $=(1-\mathrm{i})^{n}\left(1-\frac{1}{i}\right)^{\mathrm{n}}$
$=(1-\mathrm{i})^{\mathrm{n}}\left(1-\mathrm{i}^{-4^{*} 1+3}\right)^{\mathrm{n}}$
$=(1-i)^{n}\left(1-i^{3}\right)^{n}$
Since, $i^{4 n+3}=-1$
$=(1-i)^{n}(1+i)^{n}$
Applying $a^{n} b^{n}=(a b)^{n}$
$=((1-i)(1+i))^{n}$
$=\left(1-\mathrm{i}^{2}\right)^{\mathrm{n}}$
$=2^{\mathrm{n}}$
L.H.S = R.H.S
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