Which among the following expressions are equivalent $(\forall n \in I)$ .

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

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

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

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

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Solution

The correct option is D $\frac{2^{n}}{(1 + i)^{2 n}} + \frac{2^{n}}{(1 - i)^{2 n}}$
$(1 - i)^{2 n} = ((1 - i)^{2})^{n} = (1 - 1 - 2 i)^{n} = (- 2 i)^{n} = 2^{n} \times (- i)^{n} (1 + i)^{2 n} = ((1 + i)^{2})^{n} = (1 - 1 + 2 i)^{n} = (2 i)^{n} = 2^{n} \times (i)^{n}$

Also, $- i = \frac{- i . i}{i} = \frac{1}{i} = i^{- 1}$

Now from options,
$(A) \frac{2^{n}}{(1 - i)^{2 n}} + \frac{(1 + i)^{2 n}}{2^{n}} = \frac{1}{(- i)^{n}} + (i)^{n} = i^{n} + i^{n} = 2 i^{n}$

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

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

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

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