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}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

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

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

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

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

Suggest Corrections

Similar questions

{(- x)}^{\frac{1}{5}}

f (x) = sin (\frac{π}{6} sin (\frac{π}{2} sin x)), g (x) = \frac{π}{2} sin x \forall x \in R

(\exists x) (\forall y) [(a (x, y) \land (b (x, y)) \land ⇁ c (x, y)]

\frac{1}{2} y^{2} (6 x + 2 y + 12 x - 2 y)

f : R \to R, f (x^{2} + x + 3) + 2 f (x^{2} - 3 x + 5) = 6 x^{2} - 10 x + 17 \forall x \in R

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program