If n is a positive integer then $(1 + i)^{n} + (1 - i)^{n}$ =

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

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

Open in App

Solution

The correct option is A

Putting 1 = $r c o s θ, 1$ = $r s i n θ$

i.e., $r^{2}$ = 1 + 1 = 2,r = $\sqrt{2}, θ$ = $\frac{π}{4}$

$(1 + i)^{n} + (1 - i)^{n}$

= $r^{n} [(c o s θ + i s i n θ)^{n} + (c o s θ + i s i n θ)^{n}]$

= $2 r^{n} c o s n θ$ = $2. (2)^{n / 2} c o s (\frac{n π}{4})$ = $2^{\frac{(n + 2)}{2}} c o s (\frac{n π}{4})$

Suggest Corrections

Similar questions

(1 + i)^{n} + (1 + i)^{n} = 2^{k} c o s \frac{n π}{4},

I_{n} = \int_{0}^{π / 4} t a n^{n} θ d θ

n (I_{n - 1} + I_{n + 1})

2^{n} > 1 + n \sqrt{(2 n - 1)}

i = \sqrt{- 1}

n

i^{n} + i^{n + 1} + i^{n + 2} + i^{n + 3}

{(\frac{1 + i}{1 - i})}^{4 n + 1}

Join BYJU'S Learning Program