If n is a positive integer than show that
${(1 + i)}^{2 n} + {(1 - i)}^{2 n} = 2^{(n + 1)} c o s (n π / 2)$

Open in App

Solution

$(1 + i) = \sqrt{2} (c o s \frac{π}{4} + s i n \frac{π}{4})$

$1 - i = \sqrt{2} (c o s \frac{π}{4} - s i n \frac{π}{4})$

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

$(1 - i)^{2 n} = (\sqrt{2})^{2 n} [c o s (\frac{2 n π}{4}) - i s i n (\frac{2 n π}{4})]$

Adding $(1 + i)^{2 n} + (1 - i)^{2 n} = (\sqrt{2})^{2 n} [2 c o s (\frac{2 n π}{4})]$

$= 2^{\frac{2 n}{n}} {.2}^{1} c o s (\frac{n π}{2})$

$= 2^{n + 1} c o s (\frac{2 n π}{2})$

Suggest Corrections

Similar questions

n \in I

n

{(1 + i)}^{2 n} = {(1 - i)}^{2 n}

What is the smallest positive integer n for which $(1 + i)^{2 n} = (1 - i)^{2 n} ?$

n

(1 + i)^{2 n} = (1 - i)^{2 n}

n

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

Join BYJU'S Learning Program