Question

The least positive integer n such that (2i/1+i)^n is a positive integer is​

Answer 1

Answer:

To find n such that,

is positive Rationalizing

If n = 2

So n=2 is not the required result as it gives complex value. If n = 3

So n = 3 is not the required result as it gives complex value. If n = 4

So n = 4 is not the required result as it gives real but negative value. Now this value can square at n = 8 only, when the result will become positive. If n = 8

Hence, n=8 is the required result

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Answer 2

Answer:

the least positive integer n such that (2i/1+ i )^n is positive integer is

So (2i / 1 + i)^n is positive.

Rationalizing the denominator we get

(2i / 1 + i x 1 – i / 1 – i)^n

(2i (1 – i) / (1 – i^2)^n

(2i (1 – i) / 1 – (- 1))^n

(2i (1 – i) / 2)^n

(1 – i^2)^n

(i – i^2) ^n

(i + 1)^n

Now if n = 2

(i – i^2)^2 = 1 + i^2 + 2i

= 2i

So n = 2 is not satisfied.

Now if n = 3 we get

(i – i^2)^3 = (1 + i)^2(1 + i)

= 2i (1 + i)

= 2i – 2

So n= 3 is not satisfied.

Now if n = 4 is not satisfied as it gives negative value.

Now if n = 8 we get

So (i – i^2)^8

((i – i^2)^4)^2

(- 4)^2

= 16

Therefore n = 8 is satisfied.