The Least Positive Integer N Such that ( 2 I 1 + I ) N is a Positive Integer, Is. - Mathematics

MCQ

The least positive integer n such that $\left( \frac{2i}{1 + i} \right)^n$ is a positive integer, is.

•  16

• 8

• 4

• 2

Solution

$8$

$\text { Let } z = \left( \frac{2i}{1 + i} \right)$

$\Rightarrow z = \frac{2i}{1 + i} \times \frac{1 - i}{1 - i}$

$\Rightarrow z = \frac{2i\left( 1 - i \right)}{1 - i^2}$

$\Rightarrow z = \frac{2i\left( 1 - i \right)}{1 + 1} \left[ \because i^2 = - 1 \right]$

$\Rightarrow z = \frac{2i\left( 1 - i \right)}{2}$

$\Rightarrow z = i - i^2$

$\Rightarrow z = i + 1$

$\text { Now }, z^n = \left( 1 + i \right)^n$

$\text { For } n = 2,$

$z^2 = \left( 1 + i \right)^2$

$= 1 + i^2 + 2i$

$= 1 - 1 + 2i$

$= 2i . . . (1)$

$\text { Since this is not a positive integer },$

$\text { For } n = 4,$

$z^4 = \left( 1 + i \right)^4$

$= \left[ \left( 1 + i \right)^2 \right]^2$

$= \left( 2i \right)^2 \left[ \text { Using } (1) \right]$

$= 4 i^2$

$= - 4 . . . (2)$

$\text { This is a negative integer }.$

$\text { For } n = 8,$

$z^8 = \left( 1 + i \right)^8$

$= \left[ \left( 1 + i \right)^4 \right]^2$

$= \left( - 4 \right)^2 \left[ \text { Using } (2) \right]$

$= 16$

$\text { This is a positive integer } .$

$\text { Thus }, z = \left( \frac{2i}{1 + i} \right)^n\text { is positive for } n = 8 .$

$\text { Therefore, 8 is the least positive integer such that } \left( \frac{2i}{1 + i} \right)^n\text { is a positive integer } .$

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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 13 Complex Numbers
Q 13 | Page 64