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

#### Options

16

8

4

2

#### Solution

\[8\]

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

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

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

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

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

\[ \Rightarrow z = i - i^2 \]

\[ \Rightarrow z = i + 1\]

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

\[\text { For } n = 2, \]

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

\[ = 1 + i^2 + 2i\]

\[ = 1 - 1 + 2i\]

\[ = 2i . . . (1) \]

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

\[\text { For } n = 4, \]

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

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

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

\[ = 4 i^2 \]

\[ = - 4 . . . (2)\]

\[\text { This is a negative integer }. \]

\[\text { For } n = 8, \]

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

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

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

\[ = 16\]

\[\text { This is a positive integer } . \]

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

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