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Question
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
\[8\]
\[\text { Let } z = \left( \frac{2i}{1 + i} \right)\]
\[ \Rightarrow z = \frac{2i}{1 + i} \times \frac{1 - i}{1 - i}\]
\[ \Rightarrow z = \frac{2i\left( 1 - i \right)}{1 - i^2}\]
\[ \Rightarrow z = \frac{2i\left( 1 - i \right)}{1 + 1} \left[ \because i^2 = - 1 \right]\]
\[ \Rightarrow z = \frac{2i\left( 1 - i \right)}{2}\]
\[ \Rightarrow z = i - i^2 \]
\[ \Rightarrow z = i + 1\]
\[\text { Now }, z^n = \left( 1 + i \right)^n \]
\[\text { For } n = 2, \]
\[ z^2 = \left( 1 + i \right)^2 \]
\[ = 1 + i^2 + 2i\]
\[ = 1 - 1 + 2i\]
\[ = 2i . . . (1) \]
\[\text { Since this is not a positive integer }, \]
\[\text { For } n = 4, \]
\[ z^4 = \left( 1 + i \right)^4 \]
\[ = \left[ \left( 1 + i \right)^2 \right]^2 \]
\[ = \left( 2i \right)^2 \left[ \text { Using } (1) \right] \]
\[ = 4 i^2 \]
\[ = - 4 . . . (2)\]
\[\text { This is a negative integer }. \]
\[\text { For } n = 8, \]
\[ z^8 = \left( 1 + i \right)^8 \]
\[ = \left[ \left( 1 + i \right)^4 \right]^2 \]
\[ = \left( - 4 \right)^2 \left[ \text { Using } (2) \right]\]
\[ = 16\]
\[\text { This is a positive integer } . \]
\[\text { Thus }, z = \left( \frac{2i}{1 + i} \right)^n\text { is positive for } n = 8 . \]
\[\text { Therefore, 8 is the least positive integer such that } \left( \frac{2i}{1 + i} \right)^n\text { is a positive integer } .\]
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