The least positive integers n such that 2in1-in-2- i-1- is a positive integer- is

(2i)^n(1 i)^n 2= ( 2e^iπ /2 )^n ( √(e)^ iπ /4 )^n 2 = ( nbsp;√(2) )^n+2e^i(3n 2)π/4 For positive integer n should be atleast 6 = ( √(2) )^8 e^i.4π= ( √(2) )^ ...

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Standard XII

Mathematics

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The least pos...

Question

The least positive integers $n$ such that $\frac{(2 i)^{n}}{(1 - i)^{n - 2}}, i = \sqrt{- 1},$ is a positive integer, is

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Solution

$\frac{(2 i)^{n}}{(1 - i)^{n - 2}} = \frac{{(2 e^{i π / 2})}^{n}}{{({\sqrt{e}}^{- i π / 4})}^{n - 2}}$
$= {(\sqrt{2})}^{n + 2} e^{i (3 n - 2) \frac{π}{4}}$
For positive integer $n$ should be atleast $6$
$= {(\sqrt{2})}^{8} e^{i .4 π} = {(\sqrt{2})}^{8} = 16$

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The least positive integers n such that (2i)n(1−i)n−2, i=√−1, is a positive integer, is

(2i)n(1−i)n−2=(2eiπ/2)n(√e−iπ/4)n−2 =(√2)n+2ei(3n−2)π4 For positive integer n should be atleast 6 =(√2)8 ei.4π=(√2)8=16

The least positive integers $n$ such that $\frac{(2 i)^{n}}{(1 - i)^{n - 2}}, i = \sqrt{- 1},$ is a positive integer, is

$\frac{(2 i)^{n}}{(1 - i)^{n - 2}} = \frac{{(2 e^{i π / 2})}^{n}}{{({\sqrt{e}}^{- i π / 4})}^{n - 2}}$
$= {(\sqrt{2})}^{n + 2} e^{i (3 n - 2) \frac{π}{4}}$
For positive integer $n$ should be atleast $6$
$= {(\sqrt{2})}^{8} e^{i .4 π} = {(\sqrt{2})}^{8} = 16$