For a positive integer n- find the value of 1- i n 1-1- i n

Given expression = (1 i)^n ( 1 1/i )^n =(1 i)^n(i 1)^n i^ n=(1 i)^n(1 i)^n. i^ n =[(1 i)^2]^n(1 i)^n. i^ n=(1+i^2 2i)^n( 1)^n i^ n [∵ i^2 = 1] =(1 1 2 ...

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Standard XII

Mathematics

Binomial Expression

For a positiv...

Question

For a positive integer n, find the value of $(1 - i)^{n} {(1 - \frac{1}{i})}^{n}$

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Solution

Given expression = $(1 - i)^{n} {(1 - \frac{1}{i})}^{n}$

$= (1 - i)^{n} (i - 1)^{n} i^{- n} = (1 - i)^{n} (1 - i)^{n} . i^{- n}$
$= [(1 - i)^{2}]^{n} (1 - i)^{n} . i^{- n} = (1 + i^{2} - 2 i)^{n} (- 1)^{n} i^{- n}$ $[∵ i^{2} = - 1]$
$= (1 - 1 - 2 i)^{n} (- 1)^{n} i^{- n} = (- 2)^{n} . i^{n} (- 1)^{n} i^{- n}$
$= (- 1)^{2 n} {.2}^{n} = 2^{n}$

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For a positive integer n, find the value of (1 − i)n(1−1i)n

Given expression = (1 − i)n(1−1i)n =(1−i)n(i−1)n i−n=(1−i)n(1−i)n. i−n =[(1−i)2]n(1−i)n. i−n=(1+i2−2i)n(−1)n i−n [∵i2 =−1] =(1−1−2i)n(−1)n i−n=(−2)n.in(−1)n i−n =(−1)2n.2n =2n

For a positive integer n, find the value of $(1 - i)^{n} {(1 - \frac{1}{i})}^{n}$