The smallest positive integer n for which 1- i 2 n -1- i 2 n isA- 4B- 8C- 12D- 2

We have, (1+i)^2n=(1 i)^2n ⇒ ( 1+i1 i )^2n = 1 ⇒ ( (1+i)(1+i)1^2 i^2 )^2n = 1 ⇒ ( 1+i^2+2i2 )^2n = 1 ⇒ ( 2i2 )^2n = 1 ⇒ i^2n=1 For smallest positive integer i^ ...

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Standard XII

Mathematics

Complex Numbers

The smallest ...

Question

The smallest positive integer $n$ for which $(1 + i)^{2 n} = (1 - i)^{2 n}$ is

4

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2

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12

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Solution

The correct option is C $2$
We have,
$(1 + i)^{2 n} = (1 - i)^{2 n}$
$\Rightarrow {(\frac{1 + i}{1 - i})}^{2 n} = 1$
$\Rightarrow {(\frac{(1 + i) (1 + i)}{1^{2} - i^{2}})}^{2 n} = 1$
$\Rightarrow {(\frac{1 + i^{2} + 2 i}{2})}^{2 n} = 1$
$\Rightarrow {(\frac{2 i}{2})}^{2 n} = 1$
$\Rightarrow i^{2 n} = 1$

For smallest positive integer $i^{4} = 1$
$\Rightarrow n = 2$

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The correct option is C 2We have, (1+i)2n=(1−i)2n ⇒(1+i1−i)2n=1 ⇒((1+i)(1+i)12−i2)2n=1 ⇒(1+i2+2i2)2n=1 ⇒(2i2)2n=1 ⇒i2n=1 For smallest positive integer i4=1 ⇒n=2

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