The correct option is A i 1+i/1 i=(1+i)^2/1^2 i^2=1 1+2i/2=i nbsp;∴ (i)^4n+1 nbsp;=i^4ni^1 nbsp;= i ...

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Conjugate of a Complex Number

1+i/1-i 4n+1=

Question

${(\frac{1 + i}{1 - i})}^{4 n + 1} =$

i

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Solution

The correct option is A $i$
$\frac{1 + i}{1 - i} = \frac{(1 + i)^{2}}{1^{2} - i^{2}} = \frac{1 - 1 + 2 i}{2} = i$
$∴ (i)^{4 n + 1}$
$= i^{4 n} i^{1}$
$= i$

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