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4.1 complex nubers — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHS4.1 complex nubers1 Mark
MCQ
$\sqrt i = $
  • A
    $\frac{{1 \pm i}}{{\sqrt 2 }}$
  • B
    $ \pm \frac{{1 - i}}{{\sqrt 2 }}$
  • ✓
    $ \pm \frac{{1 + i}}{{\sqrt 2 }}$
  • D
    None of these
✓

Answer

Correct option: C.
$ \pm \frac{{1 + i}}{{\sqrt 2 }}$
c
(c)$\sqrt i = {(i)^{1/2}} = {\left[ {\cos \frac{\pi }{2} + i\sin \frac{\pi }{2}} \right]^{1/2}}$
$ = {\left[ {\cos \left( {2n\pi + \frac{\pi }{2}} \right) + i\sin \left( {2n\pi + \frac{\pi }{2}} \right)} \right]^{1/2}}$(where $n \in I$)
$ = \left[ {\cos \frac{1}{2}\left( {2n\pi + \frac{\pi }{2}} \right) + i\sin \frac{1}{2}\left( {2n\pi + \frac{\pi }{2}} \right)} \right]$
(Using De Moivre's theorem)
= $[\cos \frac{{4n\pi + \pi }}{4} + i\sin \frac{{4n\pi + \pi }}{4}]$
Putting $n = 0, 1$
we get $\cos \frac{\pi }{4} + i\sin \frac{\pi }{4} = \frac{1}{{\sqrt 2 }} + i\frac{1}{{\sqrt 2 }} = \frac{{1 + i}}{{\sqrt 2 }}$
and $\cos \frac{{5\pi }}{4} + i\sin \frac{{5\pi }}{4} = - \frac{1}{{\sqrt 2 }} - i\frac{1}{{\sqrt 2 }} = - \left( {\frac{{1 + i}}{{\sqrt 2 }}} \right)$
Therefore $\sqrt i = \pm \frac{{1 + i}}{{\sqrt 2 }}$
Trick : Check by squaring the options, here $(c)$  is the square root of $i$ because on squaring$\left( { \pm \frac{{1 + i}}{{\sqrt 2 }}} \right)$,we get $i$.

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