Question

Show that [√3/2+i/2]³=i​

Answer 1

We have to show that, $(\dfrac{\sqrt{3} }{2} +\dfrac{i}{2} ) ^3=i.$

Solution:

L.H.S. = $(\dfrac{\sqrt{3} }{2} +\dfrac{i}{2} ) ^3$

= $(\dfrac{\sqrt{3}+i }{2} ) ^3$

= $\dfrac{(\sqrt{3})^3+i^3+3\sqrt{3}^2i+3\sqrt{3}i^2 }{8}$

= $\dfrac{3\sqrt{3}-i+3(3)i-3\sqrt{3} }{8}$

= $\dfrac{-i+9i }{8}$

= $\dfrac{8i }{8}$

= i = R.H.S., proved.

Hence, $(\dfrac{\sqrt{3} }{2} +\dfrac{i}{2} ) ^3=i$ proved.

Answer 2

SOLUTION

TO PROVE

$\displaystyle \sf{ { \bigg( \frac{ \sqrt{3} }{2} + \frac{i}{2} \bigg)}^{3} }$

FORMULA TO BE IMPLEMENTED

$\displaystyle \sf{ {( \cos \theta + i \sin \theta)}^{n} = \cos n\theta + i \sin n\theta}$

EVALUATION

$\displaystyle \sf{ { \bigg( \frac{ \sqrt{3} }{2} + \frac{i}{2} \bigg)}^{3} }$

$\displaystyle \sf{ = { \bigg( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \bigg)}^{3} }$

$\displaystyle \sf{ = { \bigg( \cos \frac{3\pi}{6} + i \sin \frac{3\pi}{6} \bigg)} }$

$\displaystyle \sf{ = { \bigg( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \bigg)} }$

$\displaystyle \sf{ =0 + i}$

$\displaystyle \sf{ = i }$

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