Question
If $$i = \sqrt { - 1} $$ then $$4 + 5{\left( { - \frac{1}{2} + i\frac{{\sqrt 3 }}{2}} \right)^{334}} - 3{\left( {\frac{1}{2} + i\frac{{\sqrt 3 }}{2}} \right)^{365}}$$ is equal to
A.
$$1 - i\sqrt 3 $$
B.
$$ - 1 + i\sqrt 3 $$
C.
$$\sqrt {3}i $$
D.
$$ - i\sqrt 3 $$
Answer :
$$\sqrt {3}i $$
Solution :
If in a complex number $$a + ib,$$ the ratio $$a : b$$ is $$1:\sqrt 3 ,$$ then it always convert the complex number in $$\omega .$$
$$\eqalign{
& {\text{Since, }}\omega = - \frac{1}{2} + \frac{{\sqrt 3 }}{2}i \cr
& \therefore \,4 + 5{\left( { - \frac{1}{2} + \frac{{i\sqrt 3 }}{2}} \right)^{334}} + 3{\left( { - \frac{1}{2} + \frac{{i\sqrt 3 }}{2}} \right)^{365}} \cr
& = 4 + 5{\omega ^{334}} + 3{\omega ^{365}} \cr
& = 4 + 5 \cdot {\left( {{\omega ^3}} \right)^{111}} \cdot \omega + 3 \cdot {\left( {{\omega ^3}} \right)^{121}} \cdot {\omega ^2} \cr
& = 4 + 5\omega + 3{\omega ^2}\,\,\,\,\,\,\,\,\,\left( {\because \,{\omega ^3} = 1} \right) \cr
& = 1 + 3 + 2\omega + 3\omega + 3{\omega ^2} \cr
& = 1 + 2\omega + 3\left( {1 + \omega + {\omega ^2}} \right) \cr
& = 1 + 2\omega + 3 \times 0\,\,\,\,\,\,\,\,\,\,\,\,\left( {\because \,1 + \omega + {\omega ^2} = 0} \right) \cr
& = 1 + \left( { - 1 + \sqrt 3 i} \right) \cr
& = \sqrt 3 i \cr} $$