Question

If $z = \dfrac{{\sqrt 3 }}{2} + \dfrac{i}{2}(i = \sqrt { - 1} )$, then ${(1 + iz + {z^5} + i{z^8})^9}$is equal to
A. -1
B. 1
C. 0
D. ${( - 1 + 2i)^9}$

Answer 1

Hint-In this equation, complex number z is given to us. Convert this complex number to its polar form and then use this form of z to proceed and obtain the answer.

Complete step-by-step answer:
Given; $z = \dfrac{{\sqrt 3 }}{2} + \dfrac{1}{2}i$
Polar form of $z = \cos \dfrac{\pi }{6} + i\sin \dfrac{\pi }{6} = {e^{i\dfrac{\pi }{6}}}$
$
  {(1 + iz + {z^5} + i{z^8})^9} = {\left[ {1 + i{e^{i\dfrac{\pi }{6}}} + {{\left( {{e^{i\dfrac{\pi }{6}}}} \right)}^5} + i{{\left( {{e^{i\dfrac{\pi }{6}}}} \right)}^8}} \right]^9} \\
   = {\left[ {1 + i{e^{i\dfrac{\pi }{6}}} + {e^{i\dfrac{{5\pi }}{6}}} + i{e^{i\dfrac{{8\pi }}{6}}}} \right]^9}............(1) \\
    \\
$
Also $i = \cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2} = {e^{i\dfrac{\pi }{2}}}$
Put this value of i in (1)
\[
   = {\left[ {1 + {e^{i\dfrac{\pi }{2}}} \times {e^{i\dfrac{\pi }{6}}} + {e^{i\dfrac{{5\pi }}{6}}} + {e^{i\dfrac{\pi }{2}}} \times {e^{i\dfrac{{8\pi }}{6}}}} \right]^9} \\
   = {\left[ {1 + {e^{i\left( {\dfrac{\pi }{2} + \dfrac{\pi }{6}} \right)}} + {e^{i\dfrac{{5\pi }}{6}}} + {e^{i\left( {\dfrac{\pi }{2} + \dfrac{{8\pi }}{6}} \right)}}} \right]^9} \\
   = {\left[ {1 + {e^{i\dfrac{{2\pi }}{3}}} + {e^{i\dfrac{{5\pi }}{6}}} + {e^{i\dfrac{{11\pi }}{6}}}} \right]^9} \\
   = {\left[ {1 + \left( {\cos \dfrac{{2\pi }}{3} + i\sin \dfrac{{2\pi }}{3}} \right) + \left( {\cos \dfrac{{5\pi }}{6} + i\sin \dfrac{{5\pi }}{6}} \right) + \left( {\cos \dfrac{{11\pi }}{6} + i\sin \dfrac{{11\pi }}{6}} \right)} \right]^9} \\
   = {\left[ {1 - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2} - \dfrac{{\sqrt 3 }}{2} + i\dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2} - \dfrac{i}{2}} \right]^9} \\
   = {\left( {\dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^9} \\
   = {\left( {\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}} \right)^9} \\
  {\text{For any natural number n;}} \\
  {\left( {\cos x + i\sin x} \right)^n} = \left( {\cos nx + i\sin nx} \right) \\
   \Rightarrow {\left( {\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}} \right)^9} = \left( {\cos \dfrac{{9\pi }}{3} + i\sin \dfrac{{9\pi }}{3}} \right) \\
   = \cos 3\pi + i\sin 3\pi \\
   = - 1 \\
\]
Note-For these types of questions, the key concept is to apply the De Moivre’s theorem i.e. ${(\cos \theta + i\sin \theta )^n} = \cos n\theta + i\sin n\theta $ .Always remember, this theorem can be used most often in the problem of complex numbers.