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Question: If \(z = \dfrac{{\sqrt 3 }}{2} + \dfrac{i}{2}(i = \sqrt { - 1} )\), then \({(1 + iz + {z^5} + i{z^8}...

If z=32+i2(i=−1)z = \dfrac{{\sqrt 3 }}{2} + \dfrac{i}{2}(i = \sqrt { - 1} ), then (1+iz+z5+iz8)9{(1 + iz + {z^5} + i{z^8})^9}is equal to
A. -1
B. 1
C. 0
D. (−1+2i)9{( - 1 + 2i)^9}

Explanation

Solution

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=32+12iz = \dfrac{{\sqrt 3 }}{2} + \dfrac{1}{2}i
Polar form of z=cos⁡π6+isin⁡π6=eiπ6z = \cos \dfrac{\pi }{6} + i\sin \dfrac{\pi }{6} = {e^{i\dfrac{\pi }{6}}}
(1+iz+z5+iz8)9=[1+ieiπ6+(eiπ6)5+i(eiπ6)8]9 =[1+ieiπ6+ei5π6+iei8π6]9............(1)   {(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⁡π2+isin⁡π2=eiπ2i = \cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2} = {e^{i\dfrac{\pi }{2}}}
Put this value of i in (1)

=[1+eiπ2×eiπ6+ei5π6+eiπ2×ei8π6]9 =[1+ei(π2+π6)+ei5π6+ei(π2+8π6)]9 =[1+ei2π3+ei5π6+ei11π6]9 =[1+(cos⁡2π3+isin⁡2π3)+(cos⁡5π6+isin⁡5π6)+(cos⁡11π6+isin⁡11π6)]9 =[1−12+i32−32+i12+32−i2]9 =(12+i32)9 =(cos⁡π3+isin⁡π3)9 For any natural number n; (cos⁡x+isin⁡x)n=(cos⁡nx+isin⁡nx) ⇒(cos⁡π3+isin⁡π3)9=(cos⁡9π3+isin⁡9π3) =cos⁡3π+isin⁡3π =−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⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθ{(\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.