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Question: If \[i=\sqrt{-1}\] then \[4+5{{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}-3{{\left[ \...

If i=−1i=\sqrt{-1} then 4+5[−12+i32]334−3[12+i32]3654+5{{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}-3{{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}is equal to
(a) 1−i31-i\sqrt{3}
(b) −1+i3-1+i\sqrt{3}
(c) 43i4\sqrt{3}i
(d) −i3-i\sqrt{3}

Explanation

Solution

To solve this problem, we should know that concept of cube roots of unity. We know that according to cube roots of unity as 1, w=−12+i32w=\dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} and w2=−12−i32{{w}^{2}}=\dfrac{-1}{2}-i\dfrac{\sqrt{3}}{2}which also satisfies w2+w+1=0{{w}^{2}}+w+1=0 and w3=1{{w}^{3}}=1. We assume the value of 4+5[−12+i32]334−3[12+i32]3654+5{{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}-3{{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}is equal to A. We then substitute w=−12+i32w=\dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} and w2=−12−i32{{w}^{2}}=\dfrac{-1}{2}-i\dfrac{\sqrt{3}}{2} in A. Now by further simplification, we will get the value of A.

Complete step-by-step answer:
Let us assume the value of 4+5[−12+i32]334−3[12+i32]3654+5{{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}-3{{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}} is equal to A.
We know that according to cube roots of unity. If w=−12+i32w=\dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} and w2=−12−i32{{w}^{2}}=\dfrac{-1}{2}-i\dfrac{\sqrt{3}}{2}, then w2+w+1=0{{w}^{2}}+w+1=0 and w3=1{{w}^{3}}=1. So, let us substitute w=−12+i32w=\dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} and w2=−12−i32{{w}^{2}}=\dfrac{-1}{2}-i\dfrac{\sqrt{3}}{2} in A.
⇒A=4+5w334−3(−w2)365\Rightarrow A=4+5{{w}^{334}}-3{{\left( -{{w}^{2}} \right)}^{365}}.
⇒A=4+5w333.w+3w730\Rightarrow A=4+5{{w}^{333}}.w+3{{w}^{730}}.
⇒A=4+5w333.w+3w729.w\Rightarrow A=4+5{{w}^{333}}.w+3{{w}^{729}}.w.
⇒A=4+5(w3)111.w+3(w3)243.w\Rightarrow A=4+5{{\left( {{w}^{3}} \right)}^{111}}.w+3{{\left( {{w}^{3}} \right)}^{243}}.w.
We know that w3=1{{w}^{3}}=1. Now we will substitute this in the value of A.
⇒A=4+5w+3w\Rightarrow A=4+5w+3w.
⇒A=4+8w\Rightarrow A=4+8w.
We know that w=−12+i32w=\dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2}. Now we will substitute this in the value of A.
⇒A=4+8(−12+i32)\Rightarrow A=4+8\left( \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right).
⇒A=4+(−4+43)\Rightarrow A=4+\left( -4+4\sqrt{3} \right).
⇒A=43i\Rightarrow A=4\sqrt{3}i.
Hence, option C is correct.

So, the correct answer is “Option C”.

Note: Alternatively, we can solve this problem as shown below:
From the above question, it is clear that we have to find the value of 4+5[−12+i32]334−3[12+i32]3654+5{{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}-3{{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}.
First of all, let us find the value of [−12+i32]334{{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}.
We know that cos⁡(π−π3)=−12\cos \left( \pi -\dfrac{\pi }{3} \right)=\dfrac{-1}{2} and sin⁡(π−π3)=32\sin \left( \pi -\dfrac{\pi }{3} \right)=\dfrac{\sqrt{3}}{2}.
⇒[−12+i32]334=[cos⁡(π−π3)+isin⁡(π−π3)]334\Rightarrow {{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}={{\left[ \cos \left( \pi -\dfrac{\pi }{3} \right)+i\sin \left( \pi -\dfrac{\pi }{3} \right) \right]}^{334}}.
We know that cisθ=cos⁡θ+isin⁡θcis\theta =\cos \theta +i\sin \theta .
⇒[−12+i32]334=[cis(π−π3)]334\Rightarrow {{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}={{\left[ cis\left( \pi -\dfrac{\pi }{3} \right) \right]}^{334}}.
We know that cisnθ=cisnθci{{s}^{n}}\theta =cisn\theta .
⇒[−12+i32]334=[cis(334(π−π3))]\Rightarrow {{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}=\left[ cis\left( 334\left( \pi -\dfrac{\pi }{3} \right) \right) \right].
⇒[−12+i32]334=[cis(334π−334π3)]\Rightarrow {{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}=\left[ cis\left( 334\pi -\dfrac{334\pi }{3} \right) \right].
We know that cisθ=cos⁡θ+isin⁡θcis\theta =\cos \theta +i\sin \theta .
⇒[−12+i32]334=[cos⁡(334π−334π3)+isin⁡(334π−334π3)]\Rightarrow {{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}=\left[ \cos \left( 334\pi -\dfrac{334\pi }{3} \right)+i\sin \left( 334\pi -\dfrac{334\pi }{3} \right) \right].
We know that cos⁡(nπ−θ)=cos⁡θ\cos \left( n\pi -\theta \right)=\cos \theta and sin⁡(nπ−θ)=−sin⁡θ\sin \left( n\pi -\theta \right)=-\sin \theta where n is even.
⇒[−12+i32]334=[cos⁡(334π3)−isin⁡(334π3)]\Rightarrow {{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}=\left[ \cos \left( \dfrac{334\pi }{3} \right)-i\sin \left( \dfrac{334\pi }{3} \right) \right].
⇒[−12+i32]334=[cos⁡(111π+π3)−isin⁡(111π+π3)]\Rightarrow {{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}=\left[ \cos \left( 111\pi +\dfrac{\pi }{3} \right)-i\sin \left( 111\pi +\dfrac{\pi }{3} \right) \right].
We know that cos⁡(nπ+θ)=−cos⁡θ\cos \left( n\pi +\theta \right)=-\cos \theta and sin⁡(nπ+θ)=−sin⁡θ\sin \left( n\pi +\theta \right)=-\sin \theta where n is odd.
⇒[−12+i32]334=[−cos⁡(π3)+isin⁡(π3)]\Rightarrow {{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}=\left[ -\cos \left( \dfrac{\pi }{3} \right)+i\sin \left( \dfrac{\pi }{3} \right) \right].
We know that cos⁡π3=12\cos \dfrac{\pi }{3}=\dfrac{1}{2} and sin⁡π3=32\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}.
⇒[−12+i32]334=[−12+i32]........(1)\Rightarrow {{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}=\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]........(1).
Now we should find the value of [12+i32]365{{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}.
We know that cos⁡(π3)=12\cos \left( \dfrac{\pi }{3} \right)=\dfrac{1}{2} and sin⁡(π3)=32\sin \left( \dfrac{\pi }{3} \right)=\dfrac{\sqrt{3}}{2}.
⇒[12+i32]365=[cos⁡(π3)+isin⁡(π3)]365\Rightarrow {{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}={{\left[ \cos \left( \dfrac{\pi }{3} \right)+i\sin \left( \dfrac{\pi }{3} \right) \right]}^{365}}.
We know that cisθ=cos⁡θ+isin⁡θcis\theta =\cos \theta +i\sin \theta .
⇒[12+i32]365=[cis(π3)]365\Rightarrow {{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}={{\left[ cis\left( \dfrac{\pi }{3} \right) \right]}^{365}}.
We know that cisnθ=cisnθci{{s}^{n}}\theta =cisn\theta .
⇒[12+i32]365=[cis(365π3)]\Rightarrow {{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}=\left[ cis\left( \dfrac{365\pi }{3} \right) \right].
⇒[12+i32]365=[cis(121π+2π3)]\Rightarrow {{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}=\left[ cis\left( 121\pi +\dfrac{2\pi }{3} \right) \right].
We know that cisθ=cos⁡θ+isin⁡θcis\theta =\cos \theta +i\sin \theta .
⇒[12+i32]365=[cos⁡(121π+2π3)+isin⁡(121π+2π3)]\Rightarrow {{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}=\left[ \cos \left( 121\pi +\dfrac{2\pi }{3} \right)+i\sin \left( 121\pi +\dfrac{2\pi }{3} \right) \right].
We know that cos⁡(nπ+θ)=−cos⁡θ\cos \left( n\pi +\theta \right)=-\cos \theta and sin⁡(nπ+θ)=−sin⁡θ\sin \left( n\pi +\theta \right)=-\sin \theta where n is odd.
⇒[12+i32]365=[−cos⁡(2π3)−isin⁡(2π3)]\Rightarrow {{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}=\left[ -\cos \left( \dfrac{2\pi }{3} \right)-i\sin \left( \dfrac{2\pi }{3} \right) \right].
⇒[−12+i32]365=−[cos⁡(2π3)+isin⁡(2π3)]\Rightarrow {{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}=-\left[ \cos \left( \dfrac{2\pi }{3} \right)+i\sin \left( \dfrac{2\pi }{3} \right) \right].
⇒[−12+i32]365=−[cos⁡(π−π3)+isin⁡(π−π3)]\Rightarrow {{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}=-\left[ \cos \left( \pi -\dfrac{\pi }{3} \right)+i\sin \left( \pi -\dfrac{\pi }{3} \right) \right].
We know that cos⁡(nπ−θ)=−cos⁡θ\cos \left( n\pi -\theta \right)=-\cos \theta and sin⁡(nπ−θ)=sin⁡θ\sin \left( n\pi -\theta \right)=\sin \theta where n is odd.
⇒[12+i32]365=−[−cos⁡(π3)+isin⁡(π3)]\Rightarrow {{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}=-\left[ -\cos \left( \dfrac{\pi }{3} \right)+i\sin \left( \dfrac{\pi }{3} \right) \right].
We know that cos⁡π3=12\cos \dfrac{\pi }{3}=\dfrac{1}{2} and sin⁡π3=32\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}.
⇒[12+i32]365=−[−12+i32]\Rightarrow {{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}=-\left[ -\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right].
⇒[12+i32]365=[12−i32]......(2)\Rightarrow {{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}=\left[ \dfrac{1}{2}-i\dfrac{\sqrt{3}}{2} \right]......(2).
Let us assume
I=4+5[−12+i32]334−3[12+i32]365......(3)I=4+5{{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}-3{{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}......(3).
Now we will substitute equation (1) and equation (2) in equation (3), then we get
⇒I=4−52+i(532)−32+i(332)\Rightarrow I=4-\dfrac{5}{2}+i\left( \dfrac{5\sqrt{3}}{2} \right)-\dfrac{3}{2}+i\left( \dfrac{3\sqrt{3}}{2} \right).
⇒I=4−52−32+i(532+332)\Rightarrow I=4-\dfrac{5}{2}-\dfrac{3}{2}+i\left( \dfrac{5\sqrt{3}}{2}+\dfrac{3\sqrt{3}}{2} \right).
⇒I=0+i(832)\Rightarrow I=0+i\left( \dfrac{8\sqrt{3}}{2} \right).
⇒I=43i......(4)\Rightarrow I=4\sqrt{3}i......(4).
From equation (4), it is clear that the value of 4+5[−12+i32]334−3[12+i32]3654+5{{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}-3{{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}} is equal to 43i4\sqrt{3}i.
Hence, Option c is correct.