Question

If \[i=\sqrt{-1}\] then \[4+5{{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}-3{{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}\]is equal to
(a) $1-i\sqrt{3}$
(b) $-1+i\sqrt{3}$
(c) $4\sqrt{3}i$
(d) $-i\sqrt{3}$

Answer 1

Hint: 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, \[w=\dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2}\] and \[{{w}^{2}}=\dfrac{-1}{2}-i\dfrac{\sqrt{3}}{2}\]which also satisfies \[{{w}^{2}}+w+1=0\] and \[{{w}^{3}}=1\]. We assume the value of \[4+5{{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}-3{{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}\]is equal to A. We then substitute \[w=\dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2}\] and \[{{w}^{2}}=\dfrac{-1}{2}-i\dfrac{\sqrt{3}}{2}\] in A. Now by further simplification, we will get the value of A.

Complete step-by-step answer:
Let us assume the value of \[4+5{{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}-3{{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}\] is equal to A.
We know that according to cube roots of unity. If \[w=\dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2}\] and \[{{w}^{2}}=\dfrac{-1}{2}-i\dfrac{\sqrt{3}}{2}\], then \[{{w}^{2}}+w+1=0\] and \[{{w}^{3}}=1\]. So, let us substitute \[w=\dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2}\] and \[{{w}^{2}}=\dfrac{-1}{2}-i\dfrac{\sqrt{3}}{2}\] in A.
\[\Rightarrow A=4+5{{w}^{334}}-3{{\left( -{{w}^{2}} \right)}^{365}}\].
\[\Rightarrow A=4+5{{w}^{333}}.w+3{{w}^{730}}\].
\[\Rightarrow A=4+5{{w}^{333}}.w+3{{w}^{729}}.w\].
\[\Rightarrow A=4+5{{\left( {{w}^{3}} \right)}^{111}}.w+3{{\left( {{w}^{3}} \right)}^{243}}.w\].
We know that \[{{w}^{3}}=1\]. Now we will substitute this in the value of A.
\[\Rightarrow A=4+5w+3w\].
\[\Rightarrow A=4+8w\].
We know that \[w=\dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2}\]. Now we will substitute this in the value of A.
\[\Rightarrow A=4+8\left( \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right)\].
\[\Rightarrow A=4+\left( -4+4\sqrt{3} \right)\].
\[\Rightarrow A=4\sqrt{3}i\].
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 \[4+5{{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}-3{{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}\].
First of all, let us find the value of \[{{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}\].
We know that \[\cos \left( \pi -\dfrac{\pi }{3} \right)=\dfrac{-1}{2}\] and \[\sin \left( \pi -\dfrac{\pi }{3} \right)=\dfrac{\sqrt{3}}{2}\].
\[\Rightarrow {{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}={{\left[ \cos \left( \pi -\dfrac{\pi }{3} \right)+i\sin \left( \pi -\dfrac{\pi }{3} \right) \right]}^{334}}\].
We know that \[cis\theta =\cos \theta +i\sin \theta \].
\[\Rightarrow {{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}={{\left[ cis\left( \pi -\dfrac{\pi }{3} \right) \right]}^{334}}\].
We know that \[ci{{s}^{n}}\theta =cisn\theta \].
\[\Rightarrow {{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}=\left[ cis\left( 334\left( \pi -\dfrac{\pi }{3} \right) \right) \right]\].
\[\Rightarrow {{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}=\left[ cis\left( 334\pi -\dfrac{334\pi }{3} \right) \right]\].
We know that \[cis\theta =\cos \theta +i\sin \theta \].
\[\Rightarrow {{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}=\left[ \cos \left( 334\pi -\dfrac{334\pi }{3} \right)+i\sin \left( 334\pi -\dfrac{334\pi }{3} \right) \right]\].
We know that \[\cos \left( n\pi -\theta \right)=\cos \theta \] and \[\sin \left( n\pi -\theta \right)=-\sin \theta \] where n is even.
\[\Rightarrow {{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}=\left[ \cos \left( \dfrac{334\pi }{3} \right)-i\sin \left( \dfrac{334\pi }{3} \right) \right]\].
\[\Rightarrow {{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}=\left[ \cos \left( 111\pi +\dfrac{\pi }{3} \right)-i\sin \left( 111\pi +\dfrac{\pi }{3} \right) \right]\].
We know that \[\cos \left( n\pi +\theta \right)=-\cos \theta \] and \[\sin \left( n\pi +\theta \right)=-\sin \theta \] where n is odd.
\[\Rightarrow {{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}=\left[ -\cos \left( \dfrac{\pi }{3} \right)+i\sin \left( \dfrac{\pi }{3} \right) \right]\].
We know that \[\cos \dfrac{\pi }{3}=\dfrac{1}{2}\] and \[\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}\].
\[\Rightarrow {{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}=\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]........(1)\].
Now we should find the value of \[{{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}\].
We know that \[\cos \left( \dfrac{\pi }{3} \right)=\dfrac{1}{2}\] and \[\sin \left( \dfrac{\pi }{3} \right)=\dfrac{\sqrt{3}}{2}\].
\[\Rightarrow {{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}={{\left[ \cos \left( \dfrac{\pi }{3} \right)+i\sin \left( \dfrac{\pi }{3} \right) \right]}^{365}}\].
We know that \[cis\theta =\cos \theta +i\sin \theta \].
\[\Rightarrow {{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}={{\left[ cis\left( \dfrac{\pi }{3} \right) \right]}^{365}}\].
We know that \[ci{{s}^{n}}\theta =cisn\theta \].
\[\Rightarrow {{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}=\left[ cis\left( \dfrac{365\pi }{3} \right) \right]\].
\[\Rightarrow {{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}=\left[ cis\left( 121\pi +\dfrac{2\pi }{3} \right) \right]\].
We know that \[cis\theta =\cos \theta +i\sin \theta \].
\[\Rightarrow {{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}=\left[ \cos \left( 121\pi +\dfrac{2\pi }{3} \right)+i\sin \left( 121\pi +\dfrac{2\pi }{3} \right) \right]\].
We know that \[\cos \left( n\pi +\theta \right)=-\cos \theta \] and \[\sin \left( n\pi +\theta \right)=-\sin \theta \] where n is odd.
\[\Rightarrow {{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}=\left[ -\cos \left( \dfrac{2\pi }{3} \right)-i\sin \left( \dfrac{2\pi }{3} \right) \right]\].
\[\Rightarrow {{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}=-\left[ \cos \left( \dfrac{2\pi }{3} \right)+i\sin \left( \dfrac{2\pi }{3} \right) \right]\].
\[\Rightarrow {{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}=-\left[ \cos \left( \pi -\dfrac{\pi }{3} \right)+i\sin \left( \pi -\dfrac{\pi }{3} \right) \right]\].
We know that \[\cos \left( n\pi -\theta \right)=-\cos \theta \] and \[\sin \left( n\pi -\theta \right)=\sin \theta \] where n is odd.
\[\Rightarrow {{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}=-\left[ -\cos \left( \dfrac{\pi }{3} \right)+i\sin \left( \dfrac{\pi }{3} \right) \right]\].
We know that \[\cos \dfrac{\pi }{3}=\dfrac{1}{2}\] and \[\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}\].
\[\Rightarrow {{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}=-\left[ -\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]\].
\[\Rightarrow {{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}=\left[ \dfrac{1}{2}-i\dfrac{\sqrt{3}}{2} \right]......(2)\].
Let us assume
\[I=4+5{{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}-3{{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}......(3)\].
Now we will substitute equation (1) and equation (2) in equation (3), then we get
\[\Rightarrow I=4-\dfrac{5}{2}+i\left( \dfrac{5\sqrt{3}}{2} \right)-\dfrac{3}{2}+i\left( \dfrac{3\sqrt{3}}{2} \right)\].
\[\Rightarrow I=4-\dfrac{5}{2}-\dfrac{3}{2}+i\left( \dfrac{5\sqrt{3}}{2}+\dfrac{3\sqrt{3}}{2} \right)\].
\[\Rightarrow I=0+i\left( \dfrac{8\sqrt{3}}{2} \right)\].
\[\Rightarrow I=4\sqrt{3}i......(4)\].
From equation (4), it is clear that the value of \[4+5{{\left[ \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{334}}-3{{\left[ \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \right]}^{365}}\] is equal to \[4\sqrt{3}i\].
Hence, Option c is correct.