Question

If a complex equation is given by \[{{\left( \sqrt{3}+i \right)}^{100}}={{2}^{99}}\left( a+ib \right)\], then show that \[{{a}^{2}}+{{b}^{2}}=4\].

Answer 1

Hint: In order to solve this question, we should know about a few trigonometric ratios like \[\cos \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2},\sin \dfrac{\pi }{6}=\dfrac{1}{2},\cos \dfrac{2\pi }{3}=\dfrac{-1}{2},\sin \dfrac{2\pi }{3}=\dfrac{\sqrt{3}}{2}\]. Also, we need to remember that \[\left( \cos \theta +i\sin \theta \right)={{e}^{i\theta }}\] and \[\left| x+iy \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}\]. By using this, we can prove equality.

Complete step-by-step answer:

In this question, we have been given equality that is \[{{\left( \sqrt{3}+i \right)}^{100}}={{2}^{99}}\left( a+ib \right)\] and we have been asked to prove that \[{{a}^{2}}+{{b}^{2}}=4\].
So, to solve this question, we should know that \[\cos \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}\text{ and }\sin \dfrac{\pi }{6}=\dfrac{1}{2}\].
So, to prove the equality \[{{a}^{2}}+{{b}^{2}}=4\], we will first consider \[{{\left( \sqrt{3}+i \right)}^{100}}={{2}^{99}}\left( a+ib \right)\]. So, we will first multiply and divide \[{{\left( \sqrt{3}+i \right)}^{100}}\] by \[{{2}^{100}}\]. So, we get,
\[{{2}^{100}}{{\left( \dfrac{\sqrt{3}}{2}+\dfrac{i}{2} \right)}^{100}}={{2}^{99}}\left( a+ib \right)\]
Now, we know that \[\dfrac{\sqrt{3}}{2}=\cos \dfrac{\pi }{6}\text{ and }\dfrac{1}{2}=\sin \dfrac{\pi }{6}\].
So, we get,
\[{{2}^{100}}{{\left[ \cos \dfrac{\pi }{6}+i\sin \dfrac{\pi }{6} \right]}^{100}}={{2}^{99}}\left( a+ib \right)\]
Now, we know that \[\cos \theta +i\sin \theta ={{e}^{i\theta }}\]. So, we can write \[\cos \dfrac{\pi }{6}+i\sin \dfrac{\pi }{6}={{e}^{i\dfrac{\pi }{6}}}\]. So, we get,
\[{{2}^{100}}{{\left( {{e}^{i\dfrac{\pi }{6}}} \right)}^{100}}={{2}^{99}}\left( a+ib \right)\]
\[{{2}^{100}}\left( {{e}^{i100\dfrac{\pi }{6}}} \right)={{2}^{99}}\left( a+ib \right)\]
\[{{2}^{100}}\left( {{e}^{i50\dfrac{\pi }{3}}} \right)={{2}^{99}}\left( a+ib \right)\]
Now, we know that \[\dfrac{50\pi }{3}\] can be written as \[16\pi +\dfrac{2\pi }{3}\]. So, we get,
\[{{2}^{100}}\left( {{e}^{i\left( 16\pi +\dfrac{2\pi }{3} \right)}} \right)={{2}^{99}}\left( a+ib \right)\]
And we know that \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\]. So, we can write \[{{e}^{i\left( 16\pi +\dfrac{2\pi }{3} \right)}}\] as \[\left( {{e}^{i16\pi }}\times {{e}^{i\dfrac{2\pi }{3}}} \right)\]. Therefore, we get,
\[{{2}^{100}}\left( {{e}^{i16\pi }} \right)\left( {{e}^{i\dfrac{2\pi }{3}}} \right)={{2}^{99}}\left( a+ib \right)\]
Now, we know that \[{{e}^{i\theta }}=\cos \theta +i\sin \theta \]. So, we can write \[{{e}^{i16\pi }}=\cos 16\pi +i\sin 16\pi ,\text{ }{{e}^{i\dfrac{2\pi }{3}}}=\cos \dfrac{2\pi }{3}+i\sin \dfrac{2\pi }{3}\]
Therefore, we get,
\[{{2}^{100}}\left[ \cos 16\pi +i\sin 16\pi \right]\left[ \cos \dfrac{2\pi }{3}+i\sin \dfrac{2\pi }{3} \right]={{2}^{99}}\left( a+ib \right)\]
Now, we know that \[\cos 2n\pi =1\text{ and }\sin 2n\pi =0\]. We can write \[\cos 16\pi =\cos \left( 2\times 8\pi \right)=1\text{ and }\sin 16\pi =\sin \left( 2\times 8\pi \right)=0\]. Also, we know that \[\cos \dfrac{2\pi }{3}=\dfrac{-1}{2}\text{ and }\sin \dfrac{2\pi }{3}=\dfrac{\sqrt{3}}{2}\]. So, we get,
\[{{2}^{100}}\left( 1+0i \right)\left[ \dfrac{-1}{2}+\dfrac{\sqrt{3}}{2}i \right]={{2}^{99}}\left( a+ib \right)\]
\[{{2}^{100}}\left( \dfrac{-1+\sqrt{3}i}{2} \right)={{2}^{99}}\left( a+ib \right)\]
\[\dfrac{\left( -1+\sqrt{3}i \right)}{1}=\dfrac{{{2}^{99}}\left( a+ib \right)2}{{{2}^{100}}}\]
\[-1+\sqrt{3}i=a+ib\]
On comparing both the sides, we can say a = – 1 and \[b=\sqrt{3}\]. Now, we have to show that \[{{a}^{2}}+{{b}^{2}}=4\]. For that, we will put the value of a and b in \[{{a}^{2}}+{{b}^{2}}=4\], we get,
\[{{\left( -1 \right)}^{2}}+{{\left( \sqrt{3} \right)}^{2}}=4\]
\[1+3=4\]
\[4=4\]
LHS = RHS
Hence proved

Note: While solving this question, we need to remember that \[\cos \dfrac{2\pi }{3}\text{ and }\sin \dfrac{2\pi }{3}\] are calculated by the property of \[\cos \left( 180-\theta \right)=-\cos \theta \text{ and }\sin \left( 180-\theta \right)=\sin \theta \] where \[\theta =\dfrac{\pi }{3}\] because \[\dfrac{2\pi }{3}=180-\dfrac{\pi }{3}\] and therefore we get \[\cos \dfrac{2\pi }{3}=\dfrac{-1}{2}\text{ and }\sin \dfrac{2\pi }{3}=\dfrac{\sqrt{3}}{2}\]. Also, we have to remember that \[{{e}^{i\theta }}=\cos \theta +i\sin \theta \]. By these concepts, we can solve the question. The important step here is the splitting of the angle \[\dfrac{50\pi }{3}\text{ as }16\pi +\dfrac{2\pi }{3}\]. Whichever angle we get in the power, we must try to represent it as the sum or difference of \[2n\pi \pm \theta \]. This will make it easier to solve this type of question.