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.