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Complex Numbers question

2021 · 26 Aug · Shift 2 · Q34

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Complex Numbers
2021 · 26 Aug · Shift 2 · Q34

Complex Numbers question

2021 · 26 Aug · Shift 2 · Q34

JEE MainMathematicsComplex NumbersMCQ+4 / −1

If

{\left( {\sqrt 3 + i} \right)^{100}} = {2^{99}}(p + iq)

, then p and q are roots of the equation :

A
${x^2} - \left( {\sqrt 3 - 1} \right)x - \sqrt 3 = 0$
B
${x^2} + \left( {\sqrt 3 + 1} \right)x + \sqrt 3 = 0$
C
${x^2} + \left( {\sqrt 3 - 1} \right)x - \sqrt 3 = 0$
D
${x^2} - \left( {\sqrt 3 + 1} \right)x + \sqrt 3 = 0$

View written solutionFree

Correct answer: A

Write the complex number in polar form

We have $\sqrt{3}+i$ Its modulus is $r=\sqrt{(\sqrt{3})^2+1^2}=\sqrt{3+1}=2$

Its argument is $\theta=\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}$

So, $\sqrt{3}+i=2\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)$

Raise to the 100th power using De Moivre's theorem

$\left(\sqrt{3}+i\right)^{100}=2^{100}\left(\cos\frac{100\pi}{6}+i\sin\frac{100\pi}{6}\right)$

Simplify the angle: $\frac{100\pi}{6}=\frac{50\pi}{3}=16\pi+\frac{2\pi}{3}$

Hence, $\cos\frac{50\pi}{3}=\cos\frac{2\pi}{3}=-\frac12$ $\sin\frac{50\pi}{3}=\sin\frac{2\pi}{3}=\frac{\sqrt{3}}{2}$

Therefore, $\left(\sqrt{3}+i\right)^{100}=2^{100}\left(-\frac12+i\frac{\sqrt{3}}{2}\right)$

$=2^{99}(-1+i\sqrt{3})$

Given $\left(\sqrt{3}+i\right)^{100}=2^{99}(p+iq)$

So, $p=-1,\qquad q=\sqrt{3}$

Form the quadratic equation whose roots are $p$ and $q$

If roots are $-1$ and $\sqrt{3}$ , then

Sum of roots: $p+q=-1+\sqrt{3}=\sqrt{3}-1$
Product of roots: $pq=(-1)(\sqrt{3})=-\sqrt{3}$

Hence the quadratic is $x^2-(\text{sum})x+(\text{product})=0$

So, $x^2-(\sqrt{3}-1)x-\sqrt{3}=0$

Match with the options

This is exactly Option A: $x^2-(\sqrt{3}-1)x-\sqrt{3}=0$

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