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Complex Numbers question

2021 · 18 Mar · Shift 1 · Q43

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Complex Numbers
2021 · 18 Mar · Shift 1 · Q43

Complex Numbers question

2021 · 18 Mar · Shift 1 · Q43

JEE MainMathematicsComplex NumbersNumerical+4 / −1

Let z1, z2 be the roots of the equation z2 + az + 12 = 0 and z1, z2 form an equilateral triangle with origin. Then, the value of |a| is :

Numerical answer

View written solutionFree

Correct answer: 6

Let the roots of $z^2+az+12=0$ be $z_1,z_2$ .

By Vieta's formulas,

\qquad z_1z_2=12.$$ 2. Since the points corresponding to $z_1,z_2$ and the origin form an equilateral triangle, the distances from the origin to the two roots must be equal: $$|z_1|=|z_2|.$$ Also, the angle between $z_1$ and $z_2$ is $60^\circ$. So we may write $$z_2=z_1\,\omega$$ where $|\omega|=1$ and $\arg(\omega)=\pm \frac{\pi}{3}$. Thus, $$\omega=e^{\pm i\pi/3}.$$ 3. Now use the product: $$z_1z_2=z_1^2\omega=12.$$ Taking modulus on both sides, $$|z_1|^2|\omega|=12.$$ Since $|\omega|=1$, $$|z_1|^2=12 \implies |z_1|=2\sqrt{3}.$$ Similarly, $$|z_2|=2\sqrt{3}.$$ 4. Now compute $|a|$ using $$a=-(z_1+z_2).$$ Hence, $$|a|=|z_1+z_2|.$$ Since $|z_1|=|z_2|=2\sqrt{3}$ and the angle between them is $60^\circ$, $$|z_1+z_2|^2=|z_1|^2+|z_2|^2+2|z_1||z_2|\cos 60^\circ.$$ So, $$|z_1+z_2|^2=12+12+2(2\sqrt{3})(2\sqrt{3})\cdot \frac12.$$ $$=24+12=36.$$ Therefore, $$|a|=6.$$ 5. Final answer: $$\boxed{6}$$ Comparison with stored answer: the derived answer is $6$, which matches the stored correct answer.

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