Q.3. Let . Then is equal to :

Q: What is the correct answer for JEE Main Mathematics Q3?

The correct answer is A. Sol. The quadratic equation is $z^2+\sqrt{6}iz-3=0$. Using the quadratic formula, $z = \frac{-\sqrt{6}i \pm \sqrt{(\sqrt{6}i)^2 - 4(1)(-3)}}{2} = \frac{-\sqrt{6}i \pm \sqrt{-6+12}}{2} = \frac{-\sqrt{6}i \pm \sqrt{6}}{2}$. Let the roots be $z_1 = \frac{\sqrt{6} - \sqrt{6}i}{2}$ and $z_2 = \frac{-\sqrt{6} - \sqrt{6}i}{2}$. We can write these in polar form. $z_1 = \frac{\sqrt{6}}{2}(1-i) = \frac{\sqr

Q: Q.3. Let . Then is equal to :

The correct answer is A. Sol. The quadratic equation is $z^2+\sqrt{6}iz-3=0$. Using the quadratic formula, $z = \frac{-\sqrt{6}i \pm \sqrt{(\sqrt{6}i)^2 - 4(1)(-3)}}{2} = \frac{-\sqrt{6}i \pm \sqrt{-6+12}}{2} = \frac{-\sqrt{6}i \pm \sqrt{6}}{2}$. Let the roots be $z_1 = \frac{\sqrt{6} - \sqrt{6}i}{2}$ and $z_2 = \frac{-\sqrt{6} - \sqrt{6}i}{2}$. We can write these in polar form. $z_1 = \frac{\sqrt{6}}{2}(1-i) = \frac{\sqr

Q: What are the options for this JEE Main Mathematics question?

A. 162 · B. 184 · C. 262 · D. 324

Question

Q.3. Let . Then  is equal to :

FirstInTest · Accepted Answer

Sol. The quadratic equation is $z^2+\sqrt{6}iz-3=0$. Using the quadratic formula, $z = \frac{-\sqrt{6}i \pm \sqrt{(\sqrt{6}i)^2 - 4(1)(-3)}}{2} = \frac{-\sqrt{6}i \pm \sqrt{-6+12}}{2} = \frac{-\sqrt{6}i \pm \sqrt{6}}{2}$. Let the roots be $z_1 = \frac{\sqrt{6} - \sqrt{6}i}{2}$ and $z_2 = \frac{-\sqrt{6} - \sqrt{6}i}{2}$. We can write these in polar form. $z_1 = \frac{\sqrt{6}}{2}(1-i) = \frac{\sqrt{6}}{2} \sqrt{2} (\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}) = \sqrt{3} (\cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4})) = \sqrt{3}e^{-i\pi/4}$. $z_2 = \frac{\sqrt{6}}{2}(-1-i) = \frac{\sqrt{6}}{2} \sqrt{2} (-\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}) = \sqrt{3} (\cos(-\frac{3\pi}{4}) + i\sin(-\frac{3\pi}{4})) = \sqrt{3}e^{-i3\pi/4}$. Let's recheck the solution provided. The solution uses $z_1 = \frac{\sqrt{6}}{2}(1-i)$ and $z_2 = \frac{-\sqrt{6}}{2}(1+i)$. This is incorrect. The roots are $z_1 = \frac{\sqrt{6}}{2} - \frac{\sqrt{6}}{2}i$ and $z_2 = -\frac{\sqrt{6}}{2} - \frac{\sqrt{6}}{2}i$. Let's use the solution's approach for $z_1$ and $z_2$ as $\alpha$ and $\beta$. $\alpha = \frac{\sqrt{6}}{2}(1-i)$ and $\beta = \frac{-\sqrt{6}}{2}(1+i)$. This is not consistent with the quadratic formula. Let's use the roots from the quadratic formula: $z_1 = \frac{\sqrt{6}}{2}(1-i)$ and $z_2 = \frac{\sqrt{6}}{2}(-1-i)$. $z_1 = \sqrt{3}e^{-i\pi/4}$. $z_2 = \sqrt{3}e^{-i3\pi/4}$. Then $z_1^8 = (\sqrt{3})^8 e^{-i8\pi/4} = 3^4 e^{-i2\pi} = 81(1) = 81$. $z_2^8 = (\sqrt{3})^8 e^{-i8(3\pi/4)} = 3^4 e^{-i6\pi} = 81(1) = 81$. So $\sum z^8 = z_1^8 + z_2^8 = 81+81 = 162$. The solution provided has $\alpha = -\sqrt{3}e^{i\pi/4}$ and $\beta = +\sqrt{3}e^{-i\pi/4}$. Let's verify these roots. If $z = -\sqrt{3}e^{i\pi/4} = -\sqrt{3}(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}) = -\frac{\sqrt{6}}{2}(1+i)$. If $z = \sqrt{3}e^{-i\pi/4} = \sqrt{3}(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}) = \frac{\sqrt{6}}{2}(1-i)$. These are the roots if the equation was $z^2 - \sqrt{6}iz - 3 = 0$. However, the given equation is $z^2+\sqrt{6}iz-3=0$. The roots are $z = \frac{-\sqrt{6}i \pm \sqrt{6}}{2}$. So $z_1 = \frac{\sqrt{6}}{2} - \frac{\sqrt{6}}{2}i$ and $z_2 = -\frac{\sqrt{6}}{2} - \frac{\sqrt{6}}{2}i$. $z_1 = \sqrt{3}(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}) = \sqrt{3}e^{-i\pi/4}$. $z_2 = \sqrt{3}(-\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}) = \sqrt{3}e^{-i3\pi/4}$. Then $z_1^8 = (\sqrt{3})^8 (e^{-i\pi/4})^8 = 81 e^{-i2\pi} = 81$. $z_2^8 = (\sqrt{3})^8 (e^{-i3\pi/4})^8 = 81 e^{-i6\pi} = 81$. Sum is $81+81=162$. The solution's calculation for $\alpha^8 + \beta^8$ is $81(e^{i2\pi} + e^{-i2\pi}) = 81(1+1) = 162$. This is correct, even if the intermediate steps for $\alpha$ and $\beta$ were confusing or based on a slightly different interpretation of the roots.

Answer

162

Answer

184

Answer

262

Answer

324

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