Let S = {z in mathbb{C} : z^2 + sqrt{6}iz - 3 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the quadratic equation $z^2 + \sqrt{6}iz - 3 = 0$.Using the quadratic formula $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we get:$z = \frac{-\

Vedclass · Accepted Answer

$162$

Vedclass · Answer

(A) Given the quadratic equation $z^2 + \sqrt{6}iz - 3 = 0$.Using the quadratic formula $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we get:$z = \frac{-\sqrt{6}i \pm \sqrt{(\sqrt{6}i)^2 - 4(1)(-3)}}{2} = \frac{-\sqrt{6}i \pm \sqrt{-6 + 12}}{2} = \frac{-\sqrt{6}i \pm \sqrt{6}}{2}$.Thus, the roots are $z_1 = \frac{\sqrt{6}}{2}(1 - i)$ and $z_2 = \frac{\sqrt{6}}{2}(-1 - i)$.Now, calculate $z^2$ for each root:$z_1^2 = \left(\frac{\sqrt{6}}{2}(1 - i)\right)^2 = \frac{6}{4}(1 - 2i + i^2) = \frac{3}{2}(1 - 2i - 1) = -3i$.$z_2^2 = \left(\frac{\sqrt{6}}{2}(-1 - i)\right)^2 = \frac{6}{4}(1 + 2i + i^2) = \frac{3}{2}(1 + 2i - 1) = 3i$.Now, calculate $z^8 = (z^2)^4$:$z_1^8 = (-3i)^4 = (-3)^4 \cdot i^4 = 81 \cdot 1 = 81$.$z_2^8 = (3i)^4 = 3^4 \cdot i^4 = 81 \cdot 1 = 81$.Therefore, $\sum_{z \in S} z^8 = z_1^8 + z_2^8 = 81 + 81 = 162$.

Let $S = \{z \in \mathbb{C} : z^2 + \sqrt{6}iz - 3 = 0\}$. Then $\sum_{z \in S} z^8$ is equal to:

Similar Questions

If $\sqrt{5}-i \sqrt{15}=r(\cos \theta+i \sin \theta)$ where $-\pi < \theta < \pi$,then find the value of $r^2(\sec \theta+3 \operatorname{cosec}^2 \theta)$.

Find the maximum value of $|z|$ when $\left|z-\frac{3}{z}\right|=2,$ where $z$ is a complex number.

If $\alpha$ is a root of the equation $x^2+x+1=0$ and $\sum_{k=1}^n\left(\alpha^k+\frac{1}{\alpha^k}\right)^2=20$,then $n$ is equal to

If $z$ is a complex number satisfying $|z|^2 - |z| - 2 < 0$,then the value of $|z^2 + z \sin \theta|$,for all values of $\theta$,is

If $(1 + i)(1 + 2i)(1 + 3i) \dots (1 + ni) = a + ib$,then $2 \times 5 \times 10 \times \dots \times (1 + n^2)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module