Let S = {z in mathbb{C} : 4z^2 + overline{z} | vedclass

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

Question

(A) Given $4z^2 + \overline{z} = 0$. Let $z = x + iy$.Substituting $z$ into the equation: $4(x + iy)^2 + (x - iy) = 0$.$4(x^2 - y^2 + 2xyi) + x - iy =

Vedclass · Accepted Answer

$\frac{3}{16}$

Vedclass · Answer

(A) Given $4z^2 + \overline{z} = 0$. Let $z = x + iy$.Substituting $z$ into the equation: $4(x + iy)^2 + (x - iy) = 0$.$4(x^2 - y^2 + 2xyi) + x - iy = 0$.$4x^2 - 4y^2 + x + i(8xy - y) = 0$.Equating real and imaginary parts to zero:$4x^2 - 4y^2 + x = 0$ and $y(8x - 1) = 0$.Case $1$: $y = 0$. Then $4x^2 + x = 0 \Rightarrow x(4x + 1) = 0$. So $x = 0$ or $x = -1/4$.This gives $z_1 = 0$ $(|z_1|^2 = 0)$ and $z_2 = -1/4$ $(|z_2|^2 = 1/16)$.Case $2$: $x = 1/8$. Then $4(1/8)^2 - 4y^2 + 1/8 = 0$.$4/64 - 4y^2 + 1/8 = 0$ $\Rightarrow 1/16 + 1/8 = 4y^2$ $\Rightarrow 3/16 = 4y^2$ $\Rightarrow y^2 = 3/64$.So $y = \pm \sqrt{3}/8$. This gives $z_3 = 1/8 + i\sqrt{3}/8$ and $z_4 = 1/8 - i\sqrt{3}/8$.$|z_3|^2 = (1/8)^2 + (\sqrt{3}/8)^2 = 1/64 + 3/64 = 4/64 = 1/16$.$|z_4|^2 = (1/8)^2 + (-\sqrt{3}/8)^2 = 1/64 + 3/64 = 4/64 = 1/16$.Summing the squares of the moduli: $0 + 1/16 + 1/16 + 1/16 = 3/16$.

Let $S = \{z \in \mathbb{C} : 4z^2 + \overline{z} = 0\}$. Then $\sum_{z \in S} |z|^2$ is equal to:

Similar Questions

If $z=x+iy$, where $x$ and $y$ are real numbers and $i=\sqrt{-1}$, then the points $(x, y)$ for which $\frac{z-1}{z-i}$ is real, lie on

The locus of $z$ given by $\left| \frac{z - 1}{z - i} \right| = 1$ is

If complex numbers $z_1$ and $z_2$ both satisfy $z + \overline{z} = 2 |z - 1|$ and $\arg(z_1 - z_2) = \frac{\pi}{3},$ then the value of $\text{Im}(z_1 + z_2)$ is,where $\text{Im}(z)$ denotes the imaginary part of $z$.

If $|z + 4| \le 3$,then the greatest and the least value of $|z + 1|$ are

Let $z$ be a complex number satisfying $|z+5| \leq 4$ and $z(1+i)+\bar{z}(1-i) \geq -10$,where $i=\sqrt{-1}$. If the maximum value of $|z+1|^2$ is $\alpha+\beta \sqrt{2}$,then the value of $(\alpha+\beta)$ is ......

Vedclass Test Series

Exam Paper Generator

Online Exam Module