If the locus of z in mathbb{C}, such that ope | vedclass

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

Question

(C) Given $\operatorname{Re}\left(\frac{z-1}{2z+i}\right) + \operatorname{Re}\left(\frac{\bar{z}-1}{2\bar{z}-i}\right)=2$
Since $\operatorname{Re}(w)

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(C) Given $\operatorname{Re}\left(\frac{z-1}{2z+i}\right) + \operatorname{Re}\left(\frac{\bar{z}-1}{2\bar{z}-i}\right)=2$
Since $\operatorname{Re}(w)=\operatorname{Re}(\bar{w})$, we have
$\operatorname{Re}\left(\frac{\bar{z}-1}{2\bar{z}-i}\right)=\operatorname{Re}\left(\overline{\left(\frac{\bar{z}-1}{2\bar{z}-i}\right)}\right)=\operatorname{Re}\left(\frac{z-1}{2z+i}\right)$
Thus, $2\operatorname{Re}\left(\frac{z-1}{2z+i}\right)=2 \Rightarrow \operatorname{Re}\left(\frac{z-1}{2z+i}\right)=1$
Let $z=x+iy$. Then
$\frac{z-1}{2z+i}=\frac{(x-1)+iy}{2x+i(2y+1)}=\frac{((x-1)+iy)(2x-i(2y+1))}{4x^2+(2y+1)^2}$
Real part: $\frac{2x(x-1)+y(2y+1)}{4x^2+(2y+1)^2}=1$
$2x^2-2x+2y^2+y=4x^2+4y^2+4y+1$
$2x^2+2y^2+2x+3y+1=0$
$x^2+y^2+x+\frac{3}{2}y+\frac{1}{2}=0$
Comparing with $x^2+y^2-2ax-2by+c=0$, center is
$(a,b)=\left(-\frac{1}{2},-\frac{3}{4}\right)$
$r^2=a^2+b^2-c=\frac{1}{4}+\frac{9}{16}-\frac{1}{2}=\frac{5}{16}$
Now, $\frac{15ab}{r^2}=15\cdot\left(-\frac{1}{2}\right)\cdot\left(-\frac{3}{4}\right)\cdot\frac{16}{5}=18$

If the locus of $z \in \mathbb{C}$, such that $\operatorname{Re}\left(\frac{z-1}{2 z+i}\right)+\operatorname{Re}\left(\frac{\bar{z}-1}{2 \bar{z}-i}\right)=2$, is a circle of radius $r$ and center $(a, b)$, then $\frac{15 a b}{r^2}$ is equal to :

Similar Questions

If $z = \frac{3}{2 + \cos \theta + i \sin \theta}$,then the locus of $z$ is :-

Let $|z_1 - 8 - 2i| \leq 1$ and $|z_2 - 2 + 6i| \leq 2$,where $z_1, z_2 \in \mathbb{C}$. Then the minimum value of $|z_1 - z_2|$ is:

Reflection of the line $\bar{a} z+a \bar{z}=0$ in the real axis is given by

The reflection of the complex number $(3 + 2i)$ in the straight line $z = -i \bar{z}$ is-

If $|z_1|=1, |z_2|=2, |z_3|=3$ and $|9z_1z_2 + 4z_1z_3 + z_2z_3| = 12$,then the value of $|z_1 + z_2 + z_3|$ is equal to :-

Vedclass Test Series

Exam Paper Generator

Online Exam Module