Let S be the set of all complex numbers z sat | vedclass

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

Question

(B) The condition $|z-(2-i)| \geq \sqrt{5}$ represents the region outside or on the circle with center $C(2, -1)$ and radius $r = \sqrt{5}$.To maximiz

Vedclass · Accepted Answer

$-\frac{\pi}{2}$

Vedclass · Answer

(B) The condition $|z-(2-i)| \geq \sqrt{5}$ represents the region outside or on the circle with center $C(2, -1)$ and radius $r = \sqrt{5}$.To maximize $\frac{1}{|z-1|}$,we must minimize $|z-1|$,which is the distance of $z$ from the point $A(1, 0)$.The point $z_0$ on the circle closest to $A(1, 0)$ lies on the line segment connecting $A(1, 0)$ and $C(2, -1)$.The line $AC$ has the equation $y - 0 = \frac{-1-0}{2-1}(x-1)$,so $y = -x+1$,or $x+y-1=0$.The intersection of this line with the circle $(x-2)^2 + (y+1)^2 = 5$ gives $z_0$. Since $z_0$ is on the circle and on the line $y = 1-x$,we have $(x-2)^2 + (1-x+1)^2 = 5 \implies (x-2)^2 + (2-x)^2 = 5 \implies 2(x-2)^2 = 5 \implies x-2 = \pm \sqrt{2.5}$.Since $z_0$ is closer to $A(1, 0)$,$x_0 = 2 - \sqrt{2.5} \approx 0.42$,so $y_0 = 1 - x_0 = \sqrt{2.5} - 1 \approx 0.58$.Let $z_0 = x_0 + iy_0$. Then $z_0 - \bar{z}_0 = 2iy_0$ and $z_0 + \bar{z}_0 = 2x_0$.The expression becomes $\frac{4-2x_0}{2iy_0+2i} = \frac{2-x_0}{i(y_0+1)}$.Since $y_0 = 1-x_0$,$y_0+1 = 2-x_0$.Thus,the expression is $\frac{2-x_0}{i(2-x_0)} = \frac{1}{i} = -i$.The principal argument of $-i$ is $-\frac{\pi}{2}$.

Let $S$ be the set of all complex numbers $z$ satisfying $|z-2+i| \geq \sqrt{5}$. If the complex number $z_0$ is such that $\frac{1}{|z_0-1|}$ is the maximum of the set $\left\{\frac{1}{|z-1|}: z \in S\right\}$,then the principal argument of $\frac{4-z_0-\bar{z}_0}{z_0-\bar{z}_0+2i}$ is

Similar Questions

Let $z=x+iy$ and a point $P$ represent $z$ in the Argand plane. If the real part of $\frac{z-1}{z+i}$ is $1$,then a point that lies on the locus of $P$ is

If $A = \{z = x + iy : \text{real part of } \frac{\bar{z}-1}{z-i} = 2\}$,then the locus of the point $P(x, y)$ in the Cartesian plane is:

Let $a, b, c, d$ be real numbers such that $|a-b|=2$,$|b-c|=3$,and $|c-d|=4$. Then,the sum of all possible values of $|a-d|$ is

If $\omega$ is a complex cube root of unity whose imaginary part is positive and $|z - \omega| = |z + \omega|$,then $arg(z)$ can be-

If $z_1, z_2, z_3 \in \mathbb{C}$ are the vertices of an equilateral triangle,whose centroid is $z_0$,then $\sum_{k=1}^3 (z_k - z_0)^2$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module