The locus of z which lies in the shaded regio | vedclass

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

Question

(A) The shaded region is centered at $A(-1, 0)$,which corresponds to the complex number $z_0 = -1$. Thus,the distance from $A$ is represented by $|z -

Vedclass · Accepted Answer

$z: |z + 1| > 2, |\arg(z + 1)| < \pi/4$

Vedclass · Answer

(A) The shaded region is centered at $A(-1, 0)$,which corresponds to the complex number $z_0 = -1$. Thus,the distance from $A$ is represented by $|z - (-1)| = |z + 1|$. The region is outside the circle of radius $2$ centered at $A$,so $|z + 1| > 2$. The angular region is bounded by lines passing through $A$ with slopes $\pm 1$,corresponding to angles $\pm \pi/4$. Thus,the argument condition is $|\arg(z + 1)| Combining these,the locus is $z: |z + 1| > 2, |\arg(z + 1)| < \pi/4$.

The locus of $z$ which lies in the shaded region is best represented by

Similar Questions

If $z, \bar{z}, -z, -\bar{z}$ form a rectangle of area $2 \sqrt{3}$ square units,then one such $z$ is

Let ${z_1}$ and ${z_2}$ be two roots of the equation ${z^2 + az + b = 0}$,where ${z}$ is a complex number. Further,assume that the origin,${z_1}$,and ${z_2}$ form an equilateral triangle. Then:

If a complex number $z$ is such that $\frac{z-2i}{z-2}$ is a purely imaginary number and the locus of $z$ is a closed curve,then the area of the region bounded by that closed curve and lying in the first quadrant is

The area of the triangle with vertices $A(z)$,$B(iz)$,and $C(z+iz)$ is

If ${z^2} + z|z| + |z|^2 = 0$,then the locus of $z$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module