If z=x+iy is a complex number satisfying left | vedclass

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

Question

(A) We have,$\left|z+\frac{i}{2}\right|^2=\left|z-\frac{i}{2}\right|^2$. Substituting $z=x+iy$: $\left|x+i\left(y+\frac{1}{2}\right)\right|^2 = \left|

Vedclass · Accepted Answer

$x$-axis

Vedclass · Answer

(A) We have,$\left|z+\frac{i}{2}\right|^2=\left|z-\frac{i}{2}\right|^2$. Substituting $z=x+iy$: $\left|x+i\left(y+\frac{1}{2}\right)\right|^2 = \left|x+i\left(y-\frac{1}{2}\right)\right|^2$. Using the property $|a+ib|^2 = a^2+b^2$: $x^2 + \left(y+\frac{1}{2}\right)^2 = x^2 + \left(y-\frac{1}{2}\right)^2$. $x^2 + y^2 + y + \frac{1}{4} = x^2 + y^2 - y + \frac{1}{4}$. Subtracting $x^2 + y^2 + \frac{1}{4}$ from both sides: $y = -y$ $\Rightarrow 2y = 0$ $\Rightarrow y = 0$. The equation $y=0$ represents the $x$-axis.

If $z=x+iy$ is a complex number satisfying $\left|z+\frac{i}{2}\right|^2=\left|z-\frac{i}{2}\right|^2$,then the locus of $z$ is

Similar Questions

If the amplitude of $(Z-2)$ is $\frac{\pi}{2}$,then the locus of $Z$ 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:

Let $A = \{z \in \mathbb{C} : 1 \leq |z - (1 + i)| \leq 2\}$ and $B = \{z \in A : |z - (1 - i)| = 1\}$. Then,$B$ is:

If $z = x + iy$ and the point $P$ in the Argand diagram represents $z$,then the locus of the point $P$ satisfying the equation $2|z - 2 - 3i| = 3|z - 2 + i|$ is a circle with centre

The locus represented by $|z - 1| = |z + i|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module