The locus of a point z satisfying |z|^2 = ope | vedclass

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

Question

(C) Let $z = x + iy$. Then $|z|^2 = x^2 + y^2$. Given the equation $|z|^2 = \operatorname{Re}(z)$, we substitute the values: $x^2 + y^2 = x$. Rearrang

Vedclass · Accepted Answer

$\left(\frac{1}{2}, 0\right)$

Vedclass · Answer

(C) Let $z = x + iy$. Then $|z|^2 = x^2 + y^2$. Given the equation $|z|^2 = \operatorname{Re}(z)$, we substitute the values: $x^2 + y^2 = x$. Rearranging the terms, we get: $x^2 - x + y^2 = 0$. Completing the square for $x$: $\left(x - \frac{1}{2}\right)^2 + y^2 = \left(\frac{1}{2}\right)^2$. This is the equation of a circle in the form $(x - h)^2 + (y - k)^2 = r^2$, where the centre is $(h, k)$. Comparing the equations, the centre is $\left(\frac{1}{2}, 0\right)$.

The locus of a point $z$ satisfying $|z|^2 = \operatorname{Re}(z)$ is a circle with centre

Similar Questions

If $a$ is a complex number and $b$ is a real number,then the equation $\bar{a}+a+b=0$ represents $a$ as a:

If $|z_1| = |z_2|$ and $\arg\left( \frac{z_1}{z_2} \right) = \pi$,then $z_1 + z_2$ is equal to

If $z = (\lambda + 3) + i\sqrt{5 - \lambda^2}$,then the locus of $z$ is a

If $z$ is a complex number such that $\frac{z-1}{z-i}$ is purely imaginary and the locus of $z$ represents a circle with centre $(\alpha, \beta)$ and radius $r$,then $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=$

If $P(x, y)$ denotes $z = x + iy$ where $x, y \in \mathbb{R}$ and $i = \sqrt{-1}$ in the Argand plane,and $\left|\frac{z-1}{z+2i}\right| = 1$,then the locus of $P$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module