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

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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

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