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The locus of a point $z$ satisfying $|z|^2=\operatorname{Re}(z)$ is a circle with centre
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Verified Answer
The correct answer is:
$\left(\frac{1}{2}, 0\right)$
Let $z=x+i y$
$|z|=\sqrt{x^2+y^2}$
Now, $|z|^2=\operatorname{Re}(z)$
$x^2+y^2=x$
$\Rightarrow \quad x^2+y^2-x=0$
$g=1 / 2, f=0$
So, centre of circle $\left(\frac{1}{2}, 0\right)$.
$|z|=\sqrt{x^2+y^2}$
Now, $|z|^2=\operatorname{Re}(z)$
$x^2+y^2=x$
$\Rightarrow \quad x^2+y^2-x=0$
$g=1 / 2, f=0$
So, centre of circle $\left(\frac{1}{2}, 0\right)$.
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