If complex numbers z_1 and z_2 both satisfy z | vedclass

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(B) Let $z = x + iy$. The equation $z + \overline{z} = 2|z - 1|$ becomes $2x = 2\sqrt{(x-1)^2 + y^2}$.Squaring both sides,$x^2 = (x-1)^2 + y^2$,which

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$\csc \frac{\pi}{3}$

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(B) Let $z = x + iy$. The equation $z + \overline{z} = 2|z - 1|$ becomes $2x = 2\sqrt{(x-1)^2 + y^2}$.Squaring both sides,$x^2 = (x-1)^2 + y^2$,which simplifies to $x^2 = x^2 - 2x + 1 + y^2$,or $y^2 = 2x - 1$.Let $z_1 = x_1 + iy_1$ and $z_2 = x_2 + iy_2$. Since $y^2 = 2x - 1$,we can parameterize $z$ as $z = \frac{t^2 + 1}{2} + it$.Then $z_1 - z_2 = \frac{t_1^2 - t_2^2}{2} + i(t_1 - t_2) = \frac{(t_1 - t_2)(t_1 + t_2)}{2} + i(t_1 - t_2)$.Given $\arg(z_1 - z_2) = \frac{\pi}{3}$,we have $	an(\frac{\pi}{3}) = \frac{t_1 - t_2}{\frac{(t_1 - t_2)(t_1 + t_2)}{2}} = \frac{2}{t_1 + t_2} = \sqrt{3}$.Thus,$t_1 + t_2 = \frac{2}{\sqrt{3}}$.Since $	ext{Im}(z_1 + z_2) = t_1 + t_2$,the value is $\frac{2}{\sqrt{3}} = \csc \frac{\pi}{3}$.

If complex numbers $z_1$ and $z_2$ both satisfy $z + \overline{z} = 2 |z - 1|$ and $\arg(z_1 - z_2) = \frac{\pi}{3},$ then the value of $\text{Im}(z_1 + z_2)$ is,where $\text{Im}(z)$ denotes the imaginary part of $z$.

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