If complex numbers z_1 and z_2 both satisfy z | vedclass

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Question

$\frac{z+\bar{z}}{2}=|z-1|$

$\Rightarrow \mathrm{x}^{2}=(\mathrm{x}-1)^{2}+\mathrm{y}^{2}$ or $\mathrm{y}^{2}=2 \mathrm{x}-1$

Let $\mathrm{z}_{1

Vedclass · Accepted Answer

$\cos ec \frac{\pi }{3}$

Vedclass · Answer

$\frac{z+\bar{z}}{2}=|z-1|$

$\Rightarrow \mathrm{x}^{2}=(\mathrm{x}-1)^{2}+\mathrm{y}^{2}$ or $\mathrm{y}^{2}=2 \mathrm{x}-1$

Let $\mathrm{z}_{1}\left(\frac{\mathrm{t}_{1}^{2}+1}{2}, \mathrm{t}_{1}\right) $ and $\mathrm{z}_{2}\left(\frac{\mathrm{t}_{2}^{2}+1}{2}, \mathrm{t}_{2}\right)$

$\because \arg \left(\mathrm{z}_{1}-\mathrm{z}_{2}\right)=\frac{\pi}{3} \Rightarrow \frac{\mathrm{t}_{2}-\mathrm{t}_{1}}{\frac{\mathrm{t}_{2}^{2}-\mathrm{t}_{1}}{2}}=\sqrt{3}$

$\Rightarrow \frac{\mathrm{2}}{\mathrm{t}_{1}+\mathrm{t}_{2}}=\sqrt{3}$

$\Rightarrow \operatorname{Im}\left(\mathrm{z}_{1}+\mathrm{z}_{2}\right)=\frac{2}{\sqrt{3}}$

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 value of $Im (z_1 + z_2)$ is, where $Im (z)$ denotes imaginary part of $z$ -

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