Let z_{1} and z_{2} be two complex numbers su | vedclass

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(C) Given $|z-3|=\operatorname{Re}(z)$. Let $z=x+iy$.$(x-3)^{2}+y^{2}=x^{2}$$x^{2}-6x+9+y^{2}=x^{2}$$y^{2}=6x-9=6(x-\frac{3}{2})$.This is a parabola w

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$6$

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(C) Given $|z-3|=\operatorname{Re}(z)$. Let $z=x+iy$.$(x-3)^{2}+y^{2}=x^{2}$$x^{2}-6x+9+y^{2}=x^{2}$$y^{2}=6x-9=6(x-\frac{3}{2})$.This is a parabola with vertex at $(\frac{3}{2}, 0)$.Let $z_{1}=x_{1}+iy_{1}$ and $z_{2}=x_{2}+iy_{2}$.Since $\arg(z_{1}-z_{2})=\frac{\pi}{4}$, the slope of the line segment joining $z_{1}$ and $z_{2}$ is $	an(\frac{\pi}{4})=1$.Thus, $\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=1 \Rightarrow y_{1}-y_{2}=x_{1}-x_{2} \Rightarrow x_{1}-y_{1}=x_{2}-y_{2}$.From the parabola equation, $x_{1}=\frac{y_{1}^{2}}{6}+\frac{3}{2}$ and $x_{2}=\frac{y_{2}^{2}}{6}+\frac{3}{2}$.Substituting these into $x_{1}-x_{2}=y_{1}-y_{2}$:$(\frac{y_{1}^{2}}{6}+\frac{3}{2})-(\frac{y_{2}^{2}}{6}+\frac{3}{2})=y_{1}-y_{2}$$\frac{1}{6}(y_{1}-y_{2})(y_{1}+y_{2})=y_{1}-y_{2}$.Since $z_{1} 
eq z_{2}$, $y_{1} 
eq y_{2}$, so we can divide by $(y_{1}-y_{2})$:$\frac{1}{6}(y_{1}+y_{2})=1 \Rightarrow y_{1}+y_{2}=6$.The imaginary part of $z_{1}+z_{2}$ is $y_{1}+y_{2}=6$.

Let $z_{1}$ and $z_{2}$ be two complex numbers such that $\arg(z_{1}-z_{2})=\frac{\pi}{4}$ and $z_{1}, z_{2}$ satisfy the equation $|z-3|=\operatorname{Re}(z)$. Then the imaginary part of $z_{1}+z_{2}$ is equal to ..... .

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