Let z_{1}$ and $z_{2} be two complex numbers such that arg \left(z_{1}-z_{2}\right)=\frac{\pi}{4}$ 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__________.
 

Answers (1)

z= x+iy\\

\text { given }|z-3|=\operatorname{Re}(z) \\

\Rightarrow \sqrt{(x-3)^{2}+y^{2}}=x \\

\Rightarrow x^{2}-6 x+9+y^{2}=x^{2}

\Rightarrow y^{2}=6\left(x-\frac{3}{2}\right)

\Rightarrow z_{1} \: and\:z_{2}  lie on parabola y^{2}=6 \left ( x-3/2 \right )

Also \operatorname{Arg}\left(z_{1}-z_{2}\right)=\pi / 4

Let \,\,z_{1}=\left(\frac{3}{2}+\frac{3}{2} t_{1}^{2}, 2 \times \frac{3}{2} t_{1}\right) \&\: z_{2}=\left(\frac{3}{2}+\frac{3}{2} t_{2}^{2}, 2 \times \frac{3}{2} t_{2}\right)

Slope of line joining  z_{1} \& z_{2}= \tan \pi / 4=1

\Rightarrow \frac{3\left(t_{2}-t_{1}\right)}{\frac{3}{2}\left(t_{2}{ }^{2}-t_{1}{ }^{2}\right)}=1 \Rightarrow t_{1}+t_{2}=2 \text {. }

Imaginary part of z_{1}+z_{2}

= 3\left ( t_{1}+t_{2} \right )

= 3\times 2= 6

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