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

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Complex Numbers

Let

z_{1}

and

z_{2}

be two complex numbers such that

ar g (z_{1} - z_{2}) = \frac{π}{4}

and

z_{1}, z_{2}

satisfy the equation

∣ z - 3∣ = Re (z)

. Then the imaginary part of

z_{1} + z_{2}

is equal to

Answer: 6

Writing

z = x + i y

, the condition

∣ z - 3∣ = Re (z)

gives

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

, i.e.

y^{2} = 6 x - 9

, a parabola. For points

z_{1} = x_{1} + i y_{1}

and

z_{2} = x_{2} + i y_{2}

on it,

x_{j} = \frac{y _{j}^{2} + 9}{6}

, so

x_{1} - x_{2} = \frac{( y _{1} - y _{2} ) ( y _{1} + y _{2} )}{6} .

Since

ar g (z_{1} - z_{2}) = \frac{π}{4}

, the chord has slope

1

\frac{y _{1} - y _{2}}{x _{1} - x _{2}} = \frac{6}{y _{1} + y _{2}} = 1 \Rightarrow y_{1} + y_{2} = 6.

The imaginary part of

z_{1} + z_{2}

6

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Integer answer