Question

JEE Mains maths question is attached

Answer 1

Answer:

$Given: | z_1 - 1 |= Re(z_1)\:\:\: and\:\:\: |z_2 - 1| = Re(z_2)\\\\\text{Let's assume} \;z_1 = x_1 + iy_1\:\: and\:\: z_2 = x_2 + iy_2\\\\\\\rightarrow | x_1 + iy_1 - 1| = x_1\\\\\rightarrow ( x_1 - 1)^2 + y_1^2 = (x_1)^2\\\\\rightarrow x_1^2 + 1 - 2x_1 + y_1^2 = x_1^2\\\\\rightarrow y_1^2 -2x_1 + 1 = 0 \:\:\; \rightarrow Eqn.\:1\\\\\text{Similarly following the same process for Second equation we get:}\\\\\rightarrow y_1^2 + 1 - 2x_2^2 = 0 \:\;\: \rightarrow Eqn.\:2$

Solving the 2 equations we get:

$\rightarrow y_1^2 + 1 - 2x_1 = y_2^2 + 1 - 2x_2 \\\\\rightarrow y_1^2 - y_2^2 = 2x_1 - 2x_2\\\\\rightarrow ( y_1 + y_2)(y_1 - y_2) = 2 ( x_1 - x_2 )\\\\\rightarrow ( y_1 + y_2) = \dfrac{2 (x_1 - x_2)}{y_1 - y_2} \;\:\:\rightarrow Eqn.\:3$.

Also, it is given that: arg(z₁ - z₂) = π/3

$\rightarrow tan^{-1} [\frac{Im (z_1 - z_2)}{Re(z_1 - z_2)}]\\\\\rightarrow tan^{-1} ( \dfrac{ y_1 - y_2}{x_1 - x_2}) = \dfrac{ \pi}{3}\\\\\rightarrow \dfrac{ y_1 - y_2}{x_1 - x_2} = tan \dfrac{\pi}{3}\\\\\rightarrow \dfrac{ y_1 - y_2}{x_1 - x_2} = \sqrt{3}\\\\\rightarrow \dfrac{x_1 - x_2}{y_1-y_2} = \dfrac{1}{\sqrt{3}}$

Substituting this in Eqn. 3 we get:

$\boxed{\bf{( y_1 + y_2 ) = \dfrac{2}{\sqrt{3}}}}$

This is the required answer.

Answer 2

Answer:

Given:∣z

1

−1∣=Re(z

1

)and∣z

2

−1∣=Re(z

2

)

Let’s assumez

1

=x

1

+iy

1

andz

2

=x

2

+iy

2

arrow∣x

1

+iy

1

−1∣=x

1

arrow(x

1

−1)

2

+y

1

2

=(x

1

)

2

arrowx

1

2

+1−2x

1

+y

1

2

=x

1

2

arrowy

1

2

−2x

1

+1=0arrowEqn.1

Similarly following the same process for Second equation we get:

arrowy

1

2

+1−2x

2

2

=0arrowEqn.2

Solving the 2 equations we get:

\begin{gathered}arrow y_1^2 + 1 - 2x_1 = y_2^2 + 1 - 2x_2 \\\\arrow y_1^2 - y_2^2 = 2x_1 - 2x_2\\\\arrow ( y_1 + y_2)(y_1 - y_2) = 2 ( x_1 - x_2 )\\\\arrow ( y_1 + y_2) = \dfrac{2 (x_1 - x_2)}{y_1 - y_2} \;\:\:arrow Eqn.\:3\end{gathered}

arrowy

1

2

+1−2x

1

=y

2

2

+1−2x

2

arrowy

1

2

−y

2

2

=2x

1

−2x

2

arrow(y

1

+y

2

)(y

1

−y

2

)=2(x

1

−x

2

)

arrow(y

1

+y

2

)=

y

1

−y

2

2(x

1

−x

2

)

arrowEqn.3

.

Also, it is given that: arg(z₁ - z₂) = π/3

\begin{gathered}arrow tan^{-1} [\frac{Im (z_1 - z_2)}{Re(z_1 - z_2)}]\\\\arrow tan^{-1} ( \dfrac{ y_1 - y_2}{x_1 - x_2}) = \dfrac{ \pi}{3}\\\\arrow \dfrac{ y_1 - y_2}{x_1 - x_2} = tan \dfrac{\pi}{3}\\\\arrow \dfrac{ y_1 - y_2}{x_1 - x_2} = \sqrt{3}\\\\arrow \dfrac{x_1 - x_2}{y_1-y_2} = \dfrac{1}{\sqrt{3}}\end{gathered}

arrowtan

−1

[

Re(z

1

−z

2

)

Im(z

1

−z

2

)

]

arrowtan

−1

(

x

1

−x

2

y

1

−y

2

)=

3

π

arrow

x

1

−x

2

y

1

−y

2

=tan

3

π

arrow

x

1

−x

2

y

1

−y

2

=

3

arrow

y

1

−y

2

x

1

−x

2

=

3

1

Substituting this in Eqn. 3 we get:

\boxed{\bf{( y_1 + y_2 ) = \dfrac{2}{\sqrt{3}}}}

(y

1

+y

2

)=

3

2

This is the required answer.

Step-by-step explanation:

mark as BRAINLEST ANSWER please please