\(A\left(z_1\right)\) and \(B\left(z_2\right)\) are two points in the argand plane. Then, the locus of the…
\(A\left(z_1\right)\) and \(B\left(z_2\right)\) are two points in the argand plane. Then, the locus of the complex number \(z\) satisfying \(\arg \left(\frac{z-z_1}{z-z_2}\right)=0\) or \(\pi\), is
the circle with \(\overline{A B}\) as a diameter
the ellipse with \(A, B\) as extremities of the major axis
the perpendicular bisector of \(\overline{A B}\)
the straight line passing through the points \(A\) and \(B\)
Solution
Let the point \(z(x, y), z_1\left(x_1, y_1\right)\) and \(z_2\left(x_2, y_2\right)\).
\(\begin{aligned}
& \therefore \quad z-z_1=x+i y-x_1-i y_1 \\
& =x-x_1+i\left(y-y_1\right) \\
& \text {and } \quad z-z_2=x+i y-x_2-i y_2 \\
& =x-x_2+i\left(y-y_2\right)
\end{aligned}\)
Now, \(\frac{z-z_1}{z-z_2}\)
\(\begin{aligned}
& =\frac{\left[\left(x-x_1\right)+i\left(y-y_1\right)\right]}{\left[\left(x-x_2\right)+i\left(y-y_2\right)\right]} \times \frac{\left[\left(x-x_2\right)-i\left(y-y_2\right)\right]}{\left[\left(x-x_2\right)-i\left(y-y_2\right)\right]} \\
& \left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)+ \\
& =\frac{i\left[\left(x-x_2\right)\left(y-y_1\right)-\left(x-x_1\right)\left(y-y_2\right)\right]}{\left(x-x_2\right)^2+\left(y-y_2\right)^2} \\
& =\operatorname{Arg}\left(\frac{z-z_1}{z-z_2}\right)=0 \text { or } \pi \\
\Rightarrow & {\left[\frac{\left[\left(x-x_2\right)\left(y-y_1\right)\right]-\left[\left(x-x_1\right)\left(y-y_2\right)\right]}{\left[\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)\right]}\right]=0 } \\
\Rightarrow & \left(x-x_2\right)\left(y-y_1\right)=\left(x-x_1\right)\left(y-y_2\right) \\
\Rightarrow & x y-x y_1-x_2 y+x_2 y_1=x y-x y_2-x_1 y+x_1 y_2
\end{aligned}\)
\(\Rightarrow x\left(y_2-y_1\right)+y\left(x_1-x_2\right)+\left(x_2 y_1-x_1 y_2\right)=0\)
It represents a straight line passing through the points \(A\) of \(B\).