Question

The locus of $z$ satisfying the equation $arg$ $\left(\frac{z-2}{z+2}\right)=\frac{\pi }{3}$ is a portion of circle, then the length of that portion of circumference of the circle is equal to

Answer 1

$\begin{array}{c}z=x+iy\\ z-2=\left(x-2\right)+iy\\ z+2=\left(x+2\right)+iy\\ \frac{z-2}{z+2}=\frac{\left[\left(x-2\right)+iy\right]\left[\left(x-2\right)-iy\right]}{\left[\left(x+2\right)+iy\right]\left[\left(x+2\right)-iy\right]}\\ \frac{z-2}{z+2}=\frac{\left(x^2-4\right)+y^2+i\left(yx+2y\right)+i\left(-yx+2y\right)}{{\left(x+2\right)}^2+y^2}\\ \frac{z-2}{z+2}=\frac{\left(x^2+y^2-4\right)+i\left(4y\right)}{{\left(x+2\right)}^2+y^2}\\ \left(\frac{z-2}{z+2}\right)=\frac{\frac{4y}{{\left(x+2\right)}^2+y^2}}{\frac{x^2+y^2-4}{{\left(x+2\right)}^2+y^2}}=\frac{4y}{x^2+y^2-4}=\sqrt{3}\\ \left(x^2+y^2-4\right)=\frac{4y}{\sqrt{3}}\\ x^2+y^2-\frac{4y}{\sqrt{3}}-4=0\\ x^2{\left(y-\frac{2}{\sqrt{3}}\right)}^2=4+\frac{4}{3}\\ x^2+{\left(\frac{y-2}{\sqrt{3}}\right)}^2=\frac{16}{3}\\ x^2{\left(\frac{y-2}{\sqrt{3}}\right)}^2={\left(\frac{4}{\sqrt{3}}\right)}^2\\ Lengthofcircumference=2\pi \left(\frac{4}{\sqrt{3}}\right)\\ =\frac{8\pi }{\sqrt{3}}\end{array}$