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Mathematics Question on argand plane

The perimeter of the locus represented by arg (z+iz−i)=π4\left( \frac{z + i}{z-i}\right) = \frac{\pi}{4} is equal to

A

4π4 \pi

B

2π22 \pi \sqrt{2}

C

2π32 \pi \sqrt{3}

D

2π3\frac{2 \pi}{\sqrt{3}}

Answer

2π22 \pi \sqrt{2}

Explanation

Solution

We have, arg⁡(z+iz−i)=π4\arg \left(\frac{z+i}{z-i}\right)=\frac{\pi}{4}
Let Z=x+iyZ = x + i y
∴z+iz−i=x+iy+ix+iy−i\therefore \frac{ z + i }{ z - i } =\frac{ x + i y+ i }{ x + i y- i }
=x+(y+l)ix+(y−l)i=\frac{ x +( y + l ) i }{ x +( y - l ) i }
=(x+(y+l)i)(x−(y−l)i)(x+(y−1)i)(x−(y−l)i)=\frac{( x +( y + l ) i )( x -( y - l ) i )}{( x +(y-1) i )( x -( y - l ) i )}
=x2+y2−1+(−xy+x+xy+x)ix2+(y−1)2=\frac{ x ^{2}+ y ^{2}-1+(- x y+ x + x y + x ) i }{ x ^{2}+( y -1)^{2}}
=x2+y2−l+2xix2+(y−l)2=\frac{ x ^{2}+ y ^{2}- l +2 xi }{ x ^{2}+( y - l )^{2}}
x2+(y−1)2x^{2}+(y-1)^{2}
arg⁡(x2+y2−1+2xix2+(y−1)2)=tan⁡−1(2xx2+y2−1)\arg \left(\frac{x^{2}+y^{2}-1+2 x i}{x^{2}+(y-1)^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{x^{2}+y^{2}-1}\right)
∴tan⁡−1(2xx2+y2−1)=π4\therefore \tan ^{-1}\left(\frac{2 x}{x^{2}+y^{2}-1}\right)=\frac{\pi}{4}
⇒2x=x2+y2−1\Rightarrow 2 x=x^{2}+y^{2}-1
⇒(x−1)2+y2=(2)2\Rightarrow (x-1)^{2}+y^{2}=(\sqrt{2})^{2}
∴\therefore Locus of arg⁡(z+iz−i)=π4\arg \left(\frac{z+i}{z-i}\right)=\frac{\pi}{4} is circle,
whose centre (1,0)(1,0) and radius =2=\sqrt{2}
Perimetre of circle =2πr=2π(2)=2 \pi r =2 \pi(\sqrt{2})
=2π2=2 \pi \sqrt{2}