Question

The equation im(iz-2/z-i)+1=0 represents a part of circle having radius

Answer 1

im{(iz - 2)/(z - i)} + 1 = 0

let, z = a + ib

so, iz = ia + i²b = ia - b

now, (iz - 2)/(z - i) = (ia - b - 2)/(a + ib - i)

= (ia - b - 2)/{a + (b - 1)i}

= (ia - b - 2){a - (b - 1)i}/{a + (b - 1)i}{a - (b - 1)i}

= (ia - b - 2){a - (b - 1)i}/{a² + (b - 1)²}

= {ia² + (ab - a) -ab + (b² - b)i - 2a + 2(b - 1)i}/{a² + (b - 1)²}

= {(a² + b² - b + 2b - 2)i -3a}/{a² + (b - 1)²}

so, im{iz - 2/z - i} = (a² + b² + b - 2)/{a² + (b - 1)²

(a² + b² + b - 2 )/(a² + (b - 1)²) + 1 = 0

or, (a² + b² + b - 2) + a² + (b - 1)² = 0

or, 2a² + 2b² + b - 2 - 2b + 1 = 0

or, 2a² + 2b² - b - 1 = 0

or, a² + b² - b/2 - 1/2 = 0

or, a² + b² - b = 0

hence, radius is 3/4

Answer 2

Answer:

Step-by-step explanation:

$Im(\frac{iz-2}{z-i})=-1\\\\Let\:z\:=\:x+iy$

$\frac{iz-2}{z-i}\\\\=\frac{i(x+iy)-2}{x+iy-i}\\\\=\frac{ix-y-2}{x+iy-i}\\\\=\frac{-(y+2)+ix}{x+i(y-1)}\\\\=\frac{-(y+2)+ix}{x+i(y-1)}.\frac{x-i(y-1)}{x-i(y-1)}\\\\=\frac{[-(y+2)+ix][x-i(y-1)]}{x^2+(y-1)^2}\\\\=\frac{-(y+2)x+x(y-1)+i[x^2+(y-1)(y+2)]}{x^2+(y-1)^2}$

$But\:Im(\frac{iz-2}{z-i})=-1\\\\\frac{x^2+(y-1)(y+2)}{x^2+(y-1)^2}=-1\\\\x^2+(y-1)(y+2)=-x^2-(y-1)^2\\\\x^2+y^2+y-2+x^2+y^2-2y+1=0\\\\2x^2+2y^2-y-1=0\\\\x^2+y^2-\frac{1}{2}y-\frac{1}{2}=0$

This equation represents a circle.

Comparing this equation with

$x^2+y^2+2gx+2fy+c=0$

we\:get

g=0, f= -1/4, c= -1/2

Now,

$radius\:=\sqrt{g^2+f^2-c}\\\\radius\:=\sqrt{\frac{1}{16}+\frac{1}{2}}\\\\radius\:=\sqrt{\frac{9}{16}}\\\\radius\:=\frac{3}{4}$