The equation im(iz-2/z-i)+1=0 represents a part of circle having radius
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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
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
Answered by
10
Answer:
Step-by-step explanation:
This equation represents a circle.
Comparing this equation with
we\:get
g=0, f= -1/4, c= -1/2
Now,
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