Question

the equation im(iz-2/z-1)+1=0, represent a part of circle having radius

Answer 1

Step-by-step explanation:

We have,

$\tt{Im \left( \dfrac{iz - 2}{z - 1} \right) = 0}$

Put z = x + iy

$\tt{ \implies \: Im \left( \dfrac{ix - y - 2}{x + iy - 1} \right) = 0}$

$\tt{ \implies \: Im \left[ \dfrac{ - (y + 2) + ix}{(x - 1) + iy } \right] = 0}$

$\tt{ \implies \: Im \left[ \dfrac{ \big \{ - (y + 2) + ix \big \} \big \{ (x - 1) - iy\big \}}{ \big \{(x - 1) + iy \big \} \big \{ (x - 1) - iy\big \}} \right] = 0}$

$\tt{ \implies \: Im \left[ \dfrac{ - (y + 2)(x - 1) + ix (x - 1) + iy(y + 2) - ix \cdot \: iy}{ (x - 1 )^{2} + {y}^{2} } \right] = 0}$

$\tt{ \implies \: Im \left[ \dfrac{ (y + 2)(1 - x) + ix^{2} - ix + iy^{2} + 2yi- i^{2} \: x \: y}{ (x - 1 )^{2} + {y}^{2} } \right] = 0}$

$\tt{ \implies \: Im \left[ \dfrac{y - xy + 2 - 2x + ix^{2} - ix + iy^{2} + 2yi + \: x \: y}{ (x - 1 )^{2} + {y}^{2} } \right] = 0}$

$\tt{ \implies \: Im \left[ \dfrac{y + 2 - 2x + ix^{2} - ix + iy^{2} + 2yi }{ (x - 1 )^{2} + {y}^{2} } \right] = 0}$

$\tt{ \implies \: Im \left[ \dfrac{(y - 2x + 2) + \left(x^{2} + y^{2} - x + 2y \right) i}{ (x - 1 )^{2} + {y}^{2} } \right] = 0}$

$\tt{ \implies \: Im \left[ \dfrac{(y - 2x + 2) }{ (x - 1 )^{2} + {y}^{2} }+ \dfrac{ \left(x^{2} + y^{2} - x + 2y \right) i}{ (x - 1 )^{2} + {y}^{2} } \right] = 0}$

$\tt{ \implies \dfrac{ x^{2} + y^{2} - x + 2y }{ (x - 1 )^{2} + {y}^{2} } = 0}$

$\tt{ \implies x^{2} + y^{2} - x + 2y = 0}$

$\tt{ \green{Required \: \: \: radius} = \sqrt{ \left( \dfrac{1}{2} \right)^{2} + (1)^{2} - 0} }$

$\tt{ \implies\green{Required \: \: \: radius} = \sqrt{ \dfrac{1}{4} + 1} }$

$\tt{ \implies\green{Required \: \: \: radius} = \sqrt{ \dfrac{5}{4} } }$

$\tt{ \implies\green{Required \: \: \: radius} = \dfrac{\sqrt{ 5 }}{2} }$