Question

The equation $\text{Im}\left( \dfrac{iz-2}{z-i} \right)+1=0$ , $z\in C$,$z\ne i$ represents a part of a circle having radius equal to $$$$
A.\dfrac{1}{2}$$$$$ B.1 C.$2
D. $\dfrac{3}{4}$$$$$

Answer 1

Put the standard of any complex number $z=x+iy$ in the given equation. Simplify the equation with help of conjugate of a complex number and compare the simplified equation with the general equation of circle to find out the radius.

Complete step by step answer:
We know that any complex number $z=x+iy$ where $x$and $y$ are real numbers then the real part of $z$ is returned by the function Re($z$)=$x$ and imaginary part of $z$ is returned by the function Im($z$)=y. The conjugate of $z$ is given by $\overline{z}=x-iy$. We can deduce that $z\overline{z}=\left( x+iy \right)\left( x-iy \right)={{x}^{2}}+{{y}^{2}}$ which is a real positive number.
The given equation in the complex plane is
$\text{Im}\left( \dfrac{iz-2}{z-i} \right)+1=0....(1)$
We put $z=x+iy$ in the above equation to get,

& \text{Im}\left( \dfrac{i\left( x+iy \right)-2}{\left( x+iy \right)-i} \right)+1=0 \\\ & \Rightarrow \text{Im}\left( \dfrac{ix-y-2}{x+\left( y-1 \right)i} \right)+1=0 \\\ \end{aligned}$$ Multiplying the conjugate of the denominator within the bracket, $$\begin{aligned} & \Rightarrow \operatorname{Im}\left( \dfrac{ix-y-2}{x+\left( y-1 \right)i}\cdot \dfrac{x-\left( y-1 \right)i}{x-\left( y-1 \right)i} \right)+1=0 \\\ & \Rightarrow \operatorname{Im}\left( \dfrac{i{{x}^{2}}+i(y-1)(y+2)-x(y+2)+x(y-1)}{{{x}^{2}}+{{\left( y-1 \right)}^{2}}} \right)+1=0 \\\ \end{aligned}$$ The function $\operatorname{Im}\left( z \right)$ returns only the imaginary part of the complex number without $i$. So the above equation transforms to, $$\begin{aligned} & \Rightarrow \dfrac{{{x}^{2}}+(y-1)(y+2)}{{{x}^{2}}+{{\left( y-1 \right)}^{2}}}+1=0 \\\ & \Rightarrow {{x}^{2}}+{{y}^{2}}+y-2+{{x}^{2}}+{{y}^{2}}-2y+1=0 \\\ & \Rightarrow 2{{x}^{2}}+2{{y}^{2}}-y-1=0 \\\ & \Rightarrow {{x}^{2}}+{{y}^{2}}-\dfrac{1}{2}y-\dfrac{1}{2}=0...(2) \\\ \end{aligned}$$ The general equation of circle in two dimensions is given by $${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$$ Comparing the coefficients and the constant term in the equation (2) we get, $g=0,f=\dfrac{1}{4},c=\dfrac{-1}{2}$. We know that the radius of the circle from the general equation is given by $$r=\sqrt{{{g}^{2}}+{{f}^{2}}-c}=\sqrt{\dfrac{1}{16}+\dfrac{1}{2}}=\dfrac{3}{4}$$ So the length of the radius is $\dfrac{3}{4}$. **So, the correct answer is “Option D”.** **Note:** The question tests your knowledge of modulus of complex numbers and the general equation of circle in two dimensions. Careful solving of simultaneous equations, substitution and usage of formula will lead us to arrive at the correct result. You can also solve for equations involving more than one complex variable.