Question

Determine what the equation $\left| z-i \right|=\left| z-1 \right|$ represents given that $i=\sqrt{-1}$.
(a) The line through the origin with slope $-1$
(b) A circle with radius 1
(c) A circle with radius $\dfrac{1}{2}$
(d) The line through the origin with slope 1

Answer 1

In this question, in order to Determine what the equation $\left| z-i \right|=\left| z-1 \right|$ represents given that $i=\sqrt{-1}$ we have to first substitute the value of the complex number $z$ as $z=x+iy$ in the given equation $\left| z-i \right|=\left| z-1 \right|$ where $x$ is the real part of the complex number $z$ and $y$ is the it’s imaginary part . Now for any complex number $z$, modulus of $z=x+iy$ is defined as $\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$. Using this in the given equation we will get an equation in variables $x$ and $y$. We have to determine what that equation represents.

Complete step by step answer:
We are given with the equation $\left| z-i \right|=\left| z-1 \right|$ where $i=\sqrt{-1}$ and $z$ is a complex number.
We will now substitute the value of the complex number $z$ as $z=x+iy$ in the given equation $\left| z-i \right|=\left| z-1 \right|$ where $x$ is the real part of the complex number $z$ and $y$ is the imaginary part.
Then we get
$\left| \left( x+iy \right)-i \right|=\left| \left( x+iy \right)-1 \right|$
On simplifying the above equation we get
$\left| x+i\left( y-1 \right) \right|=\left| \left( x-1 \right)+iy \right|$
On squaring both sides we get,
${{\left| x+i\left( y-1 \right) \right|}^{2}}={{\left| \left( x-1 \right)+iy \right|}^{2}}...........(1)$
We will now find the value of ${{\left| x+i\left( y-1 \right) \right|}^{2}}$ .
Since we know that for any complex number $z$, modulus of $z=a+ib$ is defined as $\left| z \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}$.
On comparing the values of $\left| x+i\left( y-1 \right) \right|$ with $\left| z \right|=\left| a+ib \right|$, we will have
$a=x$ and $b=y-1$
Therefore we have

& {{\left| x+i\left( y-1 \right) \right|}^{2}}={{\left( \sqrt{{{x}^{2}}+{{\left( y-1 \right)}^{2}}} \right)}^{2}} \\\ & ={{x}^{2}}+{{\left( y-1 \right)}^{2}}.........(2) \end{aligned}$$ On comparing the values of $$\left| \left( x-1 \right)+iy \right|$$ with $$\left| z \right|=\left| a+ib \right|$$, we will have $$a=x-1$$ and $$b=y$$ Therefore we have $$\begin{aligned} & {{\left| \left( x-1 \right)+iy \right|}^{2}}={{\left( \sqrt{{{\left( x-1 \right)}^{2}}+{{y}^{2}}} \right)}^{2}} \\\ & ={{\left( x-1 \right)}^{2}}+{{y}^{2}}.........(3) \end{aligned}$$ Now on substituting the values in equation (2) and equation (3) in equation (1), we will have $${{x}^{2}}+{{\left( y-1 \right)}^{2}}={{\left( x-1 \right)}^{2}}+{{y}^{2}}$$ On expanding the above equation we get $${{x}^{2}}+{{y}^{2}}-2y+1={{x}^{2}}-2x+1+{{y}^{2}}$$ Now on cancelling out the common factors on both the right hand side and left hand side of the above equation we get $$-2y=-2x$$ On dividing the above equation by $$-2$$, we get $$y=x$$ Now if we substitute the value $$x=0$$ in the equation $$y=x$$ we get $$y=0$$ Also in order to calculate the slope of the line $$y=x$$, we have to calculate the derivative $$\dfrac{dy}{dx}$$. Here we have $$\dfrac{dy}{dx}=1$$ Which is the equation of the line passing the origin $$\left( 0,0 \right)$$ with slope 1.  **So, the correct answer is “Option D”.** **Note:** In this problem, when we arrive at an equation in variables $$x$$ and $$y$$ we have to determine what that equation represents. In our case we have a linear equation in variables $$x$$ and $$y$$ which is of the form $$y=mx+c$$ where $$m$$ represents the slope of the line.