Question

If z = x + iy, \[w=\dfrac{2-iz}{2z-i}\] and \[\left| w \right|=1\], find the locus of z and illustrate it in the Argand plane.

Answer 1

Hint:  In this question, we will put the value of z in w, and then we will simplify it and will find the modulus of w and will equate it to 1. And then simplify to find the locus of z and with the help of locus, we will illustrate it in argand plane.

Complete step-by-step answer:

In this question, we have to find the locus of z, and then we have to illustrate it in the argand plane. So, to find the locus of z, we will put the value of z, that is z = x + iy in w and then we will simplify it. So, we can write it as

\[w=\dfrac{2-i\left( x+iy \right)}{2\left( x+iy \right)-i}\]

Now, we will open the brackets to simplify. So, we will get, 

\[w=\dfrac{2-ix-{{i}^{2}}y}{2x+2iy-i}\]

Now, we know that \[{{i}^{2}}=-1\] and 2iy – i can be written as i(2y – 1). By using these values, we will get, 

\[w=\dfrac{2-ix-\left( -1 \right)y}{2x+i\left( 2y-1 \right)}\]

\[w=\dfrac{2-ix+y}{2x+i\left( 2y-1 \right)}\]

\[w=\dfrac{2+y-ix}{2x+i\left( 2y-1 \right)}\]

Now, we will take the modulus on both sides of the equation. So, we will get, 

\[\left| w \right|=\left| \dfrac{\left( 2+y \right)-ix}{2x+i\left( 2y-1 \right)} \right|\]

We have been given that \[\left| w \right|=1\] and we also know that \[\left| \dfrac{a}{b} \right|=\dfrac{\left| a \right|}{\left| b \right|}\]. Therefore, we can write the above equality as, 

\[1=\dfrac{\left| \left( 2+y \right)-ix \right|}{\left| 2x+i\left( 2y-1 \right) \right|}\]

Now, we will cross multiply the equation. So, we will get, 

\[\left| 2x+i\left( 2y-1 \right) \right|=\left| \left( 2+y \right)-ix \right|\]

Now, we know that |a + ib| is equal to \[\sqrt{{{a}^{2}}+{{b}^{2}}}\]. So, we will get, 

\[\sqrt{{{\left( 2x \right)}^{2}}+{{\left( 2y-1 \right)}^{2}}}=\sqrt{{{\left( 2+y \right)}^{2}}+{{\left( -x \right)}^{2}}}\]

Now, we will square both the sides of the equation, so we will get, 

\[{{\left( 2x \right)}^{2}}+{{\left( 2y-1 \right)}^{2}}={{\left( 2+y \right)}^{2}}+{{\left( x \right)}^{2}}\]

Now, we will open the brackets by squaring the terms and by applying \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\] and \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\]. So, we will get, 

\[4{{x}^{2}}+4{{y}^{2}}+1-4y=4+{{y}^{2}}+4y+{{x}^{2}}\]

Now, we will take all the terms on the left side of the equation. So, we will get

\[4{{x}^{2}}-{{x}^{2}}+4{{y}^{2}}-{{y}^{2}}-4y-4y+1-4=0\]

\[\Rightarrow 3{{x}^{2}}+3{{y}^{2}}-8y-3=0\]

Now, we will divide the whole equation by 3. So, we will get, 

\[{{x}^{2}}+{{y}^{2}}-\dfrac{8}{3}y-1=0\]

Now, we can see that the equation formed is similar to the general equation of the circle that is,

\[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\]

 So, we can say that the locus of z is a circle with equation \[{{x}^{2}}+{{y}^{2}}-\dfrac{8}{3}y-1=0\].

And we know that the center of the circle is given by (– g, – f) and the radius of the circle is given by \[\sqrt{{{g}^{2}}+{{f}^{2}}-c}\].

Now, on comparing both the equations, we can say, c = –1, g = 0 and \[f=\dfrac{4}{3}.\]

So, we can say that the center and radius of the circle is \[\left( 0,\dfrac{4}{3} \right)\] and \[\sqrt{0+\dfrac{16}{9}+1}=\dfrac{5}{3}\] respectively. 

On the Argand plane, we can illustrate it as

Note: The possible mistake one can make in this question is by rationalizing the denominator at the place where we took modulus of the equation. That is not wrong but that will make the solution more lengthy and complicated.