# Locate the points representing the complex number ‘z’ for which $\arg \left( \dfrac{z-1-i}{z-2} \right)=\dfrac{\pi }{3}$.

Answer

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Hint: Put ‘z = x+iy’, and now simplify the given expression, $\left( \dfrac{z-1-i}{z-2} \right)$ and convert it to standard form of complex number i.e., ‘a+ib’. Now, use the formula of argument i.e., ${{\tan }^{-1}}\left( \dfrac{b}{a} \right)$ for complex number, ‘a+ib’, to get locus of z.

Complete step-by-step answer:

Let us suppose ‘z = x+iy’ in the given expression.

$\arg \left( \dfrac{z-1-i}{z-2} \right)=\dfrac{\pi }{3}..................\left( i \right)$

Put ‘z = (x+iy)’, we get

$\Rightarrow$ $\arg \left( \dfrac{x+iy-1-i}{x+iy-2} \right)=\dfrac{\pi }{3}$

Now, let us convert the above complex number to ‘a+ib’ by multiplying the conjugate of the denominator. Hence, we get

$\Rightarrow$ $\arg \left( \dfrac{\left( x-1 \right)+i\left( y-1 \right)}{\left( x-2 \right)+iy}\times \dfrac{\left( x-2 \right)-iy}{\left( x-2 \right)-iy} \right)=\dfrac{\pi }{3}$

$\Rightarrow$ $\arg \left( \dfrac{\left( \left( x-1 \right)+i\left( y-1 \right) \right)\times \left( \left( x-2 \right)-iy \right)}{{{\left( x-2 \right)}^{2}}+{{y}^{2}}} \right)=\dfrac{\pi }{3}$

$\Rightarrow$ $\arg \left( \dfrac{\left( x-1 \right)\left( x-2 \right)+y\left( y-1 \right)+i\left( y-1 \right)\left( x-2 \right)-iy\left( x-1 \right)}{{{\left( x-2 \right)}^{2}}+{{y}^{2}}} \right)=\dfrac{\pi }{3}$

Opening the brackets, we get

$\Rightarrow$ $\arg \left( \dfrac{\left( {{x}^{2}}-3x+2+{{y}^{2}}-y \right)+i\left( xy-2y-x+2-xy+y \right)}{{{\left( x-2 \right)}^{2}}+{{y}^{2}}} \right)=\dfrac{\pi }{3}$

$\Rightarrow$ $\arg \left( \dfrac{\left( {{x}^{2}}+{{y}^{2}}-3x-y+2 \right)+i\left( -x-y+2 \right)}{{{\left( x-2 \right)}^{2}}+{{y}^{2}}} \right)=\dfrac{\pi }{3}$

$\Rightarrow$ $\arg \left( \dfrac{\left( {{x}^{2}}+{{y}^{2}}-3x-y+2 \right)}{{{\left( x-2 \right)}^{2}}+{{y}^{2}}}+i\dfrac{\left( -x-y+2 \right)}{{{\left( x-2 \right)}^{2}}+{{y}^{2}}} \right)=\dfrac{\pi }{3}$……………….(ii)

Now, we know that argument of any complex number ‘a+ib’ is given by the relation

$\arg \left( a+ib \right)={{\tan }^{-1}}\left( \dfrac{b}{a} \right)..............\left( iii \right)$

Hence, using equation (iii), we can get argument of equation (ii) as

$\Rightarrow$ ${{\tan }^{-1}}\left( \dfrac{\dfrac{-x-y+2}{{{\left( x-2 \right)}^{2}}+{{y}^{2}}}}{\dfrac{{{x}^{2}}+{{y}^{2}}-3x-y+2}{{{\left( x-2 \right)}^{2}}+{{y}^{2}}}} \right)=\dfrac{\pi }{3}$

Cancelling the like terms and transferring ${{\tan }^{-1}}$ function to other side, hence we get

$\Rightarrow$ $\dfrac{-x-y+2}{{{x}^{2}}+{{y}^{2}}-3x-y+2}=\tan \dfrac{\pi }{3}$

Substituting the value of right hand side, we get

$\Rightarrow$ $\dfrac{-x-y+2}{{{x}^{2}}+{{y}^{2}}-3x-y+2}=\sqrt{3}$

Now, on cross multiplying above relation, we get an equation as

$\Rightarrow$ $\dfrac{-x-y+2}{\sqrt{3}}={{x}^{2}}+{{y}^{2}}-3x-y+2$

${{x}^{2}}+{{y}^{2}}-3x-y+2=\dfrac{-x}{\sqrt{3}}-\dfrac{y}{\sqrt{3}}+\dfrac{2}{\sqrt{3}}$

${{x}^{2}}+{{y}^{2}}+x\left( \dfrac{1}{\sqrt{3}}-3 \right)+y\left( \dfrac{1}{\sqrt{3}}-1 \right)+2-\dfrac{2}{\sqrt{3}}=0...........\left( iv \right)$

Now, on comparing the above equation with the standard equation of circle, i.e.,

${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$

We can observe that equation (iv) is representing an equation of circle, where

$2g=\left( \dfrac{1}{\sqrt{3}}-3 \right),2f=\left( \dfrac{1}{\sqrt{3}}-1 \right),c=2-\dfrac{2}{\sqrt{3}}$

Hence, the locus of point ‘z’ by the given relation $\arg \left( \dfrac{z-1-i}{z-2} \right)=\dfrac{\pi }{3}$ is a circle.

Note: One can prove the locus of the points of ‘z’ from a given equation by using the property of a circle that is angle formed by a chord in the same segment will represent a circle. But this will be a lengthy process.

Hence, given relation can be generalize such that equation

$\arg \left( \dfrac{z-1-i}{z-2} \right)=\theta \left( \theta \ne \pi \right)$or$\left( \theta \ne 0 \right)$

Will always represent a circle where $\theta $ is less than ${{180}^{\circ }}$ .

Complete step-by-step answer:

Let us suppose ‘z = x+iy’ in the given expression.

$\arg \left( \dfrac{z-1-i}{z-2} \right)=\dfrac{\pi }{3}..................\left( i \right)$

Put ‘z = (x+iy)’, we get

$\Rightarrow$ $\arg \left( \dfrac{x+iy-1-i}{x+iy-2} \right)=\dfrac{\pi }{3}$

Now, let us convert the above complex number to ‘a+ib’ by multiplying the conjugate of the denominator. Hence, we get

$\Rightarrow$ $\arg \left( \dfrac{\left( x-1 \right)+i\left( y-1 \right)}{\left( x-2 \right)+iy}\times \dfrac{\left( x-2 \right)-iy}{\left( x-2 \right)-iy} \right)=\dfrac{\pi }{3}$

$\Rightarrow$ $\arg \left( \dfrac{\left( \left( x-1 \right)+i\left( y-1 \right) \right)\times \left( \left( x-2 \right)-iy \right)}{{{\left( x-2 \right)}^{2}}+{{y}^{2}}} \right)=\dfrac{\pi }{3}$

$\Rightarrow$ $\arg \left( \dfrac{\left( x-1 \right)\left( x-2 \right)+y\left( y-1 \right)+i\left( y-1 \right)\left( x-2 \right)-iy\left( x-1 \right)}{{{\left( x-2 \right)}^{2}}+{{y}^{2}}} \right)=\dfrac{\pi }{3}$

Opening the brackets, we get

$\Rightarrow$ $\arg \left( \dfrac{\left( {{x}^{2}}-3x+2+{{y}^{2}}-y \right)+i\left( xy-2y-x+2-xy+y \right)}{{{\left( x-2 \right)}^{2}}+{{y}^{2}}} \right)=\dfrac{\pi }{3}$

$\Rightarrow$ $\arg \left( \dfrac{\left( {{x}^{2}}+{{y}^{2}}-3x-y+2 \right)+i\left( -x-y+2 \right)}{{{\left( x-2 \right)}^{2}}+{{y}^{2}}} \right)=\dfrac{\pi }{3}$

$\Rightarrow$ $\arg \left( \dfrac{\left( {{x}^{2}}+{{y}^{2}}-3x-y+2 \right)}{{{\left( x-2 \right)}^{2}}+{{y}^{2}}}+i\dfrac{\left( -x-y+2 \right)}{{{\left( x-2 \right)}^{2}}+{{y}^{2}}} \right)=\dfrac{\pi }{3}$……………….(ii)

Now, we know that argument of any complex number ‘a+ib’ is given by the relation

$\arg \left( a+ib \right)={{\tan }^{-1}}\left( \dfrac{b}{a} \right)..............\left( iii \right)$

Hence, using equation (iii), we can get argument of equation (ii) as

$\Rightarrow$ ${{\tan }^{-1}}\left( \dfrac{\dfrac{-x-y+2}{{{\left( x-2 \right)}^{2}}+{{y}^{2}}}}{\dfrac{{{x}^{2}}+{{y}^{2}}-3x-y+2}{{{\left( x-2 \right)}^{2}}+{{y}^{2}}}} \right)=\dfrac{\pi }{3}$

Cancelling the like terms and transferring ${{\tan }^{-1}}$ function to other side, hence we get

$\Rightarrow$ $\dfrac{-x-y+2}{{{x}^{2}}+{{y}^{2}}-3x-y+2}=\tan \dfrac{\pi }{3}$

Substituting the value of right hand side, we get

$\Rightarrow$ $\dfrac{-x-y+2}{{{x}^{2}}+{{y}^{2}}-3x-y+2}=\sqrt{3}$

Now, on cross multiplying above relation, we get an equation as

$\Rightarrow$ $\dfrac{-x-y+2}{\sqrt{3}}={{x}^{2}}+{{y}^{2}}-3x-y+2$

${{x}^{2}}+{{y}^{2}}-3x-y+2=\dfrac{-x}{\sqrt{3}}-\dfrac{y}{\sqrt{3}}+\dfrac{2}{\sqrt{3}}$

${{x}^{2}}+{{y}^{2}}+x\left( \dfrac{1}{\sqrt{3}}-3 \right)+y\left( \dfrac{1}{\sqrt{3}}-1 \right)+2-\dfrac{2}{\sqrt{3}}=0...........\left( iv \right)$

Now, on comparing the above equation with the standard equation of circle, i.e.,

${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$

We can observe that equation (iv) is representing an equation of circle, where

$2g=\left( \dfrac{1}{\sqrt{3}}-3 \right),2f=\left( \dfrac{1}{\sqrt{3}}-1 \right),c=2-\dfrac{2}{\sqrt{3}}$

Hence, the locus of point ‘z’ by the given relation $\arg \left( \dfrac{z-1-i}{z-2} \right)=\dfrac{\pi }{3}$ is a circle.

Note: One can prove the locus of the points of ‘z’ from a given equation by using the property of a circle that is angle formed by a chord in the same segment will represent a circle. But this will be a lengthy process.

Hence, given relation can be generalize such that equation

$\arg \left( \dfrac{z-1-i}{z-2} \right)=\theta \left( \theta \ne \pi \right)$or$\left( \theta \ne 0 \right)$

Will always represent a circle where $\theta $ is less than ${{180}^{\circ }}$ .

Last updated date: 16th Sep 2023

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