Question

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

Hint: Put â€˜z = x+iyâ€™, 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., â€˜a+ibâ€™. Now, use the formula of argument i.e., ${{\tan }^{-1}}\left( \dfrac{b}{a} \right)$ for complex number, â€˜a+ibâ€™, to get locus of z.

Let us suppose â€˜z = x+iyâ€™ in the given expression.
$\arg \left( \dfrac{z-1-i}{z-2} \right)=\dfrac{\pi }{3}..................\left( i \right)$
Put â€˜z = (x+iy)â€™, 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 â€˜a+ibâ€™ 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
$\Rightarrow$ $\arg \left( \dfrac{\left( {{x}^{2}}-3x+2+{{y}^{2}}-y \right)+i\left( xy-2y-x+2-xy+y \right)}{{{\left( x-2 \right)}^{2}}+{{y}^{2}}} \right)=\dfrac{\pi }{3}$
$\Rightarrow$ $\arg \left( \dfrac{\left( {{x}^{2}}+{{y}^{2}}-3x-y+2 \right)+i\left( -x-y+2 \right)}{{{\left( x-2 \right)}^{2}}+{{y}^{2}}} \right)=\dfrac{\pi }{3}$
$\Rightarrow$ $\arg \left( \dfrac{\left( {{x}^{2}}+{{y}^{2}}-3x-y+2 \right)}{{{\left( x-2 \right)}^{2}}+{{y}^{2}}}+i\dfrac{\left( -x-y+2 \right)}{{{\left( x-2 \right)}^{2}}+{{y}^{2}}} \right)=\dfrac{\pi }{3}$â€¦â€¦â€¦â€¦â€¦â€¦.(ii)
Now, we know that argument of any complex number â€˜a+ibâ€™ 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 â€˜zâ€™ 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 â€˜zâ€™ 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 ${{180}^{\circ }}$ .