Question

The complex numbers \[z_1\] , \[z_2\] and \[z_3\] satisfying \[\dfrac{{\left( {z_1 - z_3} \right)}}{{\left( {z_2 - z_3} \right)}} = \dfrac{{\left( {1 - \sqrt 3 i} \right)}}{2}\] are the vertices of a triangle which is
\[\left( 1 \right)\] \[\text{of area zero}\]
\[\left( 2 \right)\] \[\text{right angled isosceles}\]
\[\left( 3 \right)\] \[\text{equilateral}\]
\[\left( 4 \right)\] \[\text{obtuse angled isosceles}\]

Answer 1

Hint: We have to find an ordered pair of \[\left( {x,{\text{ }}y} \right)\]. We solve this question using the concept of the cube root of unity . We should also have the knowledge of the identities of complex numbers . Firstly we have to make the equation in terms of one of the roots of unity and then comparing both the sides and then evaluating the value of \[x\] and \[y\] .

Complete step-by-step solution:
Given :
\[\dfrac{{\left( {z_1 - z_3} \right)}}{{\left( {z_2 - z_3} \right)}} = \dfrac{{\left( {1 - \sqrt 3 i} \right)}}{2}\]
Rationalising the numerator ,
multiplying the numerator and denominator by \[\left( {1 - \sqrt 3 i} \right)\], we get
\[\dfrac{{z_1 - z_3}}{{z_2 - z_3}} = \dfrac{{1 - \sqrt 3 i}}{2} \times \dfrac{{1 + \sqrt 3 i}}{{1 + \sqrt 3 i}}\]
\[\dfrac{{z_1 - z_3}}{{z_2 - z_3}} = \dfrac{{1 - 3{i^2}}}{{2 \times \left( {1 + \sqrt 3 i} \right)}}\]
We , know that $i = \sqrt { - 1} $
And ${i^2} = - 1$
\[\dfrac{{z_1 - z_3}}{{z_2 - z_3}} = \dfrac{{1 + 3}}{{2 \times \left( {1 + \sqrt 3 i} \right)}}\]
On simplifying , we get
\[\dfrac{{z_1 - z_3}}{{z_2 - z_3}} = \dfrac{2}{{\left( {1 + \sqrt 3 i} \right)}}\]
Taking reciprocal , we get
\[\dfrac{{z_2 - z_3}}{{z_1 - z_3}} = \dfrac{{1 + \sqrt 3 i}}{2}\]
As we know that ,
\[\cos \dfrac{\pi }{3} = \dfrac{1}{2}\]
\[\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}\]
Using these values , we get the expression as
\[\dfrac{{z_2 - z_3}}{{z_1 - z_3}} = \cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}\]
Taking magnitude of the expression , we get
\[\left| {\dfrac{{z_2 - z_3}}{{z_1 - z_3}} = \cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}} \right|\]
We know that the magnitude of the complex number \[\left| z \right| = \sqrt {{x^2} + {y^2}} \] .
Where , \[z = x + iy\]
Using this , we get the magnitude as :
\[\left| {\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}} \right| = \sqrt {{{\cos }^2}\dfrac{\pi }{3} + {{\sin }^2}\dfrac{\pi }{3}} \]
Also , we know that the value of sine function and cosine function is given as :
\[{\cos ^2}x + {\sin ^2}x = 1\]
Using the above formula , we get the value of the magnitude as :
\[\left| {\dfrac{{z_2 - z_3}}{{z_1 - z_3}}} \right| = 1\]
Also , the argument of the complex number \[\arg \left( z \right) = {\tan ^{ - 1}}\dfrac{y}{x}\] .
Using this formula we get , the value of the argument as :
\[\arg \left( {\dfrac{{z_2 - z_3}}{{z_1 - z_3}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{\sin \dfrac{\pi }{3}}}{{\cos \dfrac{\pi }{3}}}} \right)\]
\[\arg \left( {\dfrac{{z_2 - z_3}}{{z_1 - z_3}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{\sqrt 3 }}{2}}}{{\dfrac{1}{2}}}} \right)\]
Also on simplifying , we get the value as :
\[\arg \left( {\dfrac{{z_2 - z_3}}{{z_1 - z_3}}} \right) = {\tan ^{ - 1}}\sqrt 3 \]
Also , we know that the value of tangent function is given as :
\[\tan \dfrac{\pi }{3} = \sqrt 3 \]
using the value , we get the value of argument as :
\[\arg \left( {\dfrac{{z_2 - z_3}}{{z_1 - z_3}}} \right) = \dfrac{\pi }{3}\]
Thus , the triangle is an equilateral triangle as the value of the argument is \[\dfrac{\pi }{3}\] .
Hence , the correct option is \[\left( 3 \right)\].

Note: A number of the form \[a + ib\] , where a and b are real numbers , is called a complex number , a is called the real part and b is called the imaginary part of the complex number .
Every real number can be represented in terms of complex numbers but the converse is not true .
Since \[{b^2} - 4ac\] determines whether the quadratic equation a \[{x^2} + bx + c = 0\]
If \[{b^2} - 4ac < 0\] then the equation has imaginary roots .
The polar form of the complex number \[z = x + iy\] is \[r\left( {\cos t + i\sin t} \right)\], where $r = \sqrt {{x^2} + {y^2}} = |z|$ and $\cos t = \dfrac{x}{r}$ , \[\sin \dfrac{y}{r}\] . ($t$ is known as the argument of $z$) The value of $t$ , such that \[ - \pi < t \leqslant \pi \], is called the principal argument of $z$ .