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Question: The complex numbers \[z_1\] , \[z_2\] and \[z_3\] satisfying \[\dfrac{{\left( {z_1 - z_3} \right)}}{...

The complex numbers z1z_1 , z2z_2 and z3z_3 satisfying (z1−z3)(z2−z3)=(1−3i)2\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
(1)\left( 1 \right) of area zero\text{of area zero}
(2)\left( 2 \right) right angled isosceles\text{right angled isosceles}
(3)\left( 3 \right) equilateral\text{equilateral}
(4)\left( 4 \right) obtuse angled isosceles\text{obtuse angled isosceles}

Explanation

Solution

We have to find an ordered pair of (x, y)\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 xx and yy .

Complete step-by-step solution:
Given :
(z1−z3)(z2−z3)=(1−3i)2\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 (1−3i)\left( {1 - \sqrt 3 i} \right), we get
z1−z3z2−z3=1−3i2×1+3i1+3i\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}}
z1−z3z2−z3=1−3i22×(1+3i)\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=−1i = \sqrt { - 1}
And i2=−1{i^2} = - 1
z1−z3z2−z3=1+32×(1+3i)\dfrac{{z_1 - z_3}}{{z_2 - z_3}} = \dfrac{{1 + 3}}{{2 \times \left( {1 + \sqrt 3 i} \right)}}
On simplifying , we get
z1−z3z2−z3=2(1+3i)\dfrac{{z_1 - z_3}}{{z_2 - z_3}} = \dfrac{2}{{\left( {1 + \sqrt 3 i} \right)}}
Taking reciprocal , we get
z2−z3z1−z3=1+3i2\dfrac{{z_2 - z_3}}{{z_1 - z_3}} = \dfrac{{1 + \sqrt 3 i}}{2}
As we know that ,
cos⁡π3=12\cos \dfrac{\pi }{3} = \dfrac{1}{2}
sin⁡π3=32\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}
Using these values , we get the expression as
z2−z3z1−z3=cos⁡π3+isin⁡π3\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
∣z2−z3z1−z3=cos⁡π3+isin⁡π3∣\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 ∣z∣=x2+y2\left| z \right| = \sqrt {{x^2} + {y^2}} .
Where , z=x+iyz = x + iy
Using this , we get the magnitude as :
∣cos⁡π3+isin⁡π3∣=cos⁡2π3+sin⁡2π3\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⁡2x+sin⁡2x=1{\cos ^2}x + {\sin ^2}x = 1
Using the above formula , we get the value of the magnitude as :
∣z2−z3z1−z3∣=1\left| {\dfrac{{z_2 - z_3}}{{z_1 - z_3}}} \right| = 1
Also , the argument of the complex number arg⁡(z)=tan⁡−1yx\arg \left( z \right) = {\tan ^{ - 1}}\dfrac{y}{x} .
Using this formula we get , the value of the argument as :
arg⁡(z2−z3z1−z3)=tan⁡−1(sin⁡π3cos⁡π3)\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⁡(z2−z3z1−z3)=tan⁡−1(3212)\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⁡(z2−z3z1−z3)=tan⁡−13\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⁡π3=3\tan \dfrac{\pi }{3} = \sqrt 3
using the value , we get the value of argument as :
arg⁡(z2−z3z1−z3)=π3\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 π3\dfrac{\pi }{3} .
Hence , the correct option is (3)\left( 3 \right).

Note: A number of the form a+iba + 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 b2−4ac{b^2} - 4ac determines whether the quadratic equation a x2+bx+c=0{x^2} + bx + c = 0
If b2−4ac<0{b^2} - 4ac < 0 then the equation has imaginary roots .
The polar form of the complex number z=x+iyz = x + iy is r(cos⁡t+isin⁡t)r\left( {\cos t + i\sin t} \right), where r=x2+y2=∣z∣r = \sqrt {{x^2} + {y^2}} = |z| and cos⁡t=xr\cos t = \dfrac{x}{r} , sin⁡yr\sin \dfrac{y}{r} . (tt is known as the argument of zz) The value of tt , such that −π<t⩽π- \pi < t \leqslant \pi, is called the principal argument of zz .