Some More Concepts: Complex Number

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 1. \left| {\frac{{z - {z_1}}}{{z - {z_2}}}} \right| = k

a) if k = 1 , then the locus of 'z' is a straight line i.e. perpendicular bisector of line joining the point {z_1}and {z_2}.

b) if k \ne 0,1 , then the locus of 'z' is circle

By definition of circle, the ratio of distance from two points of a point is constant then the locus of that point is circle.

c) if k = 0, the locus of z is a point which is equal to {z_1}.



Ques: \left| {\frac{{z - i}}{{z + i}}} \right| = \frac{1}{3}
or, 3\sqrt {{x^2} + {{(y - 1)}^2}} = \sqrt {{x^2} + {{(y + 1)}^2}}

\Rightarrow 9{x^2} + 9{y^2} + 9 - 18y = {x^2} + {y^2} + 1 + 2y

\Rightarrow 8{x^2} + 8{y^2} + 8 - 20y = 0

\Rightarrow {x^2} + {y^2} - 5/2y + 1 = 0

Centre \equiv \left( {0,\frac{5}{2}} \right) , radius = \sqrt {\frac{{25}}{{16}} - 1} = \frac{3}{4}

and, since \left| {\frac{{z - i}}{{z + i}}} \right| = \frac{1}{3},

i.e.ratio of distance of 'z' from i (0,1) and -i (0,-1) is 1/3

by section formula,

{y_A} = \frac{{1.3 + 1( - 1)}}{{1 + 3}} ( internally, y = \frac{{m{y_2} + n{y_1}}}{{m + n}} )

{y_B} = \frac{{1.3 - 1( - 1)}}{{ - 1 + 3}} ( externally, y = \frac{{m{y_2} - n{y_1}}}{{m - n}} )

The points gained from section formula will always be the end points of the diameter of the circle.


2. \arg .\left( {\frac{{z - {z_2}}}{{z - {z_1}}}} \right) = \alpha

or, \arg .(z - {z_2}) - \arg .(z - {z_1}) = \alpha  

 or,  \arg .(z - {z_2}) = \arg .(z - {z_1}) + \alpha


The locus is an arc of a circle and line joining z1 and z2 as a chord subtending angle \alpha  and excluding the points z1 and z2. [angle should be measured along anti-clockwise direction from denominator i.e. (z - z1) to Numerator i.e. (z - z2) ]
{Please remember,


From the figure

 \arg {z_2} - \arg {z_1} = \theta

or, \arg \left( {\frac{{{z_2}}}{{{z_1}}}} \right) = \theta  }




Ques: Find the equation of corresponding circle of the \arg .\left( {\frac{{z - i}}{{z + i}}} \right) = \frac{\pi }{3} ?


Angle should be measured from denominator i.e. z + i to numerator i.e.     z - i.
OC = \frac{1}{{\sqrt 3 }}

Radius = \frac{1}{{2\sqrt 3 }}
Centre \equiv \left( { - \frac{1}{{\sqrt 3 }},0} \right)

 \left| {z + \frac{1}{{\sqrt 3 }}} \right| = \frac{2}{{\sqrt 3 }}
Ques: Find the equation of corresponding circle of the \arg .\left( {\frac{{z - 1}}{{z - i}}} \right) = \frac{\pi }{4} ?
Angle should be measured from denominator i.e. z - i to numerator i.e.    z-1.

 Radius = 1

 Centre \equiv (1,1)

 Therefore equation of circle is,

 |z - ( 1 + i )| = 1

 or, |z - 1 - i| = 1

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