Some Standard Results: Complex Number

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Standard results in Complex number (Argand Plane)
1. Section Formula :-
z = \frac{{m{z_2} + n{z_1}}}{{m + n}}


2.  Centroid:-
centroid,  z = \frac{{{z_1} + {z_2} + {z_3}}}{3}


3. Equation of a Straight line :-
a) Parametric Form:-

From triangle OBP

z = {z_2} + t({z_1} - {z_2})

or, z = t{z_1} + (1 - t){z_2}



b) \arg (z - {z_1}) - \arg ({z_2} - {z_1}) = 0 or \pi

=> \arg \left( {\frac{{z - {z_1}}}{{{z_2} - {z_1}}}} \right) = 0 or \pi

\Rightarrow \frac{{z - {z_1}}}{{{z_2} - {z_1}}} is Purely Real

\Rightarrow \frac{{z - {z_1}}}{{{z_2} - {z_1}}} = \overline {\left( {\frac{{z - {z_1}}}{{{z_2} - {z_1}}}} \right)}

\Rightarrow \frac{{z - {z_1}}}{{{z_2} - {z_1}}} = \left( {\frac{{\overline z - \overline {{z_1}} }}{{\overline {{z_2}} - \overline {{z_1}} }}} \right) ........................ (i)

{ \Rightarrow \left| {\begin{array}{ccccccccccccccc} z&{\overline z }&1\\ {{z_1}}&{\overline {{z_1}} }&1\\ {{z_2}}&{\overline {{z_2}} }&1 \end{array}} \right| = 0}

And, therefore condition of collinearity of {z_1},{z_2}\& {z_3}
because if  {z_1},{z_2}\& {z_3} are collinear then area form by these points must be 0.

{\left| {\begin{array}{ccccccccccccccc} {{z_1}}&{\overline {{z_1}} }&1\\ {{z_2}}&{\overline {{z_2}} }&1\\ {{z_3}}&{\overline {{z_3}} }&1 \end{array}} \right| = 0}


General Equation of a Straight Line:

\overline a z + a\overline z + b = 0

Where 'a' is a complex number and 'b' is a Purely Real Number.

a - complex number, b - Purely Real Number

Proof:- From (i) we have ,

z(\overline {{z_2}} - \overline {{z_1}} ) - {z_1}\overline {{z_2}} + |{z_1}{|^2} = \overline z ({z_2} - {z_1}) - \overline {{z_1}} {z_2} + |{z_1}{|^2}

\Rightarrow \overline z ({z_2} - {z_1}) - z(\overline {{z_2}} - \overline {{z_1}} ) + ({z_1}\overline {{z_2}} - \overline {{z_1}} {z_2}) = 0


multiply  by 'i' on both sides,


i\overline z ({z_2} - {z_1}) - iz(\overline {{z_2}} - \overline {{z_1}} ) + i({z_1}\overline {{z_2}} - \overline {{z_1}} {z_2}) = 0


Suppose,  i\overline z = a and {z_2} - {z_1} = z

and  - iz = \overline {i\overline z } = \overline a  ( since, \overline i = - i )

And,  {z_1}\overline {{z_2}} - \overline {{z_1}} {z_2}

= {z_1}\overline {{z_2}} - \overline {{z_1}\overline {{z_2}} }  = b (let)

(We know that, z - \overline z  = Purely Real )

therefore, {z_1}\overline {{z_2}} - \overline {{z_1}} {z_2} = {z_1}\overline {{z_2}} - \overline {{z_1}\overline {{z_2}} }  = Purely Real ( 'b' )


then equation becomes ,

\overline a z + a\overline z + b = 0

Where 'a' is a complex number and 'b' is a Purely Real Number.




4. Condition for the four points to be cyclic :
Angle A + Angle C = pi

 \Rightarrow \arg \left( {\frac{{{z_4} - {z_1}}}{{{z_4} - {z_1}}}} \right) + \arg \left( {\frac{{{z_2} - {z_3}}}{{{z_4} - {z_3}}}} \right) = \pi

\Rightarrow \arg \left[ {\left( {\frac{{{z_4} - {z_1}}}{{{z_4} - {z_1}}}} \right).\left( {\frac{{{z_2} - {z_3}}}{{{z_4} - {z_3}}}} \right)} \right] = \pi   ( since arg.z1 + arg.z2= arg.(z1.z2) )

Since argument of negative x-axis is pi therefore complex no. must be on negative side of x-axis,

{ \Rightarrow \left( {\frac{{{z_4} - {z_1}}}{{{z_4} - {z_1}}}} \right).\left( {\frac{{{z_2} - {z_3}}}{{{z_4} - {z_3}}}} \right)} = Purely Real


5. Equation of a circle:-
a) |z - {z_0}| = r



b) Diametric form:-

\arg \left( {\frac{{z - {z_2}}}{{z - {z_1}}}} \right) = \pm \frac{\pi }{2}

Therefore, {\frac{{z - {z_2}}}{{z - {z_1}}}} is purely imaginary,

for purely imaginary number , z + \overline z = 0

\Rightarrow \frac{{z - {z_2}}}{{z - {z_1}}} + \overline {\left( {\frac{{z - {z_2}}}{{z - {z_1}}}} \right)} = 0

or, \frac{{z - {z_2}}}{{z - {z_1}}} + \frac{{\overline z - \overline {{z_2}} }}{{\overline z - \overline {{z_1}} }} = 0

or, (z - {z_2})(\overline z - \overline {{z_1}} ) + (\overline z - \overline {{z_2}} )(z - {z_1}) = 0


(c) General Equation of Circle:-

z\overline z + a\overline z + \overline a z + b = 0

Centre \equiv - a, Radius = \sqrt {|a{|^2} - b}

Proof:-

 Equation of circle,|z - {z_0}| = r

 or, |z - {z_0}{|^2} = {r^2}

 or, (z - {z_0})(\overline {z - {z_0}} ) = {r^2}

 or, (z - {z_0})(\overline z - \overline {{z_0}} ) = {r^2}

 or, z.\overline z - z.\overline {{z_0}} - {z_0}\overline z + {z_0}.\overline {{z_0}} = {r^2}

 or, z.\overline z - z.\overline {{z_0}} - {z_0}\overline z + |{z_0}{|^2} = {r^2}

 or,z.\overline z - z.\overline {{z_0}} - {z_0}\overline z + |{z_0}{|^2} - {r^2} = 0

 let, a = - {z_0}, Centre is at '-a'
therefore, \overline a = - \overline {{z_0}}

and let, b = |{z_0}{|^2} - {r^2}

 \Rightarrow r = \sqrt {|{z_0}{|^2} - b} = \sqrt {|a{|^2} - b}   ( Radius)

 Equation becomes,

 z\overline z + a\overline z + \overline a z + b = 0

d) |z - {z_1}{|^2} + |z - {z_2}{|^2} = k 

will represent the circle if k \ge \frac{1}{2}|{z_1} - {z_2}{|^2}



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