A Little Note on Complex Numbers and Geometry for IITJEE Maths

Note:

In acute triangle, orthocentre (H), centroid (G), and circumcentre (O) are collinear and $HG:GO = 2:1$
Centroid of the triangle formed by points $A(z_{1})$ , $B(z_{2})$ , $C(z_{3})$ is $(z_{1}+z_{2}+z_{3})/3$ .
If the circumcentre of a triangle formed by $z_{1}$ , $z_{2}$ and $z_{3}$ is origin, then its orthocentre is $z_{1}+z_{2}+z_{3}$ (using 1).

Example 1:

Find the relation if $z_{1}$ , $z_{2}$ , $z_{3}$ , $z_{4}$ are the points of the vertices of a parallelogram taken in order.

Solution:

As the diagonals of a parallelogram bisect each other, the affix of the mid-point of AC is same as the affix of the mid-point of BD. That is,

$\frac{z_{1}+z_{3}}{2}=\frac{z_{2}+z_{3}}{2}$

or $z_{1}+z_{3}=z_{2}+z_{4}$

Example 2:

if $z_{1}$ , $z_{2}$ , $z_{3}$ are three non-zero complex numbers such that $z_{3}=(1-\lambda)z_{1}+\lambda z_{2}$ where $\lambda \in \Re - \{ 0 \}$ then prove that the points corresponding to $z_{1}$ , $z_{2}$ , and $z_{3}$ are collinear.

Solution:

$z_{3}=(1-\lambda)z_{1}+z_{2} =$ latex \frac{(1-\lambda)z_{1}+\lambda z_{2}}{1-\lambda +\lambda}$.

Hence, $z_{3}$ divides the line joining $A(z_{1})$ and $B(z_{2})$ in the ratio $\lambda : (1-\lambda)$ . Thus, the given points are collinear.

Homework:

Let $z_{1}$ , $z_{2}$ , $z_{3}$ be three complex numbers and a, b, c be real numbers not all zero, such that $a+b+c=0$ and $az_{1}+bz_{2}+cz_{3}=0$ . Show that $z_{1}$ , $z_{2}$ , $z_{3}$ are collinear.
In triangle PQR, $P(z_{1})$ , $Q(z_{2})$ , and $R(z_{3})$ are inscribed in the circle $|z|=5$ . If $H(z^{*})$ be the orthocentre of triangle PQR, then find $z^{*}$ .