A Little Note on Complex Numbers and Geometry for IITJEE Maths

Note:

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

  1. 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.
  2. 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^{*}.

More later,

Nalin Pithwa

 

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