First slide

Algebra of complex numbers

Question

Let z₁ and z₂, be two distinct complex numbers and let $z = (1 - t) z_{1} + t z_{2}$ for some real number t with 0 < t < 1. if arg (w) denotes the principal argument of a non-zero complex number , then

Moderate

A

$|z - z_{1}| + |z - z_{2}| = |z_{1} - z_{2}|$
B

$\arg (z - z_{1}) = \arg (z - z_{2})$
C

$|\begin{array}{cc} z - z_{1} & \bar{z} - {\bar{z}}_{1} \\ z_{2} - z_{1} & {\bar{z}}_{2} - {\bar{z}}_{1} \end{array}| = 0$
D

$\arg (z - z_{1}) = \arg (z_{2} - z_{1})$

Solution

Given $z = \frac{(1 - t) z_{1} + {tz}_{2}}{(1 - t) + t}$
Clearly, z divides z, and z, in the ratio of t :(1 - t), 0<t<1
$\Rightarrow AP + BP = AB i.e., |z - z_{1}| + |z - z_{2}| = |z_{1} - z_{2}| .$
and $\arg (z - z_{1}) = \arg (z_{2} - z) = \arg (z_{2} - z_{1})$
Also , $\arg (z - z_{1}) = \arg (z_{2} - z_{1})$
$\Rightarrow \arg (\frac{z - z_{1}}{z_{2} - z_{1}}) = 0$
$∴ \frac{z - z_{1}}{z_{2} - z_{1}}$ is purely real.
$\Rightarrow \frac{z - z_{1}}{z_{2} - z_{1}} = \frac{\bar{z} - {\bar{z}}_{1}}{{\bar{z}}_{2} - {\bar{z}}_{1}} or |\begin{array}{cc} z - z_{1} & \bar{z} - {\bar{z}}_{1} \\ z_{2} - z_{1} & {\bar{z}}_{2} - {\bar{z}}_{1} \end{array}| = 0$

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