Complex Numbers 3 Question 9
9.
Let $z _1$ and $z _2$ 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 $w$, then
(2010)
(a) $\left|z-z _1\right|+\left|z-z _2\right|=\left|z _1-z _2\right|$
(b) $\arg \left(z-z _1\right)=\arg \left(z-z _2\right)$
(c) $\left|\begin{array}{cc}z-z _1 & \bar{z}-\bar{z} _1 \\ z _2-z _1 & \bar{z} _2-\bar{z} _1\end{array}\right|=0$
(d) $\arg \left(z-z _1\right)=\arg \left(z _2-z _1\right)$
Show Answer
Answer:
Correct Answer: 9. $(a, c, d)$
Solution:
- Given, $z=\frac{(1-t) z _1+t z _2}{(1-t)+t}$
Clearly, $z$ divides $z _1$ and $z _2$ in the ratio of $t:(1-t)$, $0<t<1$
$\Rightarrow \quad A P+B P=A B$ i.e. $\left|z-z _1\right|+\left|z-z _2\right|=\left|z _1-z _2\right|$
$\Rightarrow$ Option (a) is true.
$ \text { and } \quad \begin{aligned} \arg \left(z-z _1\right) & =\arg \left(z _2-z\right) \\ & =\arg \left(z _2-z _1\right) \end{aligned} $
$\Rightarrow$ Option (b) is false and option (d) is true.
Also, $\quad \arg \left(z-z _1\right)=\arg \left(z _2-z _1\right)$
$\Rightarrow \quad \arg \frac{z-z _1}{z _2-z _1}=0$
$\therefore \quad \frac{z-z _1}{z _2-z _1}$ is purely real.
$\Rightarrow \quad \frac{z-z _1}{z _2-z _1}=\frac{\bar{z}-\bar{z} _1}{\bar{z} _2-\bar{z} _1}$
or $\left|\begin{array}{cc}z-z _1 & \bar{z}-\bar{z} _1 \\ z _2-z _1 & \bar{z} _2-\bar{z} _1\end{array}\right|=0$
Option (c) is correct.
