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) $\arg \left(z-z _1\right)=\arg \left(z _2-z _1\right)$

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Answer:

Correct Answer: 9. $(a, c, d)$

Solution:

  1. 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.



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