# Why are (a), (c), (d) true?

Let $z_1$ and $z_2$ be two distinct complex numbers and suppose $z=(1-t)z_1+tz_2$ for some real number $t$ with $t \in (0,1)$. If $\arg w$ denotes the principal argument of a nonzero complex number $w$, then:

(a) $\ |z-z_1|+|z-z_2|=|z_1-z_2|$
(b) $\ \arg(z-z_1)= \arg(z-z_2)$
(c) $\ \det \begin{bmatrix}z-z_1 & \bar z- \bar z_1\\ z_2- z_1& \bar z_2-\bar z_1 \end{bmatrix} = 0$

(d) $\ \arg (z-z_1)=\arg (z_2-z_1)$

The answer is given (a), (c), (d) . I am a beginner in complex numbers and would appreciate some help in learning how this answer was derived.

• The given condition tells you that $z$ lies between $z_1$ and $z_2$ on the line joining the two points – Hari Shankar May 2 '17 at 0:23
• – Servaes May 2 '17 at 14:48

## 1 Answer

Hints: $\require{cancel} z-z_1=(\bcancel{1}-t)z_1+tz_2-\bcancel{z_1}=-t(z_1-z_2)\,$, and $\;z-z_2=(1-t)(z_1-z_2)\,$, then:

• (a) $\;|z-z_1|+|z-z_2|= t\,|z_1-z_2|+(1-t)\,|z_1-z_2|=(\bcancel{t} + 1 - \bcancel{t})\,|z_1-z_2|$

• (b)  see (d) below, but mind the signs

• (c)  either work out the expansion, or note that the 2nd row is proportional to the first by (d)

• (d) $\;\cfrac{z-z_1}{z_2-z_1} = \cfrac{-t(z_1-z_2)}{-(z_1-z_2)} = t \in \mathbb{R}^+ \;\;\implies\;\;\arg(z-z_1) = \arg(z_2-z_1)$