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.


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

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.