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I am stuck with the following problem that says

If the complex numbers $z_1,z_2,z_3$ represents the three points $P,Q, R $ respectively and be such that $lz_1+mz_2+nz_3=0$ where $ l+m+n=0 $ then show that $P,Q, R$ lie on a straight line.

Here is my try: $lz_1+mz_2+nz_3=0 \implies -(m+n)z_1+mz_2+nz_3=0$ which gives $$m(z_2-z_1)=n(z_1-z_3) \tag1$$

Similarly, $lz_1+mz_2+nz_3=0 \implies -(m+l)z_3+mz_2+lz_1=0$ which gives $$m(z_3-z_2)=l(z_1-z_3) \tag2$$

Now from (1) and (2) we get ,$$ln(z_1-z_3)=lm(z_2-z_1)=mn(z_3-z_2)$$

Now, what next? Can someone help?

J. W. Tanner
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learner
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  • That statement is trivially false. Just let $z_1$, $z_2$, and $z_3$ be any three complex numbers and let $l=m=n=0$. Besides, are you assuming anything about $l$, $m$, and $n$ besides the fact that $l+m+n=0$? – José Carlos Santos Feb 24 '19 at 11:27

2 Answers2

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It's wrong for $l=m=n=0$.

If $m\neq0$ and $n\neq0$ so after writing $$m(z_2-z_1)=n(z_1-z_3)$$ we are done because it's just $$\vec{PQ}||\vec{PR}.$$

Michael Rozenberg
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Let $w=z_2-z_1$ so $z_3-z_2=\frac{l}{n}w$ and your points are $z_1,\,z_1+w,\,z_1+\frac{n+l}{n}w=z_1-\frac{m}{n}w$.

J.G.
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