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Can you prove $$|z_1|^2\cdot [(1-r)\cdot|z_1 - z_2|] + |z_2|^2\cdot(r\cdot|z_1-z_2|)= |z_1-z_2|\cdot(|(1-r)z_1+rz_2|^2+r\cdot(1-r)\cdot|z_1-z_2|^2) $$

(such that $z_1,z_2 \in \mathbb{C}$ and $r\in\mathbb{R}$, $0 \leq r \leq 1$)

without constructing a triangle.


So far I started solving the LHS and arrived at $$|z_1 - z_2| [|z_1|^2 \cdot(1-r) + |z_2|^2\cdot r]$$ At this point it started to look a little like the RHS. Not entirely sure if I should proceed by expressing $|z_1|^2$ as $z_1\cdot \bar{z_1}$ and expand further. When I continue this why, I don't know how to arrive at $|(1-r)z_1+rz_2|^2.$

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  • $\begingroup$ What are your thoughts? What have you tried? Where are you stuck? You need to provide context for your question. Right now, it just looks like you want somebody to do your homework for you; that's not what this site is for. If you add some appropriate context, we will be happy to help. $\endgroup$ – Greg Martin Mar 21 '17 at 5:07
  • $\begingroup$ @GregMartin you're right, shouldve included some of my work. Left it out before mainly because I feel it is incorrect or not the best way to approach the question. $\endgroup$ – u123435 Mar 21 '17 at 5:19
  • $\begingroup$ It seems like a good idea to expand out not only $|z_1|^2 = z_1 \bar z_1$, but also the same expansion with the complicated squares-of-modulus on the right-hand side. Perhaps things will cancel away conveniently. $\endgroup$ – Greg Martin Mar 21 '17 at 5:54
  • $\begingroup$ just expand the right hand side in terms of $|z_1|^2$ and $|z_2|^2$, and you will see the terms $z_1 \tilde{z}_2$ will cancel out and gives you exactly the expression on the left hand side. $\endgroup$ – mastrok Mar 21 '17 at 7:09

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