Let $z_1,z_2$ be distinct complex numbers such that $|z_1| = |z_2|$. Prove that $\frac{1}{2}|z_1+z_2| < |z_1|.$

Question

Let $z_1,z_2$ be distinct complex numbers such that $|z_1| = |z_2|$. Prove that $\frac{1}{2}|z_1+z_2| < |z_1|.$

What I Tried:- I tried it in 2 ways:-

First, let $z_1 = x + iy$ , $z_2 = z + iw$ .
We have, $x^2 + y^2 = z^2 + w^2$.
So, $\dfrac{1}{2}|z_1 + z_2| < |z_1|.$

$\rightarrow \dfrac{1}{2}|(x + z) + i(y + w)| < |z_1|.$ $\rightarrow \dfrac{1}{2}\sqrt{[(x + z)^2 + (y + w)^2]} < |z_1|.$
But now I am stuck.

In the next try, I squared both sides to get :-
$\rightarrow \dfrac{1}{2}|z_1 + z_2| < |z_1|$ .

$\rightarrow \dfrac{|z_1 + z_2|^2}{4} < |z_1||z_2|$.

$\rightarrow (z_1 + z_2)(\bar z_1 + \bar z_2) < 4|z_1||z_2|$.

$\rightarrow |z_1|^2 + |z_2|^2 + z_1 \bar z_2 + z_2 \bar z_1 < 4|z_1||z_2|$.
But I am still stuck.

Can someone help me? Thank You.

You need the triangle inequality $|z+z'| \leq |z|+|z'|$ for all $z,z' \in \mathbb C$, from where you can see how to finish the problem. Can you use the triangle inequality? — Sarvesh Ravichandran Iyer, Sep 06 '21 at 12:55
Do you know the triangle inequality? (That gives less equal in the result); equal holds when vectors are collinear and same direction, so that would force $z_1=z_2$ here etc — Conrad, Sep 06 '21 at 12:56
@Anonymous No problem, but basically you have to "copy" the proof of the triangle inequality. Look up the proof of the "triangle inequality for complex numbers" either on MSE or elsewhere. For example, see "Proof 2" over here and try to adapt it to your situation. — Sarvesh Ravichandran Iyer, Sep 06 '21 at 13:00
And do let me know if any answer at Triangle Inequality with Complex Numbers works for you. If one does, then I will try to close this as a duplicate, otherwise I'll retract my vote and search for another on-site duplicate. — Sarvesh Ravichandran Iyer, Sep 06 '21 at 13:00
@TeresaLisbon the answers dosen't match, but thank u for showing me the Triangle Inequality. With this the problem is totally direct. — Anonymous, Sep 06 '21 at 13:08

score 0 · Accepted Answer · answered Sep 06 '21 at 13:01

0

Consider, $A=|z_1+z_2|-2|z_1|$. As $|z_1|=|z_2|$, $ A=|z_1+z_2|-|z_1|-|z_2|$. We know, by triangle inequality that $|z_1|+|z_2| \gt |z_1+z_2|$. Therefore, $A \lt 0$. Hence, we are done.

answered Sep 06 '21 at 13:01

Ilovemath

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Let $z_1,z_2$ be distinct complex numbers such that $|z_1| = |z_2|$. Prove that $\frac{1}{2}|z_1+z_2| < |z_1|.$

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