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Let $ \displaystyle{ z_1, z_2 \in \mathbb{C} }$ where $ z_1, z_2 \neq 0$

Prove that: $\displaystyle |z_1 +z_2| \geq \frac{1}{2} \left( |z_1|+|z_2| \right) \left|\frac{z_1}{|z_1|} + \frac{z_2}{|z_2|}\right| $.

P.S I think that I have to use the inequality $ Re(z_1z_2) \leq |z_1||z_2| $ but I don't know how.

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  • $\begingroup$ We need the assumption that $z_1$ and $z_2$ are nonzero; otherwise, $\frac{z_i}{|z_i|}$ is not well-defined. $\endgroup$
    – Paul
    Apr 20, 2012 at 12:02
  • $\begingroup$ @Paul: Yes you are right! Thank's for noticing this. I edit it. $\endgroup$
    – passenger
    Apr 20, 2012 at 12:03

1 Answer 1

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Write $z_1=r_1e^{i\theta_1}$ and $z_2=r_2e^{i\theta_2}$. Since $z_1, z_2\neq 0$, we have $r_1, r_2>0$. Then $$\tag{1}\left(\frac{1}{2} \left( |z_1|+|z_2| \right) \left|\frac{z_1}{|z_1|} + \frac{z_2}{|z_2|}\right|\right)^2 =\left(\frac{1}{2}(r_1+r_2)|e^{i\theta_1}+e^{i\theta_2}|\right)^2=\frac{1}{4}(r_1+r_2)^2|e^{i\theta_1}+e^{i\theta_2}|^2$$ $$=\frac{1}{4}(r_1+r_2)^2(2+e^{i(\theta_1-\theta_2)}+e^{i(\theta_2-\theta_1)})$$ since $$\tag{2} |e^{i\theta_1}+e^{i\theta_2}|^2=(e^{i\theta_1}+e^{i\theta_2})(e^{-i\theta_1}+e^{-i\theta_2})=2+e^{i(\theta_1-\theta_2)}+e^{i(\theta_2-\theta_1)}.$$ Note also that $$\tag{3}|z_1+z_2|^2=|r_1e^{i\theta_1}+r_2e^{i\theta_2}|^2= r_1^2+r_2^2+r_1r_2e^{i(\theta_1-\theta_2)}+r_1r_2e^{i(\theta_2-\theta_1)}.$$ Subtract $(3)$ by $(1)$, we obtain $$|z_1+z_2|^2-\left(\frac{1}{2} \left( |z_1|+|z_2| \right) \left|\frac{z_1}{|z_1|} + \frac{z_2}{|z_2|}\right|\right)^2=\frac{1}{2}(r_1-r_2)^2-\frac{1}{4}(r_1-r_2)^2\big(e^{i(\theta_1-\theta_2)}+e^{i(\theta_2-\theta_1)}\big)$$ $$=\frac{1}{4}(r_1-r_2)^2(2-e^{i(\theta_1-\theta_2)}-e^{i(\theta_2-\theta_1)}) =\frac{1}{4}(r_1-r_2)^2|e^{i\theta_1}-e^{i\theta_2}|\geq 0.$$ where the last equality follows from a caluculation similar to $(2)$. So this implies that $$\displaystyle |z_1 +z_2| \geq \frac{1}{2} \left( |z_1|+|z_2| \right) \left|\frac{z_1}{|z_1|} + \frac{z_2}{|z_2|}\right|.$$

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    $\begingroup$ I think that you proved $\displaystyle |z_1 -z_2| \geq \frac{1}{2} \left( |z_1|+|z_2| \right) \left|\frac{z_1}{|z_1|} + \frac{z_2}{|z_2|}\right| $ since you calculate $ |z_1 -z_2| $ but it's o.k $\endgroup$
    – passenger
    Apr 20, 2012 at 12:37
  • $\begingroup$ @passenger: Oh yes, hahaha. I will try to edit it. $\endgroup$
    – Paul
    Apr 20, 2012 at 12:39
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    $\begingroup$ You don't need to do this. I can edit the question! It's o.k since the only thing that I have to change is to calculate $|z_1+z_2|$. $\endgroup$
    – passenger
    Apr 20, 2012 at 12:41
  • $\begingroup$ @passenger: I edited it before I read your comment. Anyway, hope that this is clearer. $\endgroup$
    – Paul
    Apr 20, 2012 at 12:46
  • $\begingroup$ Yes it very clear! Thank you very much for your time! $\endgroup$
    – passenger
    Apr 20, 2012 at 12:47

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