I am using Brown and Churchill's Complex Analysis Textbook, and on pg.11 of the eighth edition, there is a triangle inequality derivation as followed

to prove

$|z_1+z_2|\geq ||z_1|-|z_2||$




however I realized that if




which is a contradiction

does staring on $|z_1|$ implies an inherent assumption I have to made, if so what is the assumption?

moreover, when they reached the


conclusion, they say this only work for when $|z_1|\geq|z_2|$

when $|z_1|<|z_2|$ they exchange $z_1$ and $z_2$ to arrive $|z_1+z_2|\geq-1*(|z_1|-|z_2|)$ for which I don't follow as why would we care about the order of $z_1$ and $z_2$, isn't the designation of $z_1$ and $z_2$ arbitrary?

Any hint would be much appreciated

  • $\begingroup$ I would check your original inequality. z_1 = 1, z_2 = i seems to be a trivial counterexample. $\endgroup$ – Ncat Sep 2 '13 at 7:54
  • $\begingroup$ sorry, wrong signs (fixed, original question now valid) $\endgroup$ – user2654176 Sep 2 '13 at 7:55
  • 2
    $\begingroup$ How do you get $-|(z_1+z_2-z_2)|\leq-(|z_1+z_2|+|z_2|)$? Note just multiplying by $-1$ reverses the inequality. $\endgroup$ – Macavity Sep 2 '13 at 8:03
  • $\begingroup$ When multiplying by -1, you need to think carefully how that changes your inequality relations. $\endgroup$ – Ncat Sep 2 '13 at 8:04
  • $\begingroup$ Dang it, Macavity. lol $\endgroup$ – Ncat Sep 2 '13 at 8:05

First of all, let's recall that multiplying by a negative number switches inequality signs. That is, $x \leq y$ does not imply $-x \leq -y$, but rather $-x \geq -y$. In particular, your inequality $$-|z_1 + z_2 - z_2| \leq -(|z_1 + z_2| + |z_2|)$$ is generally false. For example, if you take $z_1 = z_2 = 1$, you get that $-1 \leq -3$, which is incorrect.

When the authors reach the point that $|z_1 + z_2| \geq |z_1| - |z_2|$, the proof is not yet finished. Remember that the point is to show that $|z_1 + z_2| \geq ||z_1| - |z_2||$. To get those extra absolute value bars we need an additional step. So:

If it's the case that $|z_1| \geq |z_2|$, then we have $|z_1| - |z_2| \geq 0$, and so $||z_1| - |z_2|| = |z_1| - |z_2|$. So in this case, we really are done because we've shown $$|z_1 + z_2| \geq |z_1| - |z_2| = ||z_1| - |z_2||.$$

But what if we don't have $|z_1| \geq |z_2|$? Well, then we must have $|z_1| < |z_2|$. But this is just like the previous case, but with the roles of $z_1$ and $z_2$ interchanged, meaning that we have $$|z_2 + z_1| \geq |z_2| - |z_1| = ||z_2| - |z_1||.$$ Since $|z_1 + z_2| = |z_2 + z_1|$ and $||z_1| - |z_2|| = ||z_2| - |z_1||$, we're done.

| cite | improve this answer | |
  • $\begingroup$ I think its $|z_2+z_1|$ instead of $-$ but otherwise thank you for the explanation, its very thorough, I realize that I am applying triangle inequality with -1 which is incorrect as I need to apply the -1 after I apply the inequality $\endgroup$ – user2654176 Sep 2 '13 at 8:26
  • $\begingroup$ Whoops, thanks. I'll edit. $\endgroup$ – Jesse Madnick Sep 2 '13 at 8:58

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.