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
    Commented Sep 2, 2013 at 7:54
  • $\begingroup$ sorry, wrong signs (fixed, original question now valid) $\endgroup$ Commented Sep 2, 2013 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
    Commented Sep 2, 2013 at 8:03
  • $\begingroup$ When multiplying by -1, you need to think carefully how that changes your inequality relations. $\endgroup$
    – Ncat
    Commented Sep 2, 2013 at 8:04
  • $\begingroup$ Dang it, Macavity. lol $\endgroup$
    – Ncat
    Commented Sep 2, 2013 at 8:05

1 Answer 1


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.

  • $\begingroup$ 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 $\endgroup$ Commented Sep 2, 2013 at 8:26
  • $\begingroup$ Whoops, thanks. I'll edit. $\endgroup$ Commented Sep 2, 2013 at 8:58

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