Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to prove $|z_1-z_2| \geq |z_1|-|z_2|$ in other way than this? I mean I tried to find on the internet but could not find. I ask for more straighforward way than the proof that is presented for item 3.

share|cite|improve this question
the quickest way I know to solve this is to consider the two cases z1 < z2 and z2< z1 seperately. Edit: and when z2=z1 it's obvious – Adam Rubinson Sep 6 '12 at 13:43
@AdamRubinson: when $z_1,z_2\in \Bbb C$, there is no such thing as $z_1\gt z_2$ when their imaginary parts are non-zero. – Aang Sep 6 '12 at 13:49
whoops, didn't see this was in complex analysis. not very observant of me... – Adam Rubinson Sep 6 '12 at 15:01
up vote 2 down vote accepted

You can use multiplication by conjugate.

$|Z_1 + Z_2 |^2 = (Z_1 + Z_2)\overline{(Z_1 + Z_2)}$. By expansion you will get

$| Z_1 + Z_2 |^2 = |Z_1|^2 + |Z_2|^2$ + 2 Real part of $(Z_1 \overline{Z_2})$ $\leq (|Z_1| + |Z_2|)^2$.

Then use this result, and prove remaining.

share|cite|improve this answer
$2$ Real part of $(Z_1 \bar{Z_2})$ is not so but I got from $|Z_1 +Z_2 |^2 =(Z_1 +Z_2 )\overline{(Z_1 +Z_2 )}=|Z_1 |^2 +|Z_2 |^2 - ix_1y_2 + iy_1x_2$, where $Z_1=x_1+iy_1 $ and $Z_2=x_2+iy_2 $? – alvoutila Sep 6 '12 at 14:28
Look at Z1Z2* + Z2Z1* in expansion on right side; where Z1Z2* = (Z1*Z2)* so, both are conjugates, what happens if you add 2 conjugates? imaginary parts will get cancelled, only real parts will remain. – Ram Sep 6 '12 at 14:32
No. I forgot to include $Z_1 \overline{Z_2}$ to it. – alvoutila Sep 6 '12 at 14:32
It could be said also $2$ Real part of $(Z_2 \overline{Z_1})$ instead of $2$ Real part of $(Z_1 \overline{Z_2})$ – alvoutila Sep 6 '12 at 14:46
Yes, take Z3 = Z2Z1* and Z3*; see it in Z3 = x + iy form, Re(Z3) = Re(Z3*) = x. – Ram Sep 6 '12 at 14:51

Use the triangle inequality on $|z_1| = |(z_1 - z_2) + (z_2)|$.

share|cite|improve this answer
I think this is the most straightforward proof we could get – M Turgeon Sep 6 '12 at 13:44
My cursory glance makes me curious what difference the parentheses makes? – Captain Giraffe Sep 6 '12 at 18:23
@CaptainGiraffe: The triangle inequality involves three values. The parentheses are there to make clear what those three values are supposed to be (namely $z_1$, $z_1 - z_2$, and $z_2$). From a mathematical standpoint, m. k.'s answer is equivalent to "Use the triangle inequality on $|z_1|$"; but that would have been a confusing and almost-useless answer, whereas this one is clear and useful. – ruakh Sep 6 '12 at 19:38
@ruakh Thank you. Your well written comment helped my intuition get back to normal. – Captain Giraffe Sep 6 '12 at 19:51

Let $z_1=r_1(\cos A+i\sin A)$ and $z_2=r_2(\cos B+i\sin B)$

So, $|z_1|=r_1$ and $|z_2|=r_2$


$=\sqrt{(r_1\cos A-r_2\cos B)^2+(r_1\sin A-r_2\sin B)^2}$


$≥\sqrt{r_1^2+r_2^2-2r_1r_2}$ as $\cos(A-B)≤1$


So, $|z_1-z_2|≥|z_1|-|z_2|$, the equality occurs when $\cos(A-B)=1$ i.e., when $A=B$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.