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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.

Mikko Korhonen
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  • 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 06 '12 at 13:43
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    @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 06 '12 at 13:49
  • whoops, didn't see this was in complex analysis. not very observant of me... – Adam Rubinson Sep 06 '12 at 15:01

3 Answers3

20

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

Mikko Korhonen
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  • I think this is the most straightforward proof we could get – M Turgeon Sep 06 '12 at 13:44
  • My cursory glance makes me curious what difference the parentheses makes? – Captain Giraffe Sep 06 '12 at 18:23
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    @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 06 '12 at 19:38
  • @ruakh Thank you. Your well written comment helped my intuition get back to normal. – Captain Giraffe Sep 06 '12 at 19:51
2

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 $|z_2|=r_2$

$|z_1-z_2|$

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

$=\sqrt{r_1^2+r_2^2-2r_1r_2\cos(A-B)}$

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

$=r_1-r_2$

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

lab bhattacharjee
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1

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.

abnry
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Ram
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  • $2$ 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 $? –  Sep 06 '12 at 14:28
  • Look at Z1Z2* + Z2Z1* in expansion on right side; where Z1Z2* = (Z1Z2) so, both are conjugates, what happens if you add 2 conjugates? imaginary parts will get cancelled, only real parts will remain. – Ram Sep 06 '12 at 14:32
  • No. I forgot to include $Z_1 \overline{Z_2}$ to it. –  Sep 06 '12 at 14:32
  • It could be said also $2$ Real part of $(Z_2 \overline{Z_1})$ instead of $2$ Real part of $(Z_1 \overline{Z_2})$ –  Sep 06 '12 at 14:46
  • Yes, take Z3 = Z2Z1* and Z3; see it in Z3 = x + iy form, Re(Z3) = Re(Z3) = x. – Ram Sep 06 '12 at 14:51
  • I choose this answer, because this answer I understood in the time that I have on this problem. –  Sep 07 '12 at 14:18