# Prove that |z1+z2+...zn|<=|z1|+|z2|+....+|zn| where z1,z2,...zn are complex numbers?

Also find the special case when it becomes equal.

### 5 Answers

- VamanLv 74 weeks agoFavorite Answer
Treat them like vectors. Now magnitude (z1+z2)

sqrt(z1^2 +z2^+2 z1 z2 cos theta). Magnitude z1+ z2= (sqrt z1^2 + sqrt z2^2). There is missing term = 2 z1 z2 cos theta. Both sides are equal only when cos theta =pi/2. Therefore you can safely say magnitude (z1 + z2) => magnitude z1 + magnitude z2. This you can extend it to all the terms.

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- atsuoLv 64 weeks ago
Use polar coordinates .

Let z1 = a(cosθ+i*sinθ) and z2 = b(cosφ+i*sinφ) .

Where 0 ≦ a,b and 0 ≦ θ,φ < 2π .

z1+z2 = (a*cosθ+b*cosφ) + i(a*sinθ+b*sinφ) , so

|z1+z2|^2

= (a*cosθ+b*cosφ)^2 + (a*sinθ+b*sinφ)^2

= a^2*cos^2(θ) + 2ab*cosθcosφ + b^2*cos^2(φ)

+ a^2*sin^2(φ) + 2ab*sinθsinφ + b^2*sin^2(φ)

= a^2 + 2ab(cosθcosφ + sinθsinφ) + b^2

= a^2 + 2ab*cos(θ-φ) + b^2

≦ a^2 + 2ab + b^2 ... because 2ab is non-negative .

= (a+b)^2

= (|z1|+|z2|)^2

That is , |z1+z2|^2 ≦ (|z1|+|z2|)^2 .

We know |z1+z2| and |z1|+|z2| are non-negative ,

so |z1+z2| ≦ |z1|+|z2| stands .

The equal sign stands when 2ab*cos(θ-φ) = 2ab . That is ,

1) At least 1 number is 0 (ab = 0) , or

2) The arguments of 2 numbers are equal (cos(θ-φ) = 1) .

|z1+...+zn| ≦ |z1|+...+|zn| is proved when n = 2 .

Next , assume that |z1+...+zn| ≦ |z1|+...+|zn| stands when n = k .

That is , |z1+...+zk| ≦ |z1|+...+|zk| .

Let z1+...+zk = z0 . z0 is a complex number , so

|z0+z(k+1)| ≦ |z0|+|z(k+1)| . That is ,

|z1+...+zk+z(k+1)| ≦ |z1+...+zk|+|z(k+1)|

And by the assumption ,

|z1+...+zk|+|z(k+1)| ≦ |z1|+...+|zk|+|z(k+1)|

So we can find

|z1+...+zk+z(k+1)| ≦ |z1|+...+|zk|+|z(k+1)|

That is ,

|z1+...+zn| ≦ |z1|+...+|zn| stands when n = k+1 .

Therefore

|z1+...+zn| ≦ |z1|+...+|zn| stands for any n .

Next , think the condition that the equal sign stands .

We found the condition for 2 complex numbers , so the condition

is "The arguments of all non-zero numbers are equal" . If only 1

non-zero number exists then its argument can be any value .

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- DixonLv 74 weeks ago
I don't know about a formal proof but considering the Argand diagram, clearly for the given magnitudes |z1|, |z2| ... |zn| the biggest magnitude sum possible is when all Z's are in the same direction and the magnitudes just add together. Any other direction for one Z will reduce the the net magnitude. That is what you are being asked to prove but it seems kind of axiomatic to me.

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- husoskiLv 74 weeks ago
The two-value version of that is called the "Triangle Inequality":

|z1 + z2| <= |z1| + |z2|

That's obvious (and equal) if at least one of of the values is zero.

It's also simple and easy to prove as equality when z2 = a*z1, where a>0 is a real number.

|z1 + a*z1| = |(1 + a) z1| = (1 + a) |z1| = |z1| + a|z1| = |z1| + |a*z1| = |z1| + |z2|

If a is allowed to be negative, the corresponding real number property |a + 1| <= |a| + 1, and the above becomes:

|z1 + a*z1| = |(1 + a) z1| = |1 + a| |z1| <= (1 + |a|) |z1| = |z1| + |a*z1| = |z1| + |z2|That covers all the "parallel cases". if z1, z2 are nonzero and z1/z2 is not real, then the addition w = z1 + z2 corresponds to a non-degenerate vector addition triangle with sides |z1|, |z2| and |w|. The triangle inequality theorem from Euclidean plane geometry tells you that |w| < |z1| + |z2|. (Euclid's Elements, book 1, prop. 20; see source). That proposition tells you that the inequality is strictly less than in this case.

Extending to n terms is simple, using the two-term theorem above:

|z1 + z2 + ... + zn| <= |z1| + |z2 + z3 + ... + zn|

<= |z1| + |z2| + |z3 + ... * zn|

...and so on

...and the inequality

You can stop after extracting two terms and show that the inequality is strictly less than whenever z1 and z2 are not either a zero or parallel-with-positive-multiplier case from above. Since you can reorder the original sum however you like, that means the inequality is strictly less than unless every pair of nonzero terms in the sum has one term as a positive real multiple of the other.

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- Mr.PersonaLv 54 weeks agoReport
I think he's working through Rudin: PoMA, chapter 1.

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