Question

Prove that: `|z_1+z_2+z_3+.......................+z_n|=|z_1|+|z_2|+|z_3|+...............+|z_n|?`

Answer 1

I don't think that equation is valid. I'm assuming `abs(z)` is the absolute value function

Explanation:

Try with two terms, `z_1=-1, z_2=3`

`abs(z_1+z_2) = abs(-1+3)=abs(2)=2`

`abs(z_1)+abs(z_2)=abs(-1)+abs(3)=1+3=4`

Hence

`abs(z_1+z_2)!=abs(z_1)+abs(z_2)`

`abs(z_1+...+z_n)!=abs(z_1)+...+abs(z_n)`

Answer 2

Perhaps you mean the triangle inequality for complex numbers:

`|z_1 + z_2 + ... + z_ n | \le |z_1| + |z_2| + ... + |z_n|`

We can abbreviate this

`|sum z_i | le sum |z_i|`

where the sums are `sum_{i=1}^n`

Lemma. `text{Re}(z) le |z|`

The real part is never bigger than the magnitude. Let `z=x+iy` for some real `x` and `y`. Clearly `x^2 le x^2+y^2` and taking square roots `x le sqrt{x^2+y^2}`. The magnitude is always positive; `x` may or may not be; either way it's never more than the magnitude.

I'll use the overbar for conjugate. Here we have a real number, the squared magnitude, which equals the product of the conjugates. The trick is that it equals its own real part. The real part of the sum is the sum of the real parts.

`|sum z_i | ^2 = sum_i z_i bar(sum_j z_j ) = text{Re}( sum_i z_i bar(sum_j z_j ) ) = sum_i text{Re}( z_i bar( sum_j z_j ))`

By our lemma, and the magnitude of the product being the product of magnitudes, and the magnitude of conjugates are equal,

# |sum z_i | ^2 le sum_i | z_i bar( sum_j z_j) | = sum_i | z_i| | bar( sum_j z_j) | = sum_i | z_i| | sum_j z_j | #

We can cancel one factor of the magnitude of the sum `|sum z_i |`, which is positive, preserving the inequality.

`|sum z_i | le sum| z_i|`

That's what we wanted to prove.