# Homework Help: Complex numbers, proof.

1. Mar 24, 2014

### bobby2k

1. The problem statement, all variables and given/known data
Hello

Assume that we have n complex numbers u: $u_1,u_2,...,u_n$, and n complex numbers v:$v_1,v_2,....v_n$

I would like to prove that:
$|\Sigma_{i=1}^nRe(u_i\bar{v_i})| \le |\Sigma_{i=1}^nu_i\bar{v_i}|$

I guess this can be written simpler:
$|\Sigma_{i=1}^nRe(z_i)| \le |\Sigma_{i=1}^n z_i|$

2. Relevant equations

3. The attempt at a solution
If n=1. I know that this obviously must hold. But When n is bigger than 1, I am not so sure how to show that it holds. It becomes quit messy, I tried moving the absolute value inside the sum, but id didn't work.

2. Mar 24, 2014

### jbunniii

I suggest writing each $z$ (you omitted the index by the way) in terms of its real and imaginary components.

Last edited: Mar 24, 2014
3. Mar 24, 2014

### micromass

Hint:

$$\textrm{Re}(z+w) = \textrm{Re}(z) + \textrm{Re}(w)$$

4. Mar 24, 2014

### jbunniii

By the way, when working with complex numbers, you should probably use some other letter for your index instead of $i$, which is already reserved for $\sqrt{-1}$. Unless you are one of those strange people who use $j$ for that purpose.

5. Mar 24, 2014

### bobby2k

I got this:

$|\Sigma z_j|=|\Sigma Re(z_j)+i*Im(z_j)|=|\Sigma Re(z_j)+i*\Sigma Im(z_j)|$

Now I would use the triangle inequality if I did not have the i. But I can still say that there is always so that the absolute value of the real part of a complex number must be less than the absolute value of the total number?

so I get

$\ge |\Sigma Re(z_j)|$?

6. Mar 24, 2014

### jbunniii

Consider working with the square of the magnitude. $|a + ib|^2 = ???$

7. Mar 24, 2014

### bobby2k

Do you mean that I should write $c=|\Sigma z_j|$, $a=\Sigma Re(z_j)$, $b=\Sigma Im(z_j)$
So then I get $c=|a+bi|$, squaring I get $c^2=|a+ib|^2=a^2+b^2$, then I get:

$|\Sigma z_j|^2 = (\Sigma Re(z_j))^2+(\Sigma Im(z_j))^2$, and since all are positive we must have that:

$|\Sigma z_j| \ge |\Sigma Re(z_j)|$

8. Mar 24, 2014

### jbunniii

Yes, the key is that $a^2 + b^2 \geq a^2$, so taking square roots of each side, we get $\sqrt{a^2 + b^2} \geq \sqrt{a^2}$. The LHS of this inequality is $|a + ib|$ and the RHS is $|a|$.

9. Mar 24, 2014