# Homework Help: Imaginary Numbers Derivation

1. Jan 18, 2013

### ElijahRockers

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

Derive the following relation, where z1 and z2 are arbitrary complex numbers

|z1z2*+z1*z2| ≤ 2|z1z2|

3. The attempt at a solution

I found the expression |z1z2*+z1*z2| = |2(a1a2+b1b2)| = √(4[a12a22 + 2a1a2b1b2 +b12b22])

But that is where I get stuck. How does the equal sign become an inequality? Also, there are no imaginary terms in my expression, but there ARE imaginary terms in the right hand side, so I think I am doing something wrong.

I also tried doing the derivation using polar forms but it got complicated very quickly.

Thanks.

2. Jan 18, 2013

### Staff: Mentor

I would try to keep z1 and z2 as they are (so don't split them in a+ib) as long as possible. As both sides are always positive (or zero): Have you tried squaring them?

In your expression, there is no |z1z2|.

3. Jan 21, 2013

### ElijahRockers

Well as far as I understand, I have to start with the expression |z1z2*+z1*z2| and somehow find that it is less or equal to 2|z1z2|.

The only way I can think to manipulate the first expression is by taking the magnitude.

In an earlier problem I derived:

|z1z2*+z1*z2| = 2$\Re${z1z2*} = 2$\Re${z1*z2} which is similar to what i'm trying to do, but where does the inequality come from?

4. Jan 21, 2013

### LCKurtz

Have you thought about the triangle inequality?

5. Jan 21, 2013

### ElijahRockers

No, in all honesty I had never heard of it. I looked it up, and I think I see what you are getting at, but I'm not sure how to go about using that.

I tried graphing some hypothetical complex numbers and their conjugates, and visually multiplying them (adding the angles, multiplying the magnitudes) and it's pretty clear that the triangle inequality proves that statement, since 2|z1z2| will always be larger than the 'side' resulting from adding z1*z2 and z1z2* together... i'm just not sure how to show that algebraically. I even have a review sheet of complex number properties next to me but I don't really see anything that lets me get from one side to the other.

6. Jan 21, 2013

### Dick

|z1*z2+z1z2*|<=|z1*z2|+|z1z2*| is the triangle inequality, yes? What's the relation between |z1*z2| and |z1z2|?

7. Jan 21, 2013

### ElijahRockers

can I argue that since |z1z2*+z1*z2| = 2R{z1z2*}, then it must be less than or equal to 2|z1z2| since 2|z1z2| can have imaginary components? where if 2|z1z2| only has real components they are equal, otherwise, 2|z1z2| is greater?

8. Jan 21, 2013

### ElijahRockers

ohhh... they are equal right? |z1*z2| = |z1z2*| = |z1z2|

so directly from that and the trianlge inequality,

|z1*z2+z1z2*| <= |z1*z2| + |z1z2*|

|z1*z2+z1z2*| <= |z1z2| + |z1z2|

|z1*z2+z1z2*| <= 2|z1z2|

9. Jan 21, 2013

### Dick

Yes, that's it.

10. Jan 21, 2013

### Dick

I think what you actually proved is that |z1z2*+z1*z2|=2*|Re(z1z2*)| or the same thing without the absolute values. But it is true that |Re(a)|<=|a|, so you could go that way too.

11. Jan 21, 2013

### ElijahRockers

That's what I was trying to get at... my brain stopped working during the christmas break. :P Thanks for your help