# Complex numbers and modulus problem

1. Jun 7, 2005

### TomMe

Suppose z1 = a + bi, z2 = c + di are complex numbers.

When does |z1 + z2| = |z1| - |z2|? (with || is modulus)

It seems obvious that this is the case when z2 = 0, but are there other solutions? According to the book, no. But after 2 days, I still cannot solve it!

Here is what I did:

$$\sqrt{(a+c)^2 + (b+d)^2} = \sqrt{a^2 + b^2} - \sqrt{c^2 + d^2}$$

Then I squared both sides, remembering that $$\sqrt{a^2 + b^2} - \sqrt{c^2 + d^2} > 0$$

If I work this out I get 3 conditions that need to be satisfied:

$$1) \sqrt{a^2 + b^2} - \sqrt{c^2 + d^2} > 0$$
$$2) ac + bd \geq 0$$
$$3) ad - bc = 0$$

I do not see how this can be equivalent with z2 = 0, so I tried another way. Instead of squaring both sides immediately, I made both sides positive:

$$\sqrt{(a+c)^2 + (b+d)^2} + \sqrt{c^2 + d^2} = \sqrt{a^2 + b^2}$$

When I work this out, I get 2 conditions:

$$1) ac + bd + c^2 + d^2 \leq 0$$
$$2) ad - bc = 0$$

I still don't see how this means that z2 = 0, furthermore I suspect that both sets of conditions need to be equivalent but I cannot prove this.

So I actually have 2 requests:
1. Can someone tell me how to solve this?
2. Can someone show me how both sets of conditions are equivalent to each other, if they are? And if they are not, did I make a mistake?

Thanks.

2. Jun 7, 2005

### Galileo

Do you know the proof for the triangle inequalities?

$$||z_1|-|z_2||\leq |z_1+z_2|\leq |z_1|+|z_2|$$

If you go through these proofs, you'll might get an idea when equality occurs.

3. Jun 7, 2005

### mathman

How about z2=-x*z1, where x is real and 0<x<1?

4. Jun 7, 2005

### TomMe

@Galileo: Actually, proving triangle inequality is one of the excercises that come after this one, so I'd like to keep that aside for now..although I'll look at it first thing tomorrow.

@mathman: You're right. If I did it correctly, that gives me $$(1-x)\sqrt{a^2 + b^2} = \sqrt{a^2 + b^2} - x\sqrt{a^2 + b^2}$$. So that means the solution in my book is wrong, right?
How did you come up with that solution anyway?

5. Jun 7, 2005

### Hurkyl

Staff Emeritus
Think geometrically.

6. Jun 8, 2005

### TomMe

Yes, I got it now. I also found a way to point it out algebraically, just write the complex numbers in polar form and go from there.

One last question though, suppose I get this question on a test or something..does it suffice to draw a picture of the complex numbers as vectors in the complex plane and point out what's happening, or do I need to do it the algebra way?
Somehow just making a drawing doesn't feel like it's complete..

Thanks for the help guys.

7. Jun 8, 2005

### mathman

Hurky is right. In fact by thinking geometrically, it is easy to see that the solution I gave you is the only solution.

8. Jun 8, 2005

### TomMe

Perhaps you're right (most probably), but I still feel a lot more at ease if I can solve it with equations though.