# Homework Help: Should be a quick one

1. Aug 4, 2013

### Jbreezy

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

How is arg(z1/z2) = arg(z1z2) ???Where the bold z2 represents the conjugate.

2. Relevant equations

3. The attempt at a solution

2. Aug 4, 2013

### SammyS

Staff Emeritus
What have you tried?

Where are you stuck ?

3. Aug 4, 2013

### Jbreezy

Yeah I tried. I just took an example say z1= 2+2i and z2= 3+4i

So arg (z1/z2) = .56-.08i
now arg(z1z2) = 14-2i
I don't get how this person wrote that arg(z1/z2) = arg(z1z2)

4. Aug 4, 2013

### SammyS

Staff Emeritus
Actually:

arg (z1/z2) = arg(.56-.08i)

and

arg(z1z2) = arg(14-2i) .

Are not those the same; arg(.56-.08i) and arg(14-2i) ?

5. Aug 4, 2013

Opps. Thanks

6. Aug 4, 2013

### skiller

So you are happy that your particular example satisfies the identity $arg(z_1/z_2)=arg(z_1 \bar{z_2})$, but do you understand why it is an identity - ie it is satisfied by any $z_1$ and $z_2$ (with $z_1,z_2 \neq 0$)?

If not, think about how you can represent each of the following:
$$arg(z_1/z_2)$$$$arg(z_1 z_2)$$$$arg(\bar{z_2})$$ in terms of $arg(z_1)$ and $arg(z_2)$.

Last edited: Aug 4, 2013
7. Aug 4, 2013

### LCKurtz

Of course, that just verifies it for those two particular values in that example. What happens if you write for general $z_1$ and $z_2$ in polar form and try it?

8. Aug 4, 2013

### Jbreezy

$$arg(z_1/z_2)= arg(z1)-arg(z2)$$
$$arg(z_1 z_2)= arg(z1) +arg(z2)$$
$$arg(\bar{z_2})= -arg(z2)$$but you said in terms of $arg(z1)$and $arg(z2)$
So I don;t know about the last one.

What do you mean?

9. Aug 4, 2013

### LCKurtz

$z = re^{i\theta}$ form.

10. Aug 4, 2013

### skiller

That's right, you've answered all three correctly.

So,
$$arg(z_1/z_2)=arg(z_1)-arg(z_2)$$$$=arg(z_1)+arg(\bar{z_2})$$
$$=arg(z_1 \bar{z_2})$$

11. Aug 4, 2013

### LCKurtz

But didn't he say he didn't understand the third one?

12. Aug 4, 2013

### skiller

I took that to mean that he/she was unsure about it simply because I'd said "in terms of $arg(z_1)$ and $arg(z_2)$" and only one of these terms was necessary. A problem of the wording really, rather than not understanding the answer, IMO.

EDIT: @Jbreezy When asked to represent an expression "in terms of A, B and C" (for example), you are not obliged to use all of A, B and C in your answer.

Last edited: Aug 4, 2013
13. Aug 4, 2013

### Jbreezy

I don't know this form. You represent a complex number like a + ib like this? what?

14. Aug 4, 2013

### skiller

Any complex number can be represented this way.

$r$ represents the modulus (ie $\sqrt{a^2+b^2}$) and $\theta$ represents the argument.

15. Aug 5, 2013

### Mentallic

It's equivalent to
$$z=r \left(\cos(\theta)+i\sin(\theta)\right)=re^{i\theta}$$

16. Aug 5, 2013

### Jbreezy

Yeah I will answer this question when I get to that part in my text.Thx

17. Aug 5, 2013

### SammyS

Staff Emeritus
Multiply z1/z2 by z2/z2 .

This gives (z1z2)/(z2z2) .

However, (z2z2) is a purely real number, so dividing by (z2z2) has no effect on the argument.