Understanding arg(z1/z2) = arg(z1z2) with Conjugate z2

[Jbreezy]

Homework Statement

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

Homework Equations

The Attempt at a Solution

 

[SammyS]

Homework Statement

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

Homework Equations

The Attempt at a Solution

What have you tried?

Where are you stuck ?

 

[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)

 

[SammyS]

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)

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) ?

 

[Jbreezy]

Opps. Thanks

 

[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).

 

[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?

 

[Jbreezy]

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).

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.

Of course, that just verifies it for those two particular values in that example. What happens if you write for general z1 and z2 in polar form and try it?

What do you mean?

 

[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?

What do you mean?

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

 

[skiller]

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.

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})

Spoiler

=arg(z_1 \bar{z_2})

 

[LCKurtz]

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

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

 

[skiller]

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

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.

 

[Jbreezy]

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

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

 

[skiller]

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

Any complex number can be represented this way.

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

 

[Mentallic]

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

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

 

[Jbreezy]

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

 

[SammyS]

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.