Solving Complex Identity: Help with z in Polar and Exponential Form

[theaterfreak]

Hi there.
I need help simplifying the following:
|[itex]\sqrt{z^2 -1}[/itex] +z| + |[itex]\sqrt{z^2 -1}[/itex] -z|

What I did was I rewrote z in polar coordinates, but I ran into some difficulties taking the square root of [itex]r^2(cos2θ-sin2θ)-1[/itex].
I also tried rewritting z in exponential form, but also had problems. Help?

 

[Alpha Floor]

In a first approach to the problem, I'd suggest you try to use the triangle inequality, both for the complete expression and for each of the terms.

You might end up with 2 inequalities in the form: "expression" less or equal to "a + b", and "expression" greater or equal to "a + c", hence "expression" equal to "a"

EDIT: I've tried and that doesn't work, sorry :P

 

[Alpha Floor]

I think I've got it!

If you square the whole expression and use a couple of identities on conjugates you end up with this:

|[itex]\sqrt{z^2 -1}[/itex] +z| + |[itex]\sqrt{z^2 -1}[/itex] -z|= [itex]\sqrt{4z^2-1}[/itex]

 

[theaterfreak]

I got 4z^2-4 not 4z^2-1

 

[Alpha Floor]

Actually I've just discovered a mistake in my calculation,

The result is: 2z

 

[theaterfreak]

Oh, ok. I ended up getting 2z as well.
But what doesn't make sense to me is that the original value should have no i (imaginary number) in it since we're taking the modulus. But after simplifying it, we get 2zm which would still have i in it.

 

[Alpha Floor]

I don't see why that doesn't make sense to you.

You can take the modulus of any complex number, and "z" is a complex number, it has an "i" in it: z=x+iy with real part x and imaginary part y.

The modulus of a complex number is its distance to the origin defined as |z|= sqrt(x^2 + y^2) = r

 

[AlephZero]

I don't see why that doesn't make sense to you.

It doen't make sense because the value of the |...| function is a real number, but your result of 2z is a complex number.

You can't just ignore the imaginary part of 2z, without a proper mathmatical reason for doing so.

 

[Alpha Floor]

Oh you're completely right, when I said 2z I meant 2|z| of course

But still I think the solution is not completely right! (Sorry, this is what happens when you try to solve problems quickly on a paper napkin). Let's review all the steps properly:

[tex]A=\left |{\sqrt{z^2-1}+z}\right |+ \left |{\sqrt{z^2-1}-z}\right |[/tex]

[tex]A^2 = \left |{\sqrt{z^2-1}+z}\right |^2 + \left |{\sqrt{z^2-1}-z}\right |^2 + 2\left |{\sqrt{z^2-1}+z}\right |\left |{\sqrt{z^2-1}-z}\right |[/tex]

[tex]A^2 = (\sqrt{z^2-1}+z)\overline{(\sqrt{z^2-1}+z)}+ (\sqrt{z^2-1}-z)\overline{(\sqrt{z^2-1}-z)}+2[/tex]

[tex]A^2= (\sqrt{z^2-1})(\overline{\sqrt{z^2-1}}) + \overline{z}z + \overline{z}\sqrt{z^2-1} + z\overline{\sqrt{z^2-1}}+ [/tex]

[tex]+ (\sqrt{z^2-1})(\overline{\sqrt{z^2-1}}) + \overline{z}z - \overline{z}\sqrt{z^2-1} - z\overline{\sqrt{z^2-1}} + 2[/tex]

[tex]A^2 = 2\left |{\sqrt{z^2-1}}\right |^2+2\left |{z}\right |^2 +2[/tex]

[tex]A^2= 2\left |{z^2-1}\right |+2\left |{z^2}\right |+2[/tex]

In conclusion:

[tex]z^2 \geq{1} \rightarrow{A^2=4z^2}\Longrightarrow{A=2\left |{z}\right |}[/tex]

[tex]z^2<1 \rightarrow{A^2=4}\Longrightarrow{A=2}[/tex]

 

[theaterfreak]

I solved this a little earlier and got the same thing. Thanks for the help!