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

In summary, the author is trying to solve a problem involving taking the square root of a complex number and ran into some difficulties. They ended up simplifying the problem to get 2z as a result. However, the imaginary part of 2z doesn't make sense to them and they need help to figure out why.
  • #1
theaterfreak
4
0
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?
 
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  • #2
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
 
  • #3
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]
 
  • #4
I got 4z^2-4 not 4z^2-1
 
  • #5
Actually I've just discovered a mistake in my calculation,

The result is: 2z
 
  • #6
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.
 
  • #7
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
 
  • #8
Alpha Floor said:
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.
 
  • #9
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]
 
Last edited:
  • #10
I solved this a little earlier and got the same thing. Thanks for the help!
 

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

1. What is the difference between polar and exponential form?

Polar form is a way of representing a complex number using its magnitude (or modulus) and angle (or argument). Exponential form, on the other hand, is a way of writing a complex number using its magnitude and a complex exponential term. In polar form, the complex number is represented as r(cosθ + isinθ), while in exponential form it is written as re^(iθ).

2. How do I convert a complex number from polar to exponential form?

To convert a complex number from polar to exponential form, you can use the following formula: re^(iθ) = r(cosθ + isinθ). Simply plug in the values of r and θ from the polar form into the formula to get the exponential form.

3. How do I convert a complex number from exponential to polar form?

To convert a complex number from exponential to polar form, you can use the following formula: r = √(a^2 + b^2) and θ = tan^-1(b/a). Simply plug in the values of a and b from the exponential form into the formula to get the polar form.

4. Why is it important to be able to solve complex identities in both polar and exponential form?

Being able to solve complex identities in both polar and exponential form allows for flexibility and ease in calculations involving complex numbers. Depending on the problem, one form may be more useful or easier to work with than the other, so having the ability to convert between the two can be very beneficial.

5. Can you provide an example of solving a complex identity using both polar and exponential form?

Sure, let's take the complex number z = 3(cos(π/4) + isin(π/4)). To convert this to exponential form, we use the formula re^(iθ) = r(cosθ + isinθ), so we have z = 3e^(iπ/4). To convert back to polar form, we use the formulas r = √(a^2 + b^2) and θ = tan^-1(b/a), so we have z = 3(cos(π/4) + isin(π/4)).

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