Confused by the behavior of sqrt(z^2+1)

  • #1
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(mentor note: this is a homework problem with a solution that the OP would like to understand better)

In Taylor's Complex Variables,

Example 1.4.10
WSK46H9.jpg


Can someone help me understand this? I don't know what they mean by (i, i inf), or how they got it and -it
 

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  • #2
FactChecker
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##(i, i\infty)## is the vertical line in the complex plane along the imaginary axis from ##i## upward to ##i\infty##.
 
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##(i, i\infty)## is the vertical line in the complex plane along the imaginary axis from ##i## upward to ##i\infty##.
I see - how does that lead to -it and it? or rather how do you find the discontinuities of sqrt(1 + z^2)?
 
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  • #4
RPinPA
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OK, I'm a little rusty on branch cuts and complex analysis but let me try to reconstruct what that answer is saying. Unfortunately I don't have whatever definition they refer to in terms of the log function.

First, what are they saying about ##\sqrt{z}## as z goes counterclockwise? ##\sqrt{z}## is a solution w to ##w^2 = z = re^{i\theta}## so ##\sqrt z = \sqrt r e^{i(\theta/2)}##. If ##\theta## is just under ##\pi##, ##\theta = \pi - \epsilon##, then we choose as the principal value for ##\sqrt z## the solution with argument, i.e. phase, ##(\pi/2) - (\epsilon/2)##. That makes it a value just to the right of the positive imaginary axis.

But as we increase the angle to something above ##\pi## to ##\theta = \pi + \epsilon##, dividing that by two would give an argument for ##\sqrt z## of just over ##\pi/2##. We choose instead to use the equivalent angle ##\theta = \pi + \epsilon - 2\pi = -\pi + \epsilon##. As a result, the argument of ##\sqrt z## is ##-(\pi/2) + (\epsilon/2)## and the square root is just to the right of the negative imaginary axis.

This accounts for the sentence in the solution "The function ##\sqrt z## jumps from values on the positive imaginary axis to their negatives as z crosses this line [the negative real axis] in the counterclockwise direction".

##\sqrt z## has a discontinuity when z crosses that line, when z goes from a negative real number with a small positive imaginary part to one with a small negative imaginary part. The cut line is where z is a negative real number. So ##\sqrt {z^2 + 1}## is going to have a jump where ##(z^2 + 1)## does that, where ##z^2 + 1## is a negative real number.

##z^2 + 1## has real values ##< 0## when ##z^2 < -1## which means z is an imaginary number ##i \alpha## with either ##\alpha > 1## or ##\alpha < -1##. So the cut lines for ##\sqrt{z^2 + 1}## are those locations on the imaginary axis, the part going from i up and from -i down. As before, when you cross those lines, the square root jumps from values with positive imaginary parts and values with negative imaginary parts.
 
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  • #5
WWGD
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One thing that may help is considering this other way of seeing what happens when you wind once, twice around ( Sorry if this is what you did)

## \sqrt {e^{i\theta}}=e^{i(\theta/2)} \sqrt{e^(i(\theta +2\pi)}= e^ (i(\theta/2+ \pi)}= e^{(\theta/2)} e^{i\pi} = -e^{i\theta/2}##

So that the square root changes sine when you wind around a second time. The branch cut is intended to prevent that from happening, i.e., to prevent a curve having that property. If you remove the x-axis in above case, no curve will wind twice around.
Notice that if you go around a third time you return to the original value of the square root. In some cases , like that of Logz , you will never return to your initial value. For n-th root, you go around n times.
 
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  • #6
nrqed
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One thing that may help is considering this other way of seeing what happens when you wind once, twice around ( Sorry if this is what you did)

$$ \sqrt{e^{(i\theta)}}=e^{i(\theta/2)} \sqrt{e^{(i(\theta +2\pi)}}= e^{(i(\theta/2+ \pi)}= e^{i\theta/2} e^{i\pi} = e^{-i\theta/2} $$

So that the square root changes sine when you wind around a second time. The branch cut is intended to prevent that from happening, i.e., to prevent a curve having that property. If you remove the x-axis in above case, no curve will wind twice around.
Notice that if you go around a third time you return to the original value of the square root. In some cases , like that of Logz , you will never return to your initial value. For n-th root, you go around n times.
I just corrected the LaTeX mistakes.
 
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  • #7
WWGD
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OK, I'm a little rusty on branch cuts and complex analysis but let me try to reconstruct what that answer is saying. Unfortunately I don't have whatever definition they refer to in terms of the log function.

First, what are they saying about ##\sqrt{z}## as z goes counterclockwise? ##\sqrt{z}## is a solution w to ##w^2 = z = re^{i\theta}## so ##\sqrt z = \sqrt r e^{i(\theta/2)}##. If ##\theta## is just under ##\pi##, ##\theta = \pi - \epsilon##, then we choose as the principal value for ##\sqrt z## the solution with argument, i.e. phase, ##(\pi/2) - (\epsilon/2)##. That makes it a value just to the right of the positive imaginary axis.

But as we increase the angle to something above ##\pi## to ##\theta = \pi + \epsilon##, dividing that by two would give an argument for ##\sqrt z## of just over ##\pi/2##. We choose instead to use the equivalent angle ##\theta = \pi + \epsilon - 2\pi = -\pi + \epsilon##. As a result, the argument of ##\sqrt z## is ##-(\pi/2) + (\epsilon/2)## and the square root is just to the right of the negative imaginary axis.

This accounts for the sentence in the solution "The function ##\sqrt z## jumps from values on the positive imaginary axis to their negatives as z crosses this line [the negative real axis] in the counterclockwise direction".

##\sqrt z## has a discontinuity when z crosses that line, when z goes from a negative real number with a small positive imaginary part to one with a small negative imaginary part. The cut line is where z is a negative real number. So ##\sqrt {z^2 + 1}## is going to have a jump where ##(z^2 + 1)## does that, where ##z^2 + 1## is a negative real number.

##z^2 + 1## has real values ##< 0## when ##z^2 < -1## which means z is an imaginary number ##i \alpha## with either ##\alpha > 1## or ##\alpha < -1##. So the cut lines for ##\sqrt{z^2 + 1}## are those locations on the imaginary axis, the part going from i up and from -i down. As before, when you cross those lines, the square root jumps from values with positive imaginary parts and values with negative imaginary parts.
I just corrected the LaTeX mistakes.
Thanks and sorry, I was just leaving the coffee shop and did not have internet access till just now.
 

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