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

[Measle]

(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

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

 

[FactChecker]

$(i, i\infty)$ is the vertical line in the complex plane along the imaginary axis from $i$ upward to $i\infty$.

 

[Measle]

$(i, i\infty)$ 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)?

 

[RPinPA]

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.

 

[WWGD]

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.

 

[nrqed]

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.

 

[WWGD]

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.