Complex Analysis - sqrt(z^2 + 1) function behavior

• Measle
In summary, the relevant equation for this problem is sqrt(z) = e^(1/2 log z) with the principal branch from (-pi, pi]. The argument of z^2 + 1 crosses the negative real half-line in the counterclockwise direction as z crosses (i, i inf) because when z = i, z^2 + 1 = 0. However, the argument of z^2 + 1 should be around pi in order to cross the negative real half-line. The value jumps from it to -it when the half lines are crossed, with the argument being close to pi but smaller for z=2i+epsilon and close to pi but larger for z'=2i-epsilon.
Measle

Homework Equations

The relevant equation is that sqrt(z) = e^(1/2 log z) and the principal branch is from (-pi, pi]

The Attempt at a Solution

The solution is provided, since this isn't a homework problem (I was told to post it here anyway). I don't understand why the number z^2 + 1 crosses the negative real half-line in the counterclockwise direction as z crosses (i, i inf). Is it because z^2 + 1 where z = i is 0? I would think the argument of z^2 + 1 should be around pi in order to cross the negative real half line, but I'm not sure. Also, I'm not sure how they said the value jumps from it to -it when the half lines are crossed.

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Let ##z=2i+\epsilon## for example, then ##z^2+1 = -3+\epsilon^2 + 2 i \epsilon##, just a little bit above the discontinuity in the square root. The argument is close to pi but smaller, the square root will be close to the positive imaginary axis.
For ##z'=2i-\epsilon## we get ##z'^2+1 = -3+\epsilon^2 - 2 i \epsilon##, just a little bit below the discontinuity in the square root. The argument is close to pi but larger, the square root will be close to the negative imaginary axis.

1. What is the domain of the function sqrt(z^2 + 1)?

The domain of the function sqrt(z^2 + 1) is all complex numbers except for those that make the expression under the square root equal to a negative number. In other words, the domain is all complex numbers except for those that make z^2 + 1 < 0.

2. What is the range of the function sqrt(z^2 + 1)?

The range of the function sqrt(z^2 + 1) is all complex numbers. This is because for every complex number z, there exists a complex number w such that w^2 = z^2 + 1. Therefore, the range is all complex numbers.

3. Is the function sqrt(z^2 + 1) continuous?

Yes, the function sqrt(z^2 + 1) is continuous. This means that as z approaches any value in its domain, the corresponding values of the function also approach a specific value. In other words, there are no abrupt changes or breaks in the graph of the function.

4. Does the function sqrt(z^2 + 1) have any singularities?

Yes, the function sqrt(z^2 + 1) has a singularity at z = i, where i is the imaginary unit. This means that the function is not defined at this point and becomes infinite.

5. What is the behavior of the function sqrt(z^2 + 1) as z approaches infinity?

The function sqrt(z^2 + 1) has two branches, one for positive values of z and one for negative values of z. As z approaches infinity on the positive branch, the function approaches infinity. On the negative branch, the function approaches negative infinity. Therefore, the behavior of the function is different depending on the direction in which z approaches infinity.

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