How ought to we outline √(z² − 1)? Properly, you possibly can sq. z, subtract 1, and take the sq. root. What else would you do?!
The query seems to be extra refined than it appears to be like.
When x is a non-negative actual quantity, √x is outlined to be the non-negative actual quantity whose sq. is x. When x is a posh quantity √x is outlined to be a operate that extends √x from the true line to the complicated aircraft by analytic continuation. However we will’t prolong √x as an analytic operate to the complete complicated aircraft ℂ. We have now to decide on to make a “reduce” someplace, and the standard alternative is to make a reduce alongside the unfavorable actual axis.
Utilizing the precept department
The “precept department” of the sq. root operate is outlined to be the distinctive operate that analytically extends √x from the constructive reals to ℂ (−∞, 0].
Assume for now that by √x we imply the precept department of the sq. root operate. Now what does √(z² − 1) imply? It might imply simply what we stated on the prime of the publish: we sq. z, subtract 1, and apply the (precept department of the) sq. root operate. If we try this, we should exclude these values of z such that (z² − 1) is unfavorable. This implies we’ve got to chop out the imaginary axis and the interval [−1, 1].
That is what Mathematica will do when requested to guage Sqrt[z^2 - 1]. The command
ComplexPlot[Sqrt[z^2 - 1], {z, -2 - 2 I, 2 + 2 I}]
makes the department cuts clear by abrupt modifications in colour.
A distinct method
Now let’s take a special method. Think about the operate √(z² − 1) as a complete. Don’t consider it procedurally as above, first squaring z and so forth. As an alternative, consider a it as a black field that takes in z and returns a posh quantity whose sq. is z² − 1.
This operate has an apparent definition for z > 1. And we will prolong this operate, through analytic continuation, to extra of the complicated aircraft. We are able to do that immediately, not by extending the sq. root operate. However as earlier than, we can’t prolong the operate analytically to all of ℂ. We have now to chop one thing out. A typical alternative is to chop out [−1, 1]. This eliminates the necessity for a department reduce alongside the imaginary axis.
The operate
extends √(z² − 1) the best way we would like.
The Mathematica code
ComplexPlot[Exp[(1/2) (Log[z - 1 ] + Log[z + 1])], {z, -2 - 2 I, 2 + 2 I}]
reveals that our operate is now steady throughout the imaginary axis, although there’s nonetheless a discontinuity as you cross [−1, 1].
We used this analytic extension of √(z² − 1) within the earlier publish to eradicate department cuts in an id.
