What will the process be if a singularity number directly interacts with under root which is divisible???
Answers
Answer:
Further Applications
An important application of Cauchy's integral formula is the following Cauchy inequality. If is analytic and bounded, |f (z)| ≤ M on a circle of radius r about the origin, then
(11.37)
gives upper bounds for the coefficients of its Taylor expansion. To prove Eq. (11.37) let us define M(r) = max|z|=r|f (z)| and use the Cauchy integral for an = f(n)(z)/n!,
An immediate consequence of the inequality, Eq. (11.37), is Liouville's theorem: If f (z) is analytic and bounded in the entire complex plane it is a constant. In fact, if |f (z)| ≤ M for all z, then Cauchy's inequality Eq. (11.37), applied for |z| = r, gives |an| ≤ Mr−n. If now we choose to let r approach ∞, we may conclude that for all n > 0, |an| = 0. Hence f (z) = a0.
Conversely, the slightest deviation of an analytic function from a constant value implies that there must be at least one singularity somewhere in the infinite complex plane. Apart from the trivial constant functions then, singularities are a fact of life, and we must learn to live with them. As pointed out when introducing the concept of the point at infinity, even innocuous functions such as f (z) = z have singularities at infinity; we now know that this is a property of every entire function that is not simply a constant. But we shall do more than just tolerate the existence of singularities. In the next section, we show how to expand a function in a Laurent series at a singularity, and we go on to use singularities to develop the powerful and useful calculus of residues in a later section of this chapter.
A famous application of Liouville's theorem yields the fundamental theorem of algebra (due to C. F. Gauss), which says that any polynomial with n > 0 and an ≠ 0 has n roots. To prove this, suppose P(z) has no zero. Then 1/P(z) is analytic and bounded as |z| → ∞, and, because of Liouville's theorem, P(z) would have to be a constant. To resolve this contradiction, it must be the case that P(z) has at least one root λ that we can divide out, forming P(z)/(z − λ), a polynomial of degree n − 1. We can repeat this process until the polynomial has been reduced to degree zero, thereby finding exactly n roots.
Exercises
Unless explicitly stated otherwise, closed contours occurring in these exercises are to be understood as traversed in the mathematically positive (counterclockwise) direction.
11.4.1
Show that
(with the contour encircling the origin once), is a representation of the Kronecker δmn.
11.4.2
Evaluate
where C is the circle |z − 1| = 1.
11.4.3
Assuming that f (z) is analytic on and within a closed contour C and that the point z0 is within C, show that
11.4.4
You know that f (z) is analytic on and within a closed contour C. You suspect that the nth derivative f(n)(z0) is given by
Using mathematical induction (Section 1.4), prove that this expression is correct.
11.4.5
(a)
A function f (z) is analytic within a closed contour C (and continuous on C). If f (z) ≠ 0 within C and |f (z)| ≤ M on C, show that
for all points within C.
Hint. Consider w(z) = 1/f (z).
(b)
If f (z) = 0 within the contour C, show that the foregoing result does not hold and that it is possible to have |f (z)| = 0 at one or more points in the interior with |f (z)| > 0 over the entire bounding contour. Cite a specific example of an analytic function that behaves this way.
11.4.6
Evaluate
for the contour a square with sides of length a > 1, centered at z = 0.
11.4.7
Evaluate
where the contour encircles the point z = a.
11.4.8
Evaluate
for the contour the unit circle.
11.4.9
Evaluate
for the contour the unit circle.
Hint. Make a partial fraction expansion.