# Exercises: Chapter 4, Section 4

1. Independence of parametrization). Let be a singular -cube and a 1-1 function such that and for . If is a -form, show that

Suppose . Using the definition of the integral, Theorem 4-9, the chain rule, and Theorem 3-13 augmented by Problem 3-39:

2. Show that , and use Stokes Theorem to conclude that for any 2-chain in (recall the definition of in Problem 4-23).

One has

If , then Stokes Theorem gives

because is closed. So .

Note however that no curve is the boundary of any two chain -- as the sum of the coefficients of a boundary is always 0.

3. Show that the integer of Problem 4-24 is unique. This integer is called the winding number of around 0.

If and where and and are 2-chains, the letting , one has . Using Stokes Theorem, one gets , which is a contradiction.

4. Recall that the set of complex numbers is simply with . If let be . Define thee singular 1-cube by , and the singular 2-cube by .
1. Show that , and that if is large enough.

The problem statement is flawed: the author wants to be defined to be . This would make the boundary . We assume these changes have been made.

When or , is the curve . When , it is the curve , and when , it is the curve . So . Let . Then, if , we have for all and all with . Since where , we see that cannot be zero since it is the sum of a number of length and one which is smaller in absolute value.

2. Using Problem 4-26, prove the Fundamental Theorem of Algebra: Every polynomial with has a root in .

Suppose that as above has no complex root. Letting be sufficiently large, we see by part (a) and Stokes' Theorem that , and so .

Now consider the 2-chain defined by . Now, when , we get the constant curve with value ; when , we get the curve ; and when or , we get the curve . So the boundary of is . Further, we have assumed that has no complex root, and so is a 2-chain with values in . Again, applying Stokes' Theorem, we get , and so . This contradicts the result of the last paragraph.

5. If is a 1-form on with , show that theere is a unique number such that for some function with .

Following the hint, implies and so is unique. On the other hand, if we let be this value and , then and .

6. If is a 1-form on such that , prove that

for some and . The differential is of the type considered in the last problem. So there is a unique for which there is a such that .

For positive and , define the singular 2-cube by . By Stokes' Theorem, we have . So . By the proof of the last problem, it follows that . Henceforth, let denote this common value. Note that ; and in particular, .

Let be a singular 1-cube with . By Problem 4-24, there is a 2-chain and an such that . By Stokes' Theorem, . So .

From the result of the last paragraph, integrating is independent of path. In fact, if you have two singular 1-cubes and with and , then prepend a curve from (1,0) to and postpend a path from to (1,0) to get two paths as in the last paragraph. The two integrals are both 0, and so the integrals over and are equal.

Now the result follows from Problem 4-32 below.

7. If , show that there is a chain such that . Use this fact, Stokes' theorem and to prove

One has . Suppose for some and some choice of . Then in a closed rectangle of positive volume centered at . Take for the k-cube defined in an obvious way so that its image is the part of the closed rectangle with for all different from the for . Then since the integrand is continuous and of the same sign throughout the region of integration.

Suppose . Let be a chain such that . By Stokes' Theorem, we would have: because . This is a contradiction.

1. Let be singular 1-cubes in with and . Show that there is a singular 2-cube such that , where and are degenerate, that is, and are points. Conclude that if is exact. Give a counter-example on if is merely closed.

Let be defined by . Then where is the curve with constant value and similarly for .

Suppose is exact, and hence closed. Then by Stokes' Theorem, we have (since is closed), and so .

The example: , , and shows that there is no independence of path in for closed forms.

2. If is a 1-form on a subset of and for all and with and , show that is exact.

Although it is not stated, we assume that the subset is open. Further, by treating each component separately, we assume that the subset is pathwise connected.

Fix a point in the subset. For every in the set, let be any curve from to , and set . Because of independence of path, is well defined. Now, if , then because is in the interior of the subset, we can assume that is calculated with a path that ends in a segment with constant. Clearly, then . Similarly, . Note that because and are continuously differentible, it follows that is closed since .

We want to check differentiability of . One has

The first pair of terms is because ; similarly the second pair of terms is . Finally, continuity of implies that the integrand is , and so the last integral is also . So is differentiable at . This establishes the assertion.

8. (A first course in complex variables.) If , define to be differentiable at if the limit

exists. (This quotient involves two complex numbers and this definition is completely different from the one in Chapter 2.) If is differeentiable at every point in an open set and is continuous on , then is called analytic on .

1. Show that is analytic and is not (where ). Show that the sum, product, and quotient of analytic functions are analytic.

and so . On the other hand, does not have a limit as because , but .

It is straightforward to check that the complex addition, subtraction, multiplication, and division operations are continuous (except when the quotient is zero). The assertion that being analytic is preserved under these operations as well as the formulas for the derivatives are then obvious, if you use the identities:

2. If is analytic on , show that and satisfy the Cauchy-Riemann} equations:

(The converse is also true, if and are continuously differentiable; this is more difficult to prove.)

Following the hint, we must have:

Comparing the real and imaginary parts gives the Cauchy-Riemann equations.

3. Let be a linear transformation (where is considered as a vector space over ). If the matrix of with respect to the basis is , show that is multiplication by a complex number if and only if and . Part (b) shows that an analytic function , considered as a function , has a derivative which is multiplication by a complex number. What complex number is this?

Comparing and gives , , , and . So, and exist if and only if and .

From the last paragraph, the complex number is where and .

4. Define

and

Show that if and only if satisfies the Cauchy-Riemann equations.

One has for that

Clearly this is zero if and only if the Cauchy-Riemann equations hold true for .

5. Prove the Cauchy Integral Theorem: If is analytic in , then for every closed curve (singular 1-cube with ) such that for some 2-chain in .

By parts (b) and (d), the 1-form is closed. By Stokes' Theorem, it follows that .

6. Show that if , then (or in classical notation) equals for some function . Conclude that .

One has

if is defined by .

This then gives .

7. If is analytic on , use the fact that is analytic in to show that

if for . Use (f) to evaluate and conclude:

Cauchy Integral Formula: If is analytic on and is a closed curve in with winding number around 0, then

The first assertion follows from part (e) applied to the singular 2-cube defined by .

By a trivial modification of Problem 4-24 (to use ) and Stokes' Theorem, for with .

Further,

if is chosen so that for all with . It follows that . Using part (f), we conclude that . The Cauchy integral formula follows from this and the result of the last paragraph.

9. If and , define by If each is a closed curve, is called a homotopy between the closed curve and the closed curve . Suppose and are homotopies of closed curves; if for each the closed curves and do not intersect, the pair is called a homotopy between the non-intersecting closed curves and . It is intuitively obvious that there is no such homotopy with the pair of curves shown in Figure 4-6 (a), and the pair of (b) or (c). The present problem, and Problem 5-33 prove this for (b) but the proof for (c) requires different techniques.
1. If are nonintersecting closed curves, define by

If is a homotopy of nonintersecting closed curves define by

Show that

When , one gets the same singular 2-cube ; similarly, when , one gets the same singular 2-cube . When (respectively ), one gets the singular 2-cube (respectively ). So which agrees with the assertion only up to a sign.

2. If is a closed 2-form on , show that

By Stokes' Theorem and part (a), one has .