Suppose . Using the definition of the integral, Theorem 4-9, the chain rule, and Theorem 3-13 augmented by Problem 3-39:
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.
If and where and and are 2-chains, the letting , one has . Using Stokes Theorem, one gets , which is a contradiction.
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.
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.
Following the hint, implies and so is unique. On the other hand, if we let be this value and , then and .
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.
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.
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.
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.
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 .
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:
(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.
Comparing and gives , , , and . So, and exist if and only if and .
From the last paragraph, the complex number is where 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 .
By parts (b) and (d), the 1-form is closed. By Stokes' Theorem, it follows that .
if is defined by .
This then gives .
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 .
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.
If is a homotopy of nonintersecting closed curves define by
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.
By Stokes' Theorem and part (a), one has .