226f03ghw2s.htm

Math 226

November 11, 2003

More graded homework

Due November 18, 2003

p. 163, #5 from our textbook.

Solution: We are to show that

where is analytic on and inside the simple closed contour and $z_{0}$ is not on But if $z_{0}$ is inside then by the Cauchy integral formulas, both of these integrals give the value The first of these gives this value since the function $f^{\prime }$ is itself analytic wherever is, (so in particular on and inside so the first Cauchy integral formula applies. The second integral gives as a direct application of the Cauchy formula for the first derivative.

If $z_{0}$ is outside the contour then both integrands are analytic on and inside and so both integrals equal zero, by the Cauchy-Goursat theorem.
Let be the circle traversed counterclockwise once. Evaluate
1. Solution: We write:
  
  This integral now has the form of the Cauchy formula:
  
  Where is a simple close contour on and in which is analytic, and $z_{0}$ is inside Here and so
2. Solution: We can use the Cauchy formula
  
  where is a simple close contour on and in which is analytic, and $z_{0}$ is inside Here and $z_{0}=i,$ so since we get
3. Solution: This integral can be written
  
  so that we can use the formula
  
  where is a simple close contour on and in which is analytic, and $z_{0}$ is inside Here and $z_{0}=0.$ (Note that, though the numerator $\frac{\sin z}{z-4}$ is not entire, it is still analytic on and inside ). Since
  
  we get
Evaluate the integral

where is the circle traversed counterclockwise once. Do this by following this plan:
1. Rewrite the integrand in the form
  
  for suitable numbers and
  
  Solution: If and are to be chosen so that
  
  we must have
  
  so that
  
  which leads to
  
  Thus
  
  and so
2. Replace the contour by a suitable pair of contours inside it, as in theorem 2 of section 46.
  
  Solution: If we let $C_{1}$ and $C_{2}$ be small circles (traversed counter-clockwise) centered at and repectively, say of radius (so that they do not overlap and are entirely inside ), then by the Cauchy-Goursat theorem, we have
  
  since, by the Cauchy integral theorem,
  
  since the integrands here are analytic on and inside the corresponding contours. But the remaining integrals can be evaluated using the Cauchy integral formula (or directly!), to give
  
  Solution 2: We could also evaluate this integral without changing contours, again by applying the Cauchy integral formula:
Let be an entire function and suppose that for all Prove that must be a constant function. (Hint: Compose with a function so that $g\circ f$ is entire and bounded. Apply Liouville's theorem to $g\circ f).$

Solution 1: Compose with the exponential function, so that for all

But since $e^{f}$ is entire, this tells us that $e^{f}$ is constant, by Liouville's theorem. This means that

for all But by the chain rule

and since, for all this tells us that for all

This implies that is constant.

Solution 2: After noting that $e^{f}$ is constant, we can write

where $w_{0}$ is the (non-zero) constant value of $e^{f},$ so that the only possible values of $f\left( z\right)$ are those in the logarithm of $w_{0}.$ But these values all lie in the line In the previous problem set we proved that every entire function whose values all lie on a single line must be constant.

Solution 3: Here's yet another way to finish the first solution, but not really; this argument makes use of topological ideas we haven't discussed. Anyway, after noting that $e^{f}$ is constant, we can write

where $w_{0}$ is the (non-zero) constant value of $e^{f},$ so that the only possible values of $f\left( z\right)$ are those in the logarithm of $w_{0}.$ But $\log w_{0}$ is the set of points This set of points is not connected, more than that, the only non-empty connected subsets of this set are the subsets consisting of a single point. ("Connected" is a technical term, which I won't define here, but you can imagine what it means). However the plane is connected, and being entire, is continuous on the whole plane, which implies that the range of is also connected. This can only be the case if the range of is a single point, which is to say, is constant.

Solution 4: We could also compose with the function Since for all we get, for all and so and so

Therefore, by Liouville's theorem,

is constant. That is, for some number $w_{0},$ it is the case that for all

This says that for all

and so is constant.

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