# Cauchys Theorem

• June 13th 2007, 05:45 AM
moolimanj
Cauchys Theorem
Hi Any help with this question would be greatly appreciated - I just cant get my head around the squared term in the denominator:

(a) Evaluate the integral:

Int (e^zpi)/((z+i)(z-3i)^2) when

(i) C= {z:|z|=2}
(ii) C = {z:|z-1|=1}
(iii) C = {z:|z-2i|=2}

(b) Use liouvilles theorem to show that there is at least one value z subset C for which |sinz|>2007
(c) Show that if f is an entire function that satisfies |2007i+f(z)|>2007 then f is constant
(d) Deduce from the result in (b) that if f is an entire function that satisfies Im(f(z))>0 for all z subset c, then f is constant

Many thanks
• June 13th 2007, 07:44 AM
ThePerfectHacker
Quote:

Originally Posted by moolimanj
(b) Use liouvilles theorem to show that there is at least one value z subset C for which |sinz|>2007

Assume not! Then $|\sin z| \leq 2007$ for all $z\in \mathbb{C}$. Now since $f(z) = \sin z$ is an entire function it must mean that by Liouville's theorem that $\sin z = K$ for some complex number $K$, which is not true, i.e. $\sin \pi = 0 \mbox{ and }\sin \frac{
\pi}{2} = 1$
. Thus by contradiction $|\sin z| > 2007$ for some $z\in \mathbb{C}$.
Quote:

(c) Show that if f is an entire function that satisfies |2007i+f(z)|>2007 then f is constant
Hint: $||f(z)|-|2007i|| \leq |f(z) - 2007i| \leq |f(z)|+|2007i|$
$|2007i + f(z)| > 2007$
• June 13th 2007, 09:54 AM
ThePerfectHacker
Quote:

Originally Posted by moolimanj
(a) Evaluate the integral:

Int (e^zpi)/((z+i)(z-3i)^2) when

This is done by Cauchy's Integral Formula:
First we write,
$\oint_C \frac{e^{\pi z}}{(z+i)(z-3i)^2} dz = \oint_C e^{\pi z}\cdot \left( \frac{-1/16}{z+i} + \frac{1/16}{z-3i} + \frac{1/(4i)}{(z-3i)^2} \right) dz$
That was partial fractions.

Now, seperate the contour integrals over "Western Europe":
$\boxed{-\frac{1}{16}\oint_C \frac{e^{\pi z}}{z+i} dz + \frac{1}{16}\oint_C \frac{e^{\pi z}}{z-3i} dz + \frac{1}{4i} \oint_C \frac{e^{\pi z}}{(z-3i)^2} dz}$

Let me state the two theorems we need to know:

Theorem 1: Let $\gamma$ be a contour and let $f(z)$ be holomorphic in an open set containing the contour and its interior. Then for any point $a$ inside $\gamma$ we have:
$f(a) = \frac{1}{2\pi i} \oint_{\gamma} \frac{f(z)}{z-a} dz$

Theorem 2: Let $\gamma$ be a contour and let $f(z)$ be holomorphic in an open set containing the contour and its interior. Then for any point $a$ inside $\gamma$ we have:
$f'(a) = \frac{1}{2\pi i} \oint_{\gamma} \frac{f(z)}{(z-a)^2} dz$

Now we know enought to do this problem.

Quote:

(i) C= {z:|z|=2}
I will do this one and leave (ii) and (iii) for you to think about. Look below. The circle is shown and the red points are the singularities.

We have three contour integral we need to evaluate.
The second and third one are simple.
It is zero because "all poles are in Eastern Europe".
By that I mean that the function we are integrating are holomorphic in that circle. And Cauchy's Theorem (not the integral formula, the one with the contours) tells us that the integrals are zero.

So the problem reduces to find,
$-\frac{1}{16}\oint_C \frac{e^{\pi z}}{z-3i} dz$
The function here is clearly $f(z) = e^{\pi z}$ and it indeed satisfies Theorem 1.
So the Contour Integral is equal to:
$- \frac{1}{16} \cdot 2\pi i \cdot f(3i) = -\frac{\pi i e^{3\pi i}}{8} = - \frac{\pi i \cos 3\pi - \pi \sin 3\pi }{8} = \frac{\pi i}{8}$
• June 15th 2007, 05:42 AM
moolimanj
Cauchys
Thanks for you help on this perfecthacker

However, I cant figure out how to do part (a)(ii) and (iii).

if C={z:|z-1|=1 then 3i is again outside the circle. Does this mean that the answer is the same? -i in this instance is on the circle. I'm confused - if its on the circle, should the answer be 0, or would the answer be the same as part (i)?

Also for part (iii) if C={z:|z-2i|=2} then -i is excluded but 3i in included in the contour. So how can the answer be 0 by Cauchys theorem when at least one is contained within the circle. Please help
• June 15th 2007, 06:28 AM
ThePerfectHacker
Problem (ii): The contour integral is still the same:
$
\boxed{-\frac{1}{16}\oint_C \frac{e^{\pi z}}{z+i} dz + \frac{1}{16}\oint_C \frac{e^{\pi z}}{z-3i} dz + \frac{1}{4i} \oint_C \frac{e^{\pi z}}{(z-3i)^2} dz}
$

If it is on the circle, the contour integral is zero. But you made a mistake it is not on the circle. Both points are clearly outside the circle and hence by Cauchy's theorem its integral is zero.

EDIT: My picture is wrong. You were right, $i$ is on the circle. But anyways still the integral is zero.
• June 15th 2007, 06:39 AM
ThePerfectHacker
Problem (iii): Look at the picture below. As you say $3i$ is inside the contour and $-i$ is outside. That means the first integral in:
$
\boxed{-\frac{1}{16}\oint_C \frac{e^{\pi z}}{z+i} dz + \frac{1}{16}\oint_C \frac{e^{\pi z}}{z-3i} dz + \frac{1}{4i} \oint_C \frac{e^{\pi z}}{(z-3i)^2} dz}
$

Is zero by Cauchy's theorem.
However, the second and third one are not necessarily zero.

If you look at Theorem 1 and Theorem 2 that I posted. You will note that the function that we need here is $f(z)=e^{\pi z}$ to determine the value of each integral.
The first integral has $z-3i$ in the denominator so:
$\frac{1}{2\pi i}\oint_C \frac{e^{\pi z}}{z-3i} dz = f(3i) = e^{3\pi i} = \cos 3\pi + i \sin 3\pi = -1$.

To do the second integral we need the derivative: $f'(z) = \pi e^{\pi z}$ because it has $(z-3i)^2$ in the denominator. Thus:
$\frac{1}{2\pi i} \oint_C \frac{e^{\pi z}}{(z-3i)^2} dz = f'(3i) = \pi e^{3\pi i} = \pi \cos 3\pi + \pi i \sin 3\pi = - \pi$

Now you just have to add and simplify to get the answer which is trivial.