Cauchy Integral Formula Interpretation Problem:

**mathfied** · July 16th 2008, 09:58 AM

Hi having problems interpreting some conditions as usual. It's usually the interpretation of the conditions that are letting me down. Anyhow, its to do with Cauchy's Integral Formula.

Heres the question followed by my solution:

(a)
Let: $ \begin{array}{*{20}c} {\gamma _1 = \{ z \in C:\left| {z - 3i} \right| = 2\} ,} & {\gamma _2 = \{ z \in C:\left| z \right| = 5\} } \\ \end{array}$ be 2 positively oriented contours. Evaluate the following integrals clearly stating which results are being used:

$ \begin{array}{*{20}c} {I_1 = \int\limits_{\gamma _1 } {\frac{z} {{z^2 + 4}}dz,} } & {I_2 = \int\limits_{\gamma _2 } {\frac{z} {{z^2 + 4}}} } \\ \end{array} dz$

(b)
Using Cauchy Riemann prove that the function h(z)=sin(Imz) is not differentiable at any point of the strip $\{ z:\frac{{ - \pi }} {2} < {Im} z < \frac{\pi } {2}\}$

solution to (a):
----------------------
for contour 1: $ {\gamma _1 = \{ z \in C:\left| {z - 3i} \right| = 2\} }$ . I interpret this as the contour being centred at 3i and having radius 2?

So using the Cauchy Integral formula: $f(z_0 ) = \frac{1} {{2\pi i}}\int\limits_\gamma {\frac{{f(z)}} {{z - z_0 }}} dz$

using partial fractions we have,
$I_1 = \int\limits_{\gamma _1 (3i,2)} {\frac{z} {{z^2 + 4}}dz} = \int\limits_{\gamma _1 (3i,2)} {\frac{A} {{z + 2i}}dz} + \int\limits_{\gamma _1 (3i,2)} {\frac{B} {{z - 2i}}dz}$

so: $z = A(z - 2i) + (z + 2i)$
$ A = \frac{1} {2},B = \frac{1} {2}$

so now we have:
$ I_1 = \int\limits_{\gamma _1 (3i,2)} {\frac{1} {{2(z + 2i)}}dz} + \int\limits_{\gamma _1 (3i,2)} {\frac{1} {{2(z - 2i)}}dz} {\text{ }} = {\text{ }}\frac{1} {2}\int\limits_{\gamma _1 (3i,2)} {\frac{1} {{(z + 2i)}}} + \frac{1} {{(z - 2i)}}dz$

again, the cauchy integral formula says:
$f(z_0 ) = \frac{1} {{2\pi i}}\int\limits_\gamma {\frac{{f(z)}} {{z - z_0 }}} dz$
so we let f(z)=z,

so now we get
$ (2i + ( - 2i)) = \frac{1} {{2\pi i}}\frac{1} {2}\int\limits_{\gamma _1 (3i,2)} {\frac{1} {{(z + 2i)}}} + \frac{1} {{(z - 2i)}}dz$

THE LHS GOES TO 0 SO THE WHOLE INTEGRAL IS THEREFORE 0?
----------------------------------------------------------------------

For the 2nd integral: ${\gamma _2 = \{ z \in C:\left| z \right| = 5\} }$ , I interpret the contour being centred at 0 and having radius 5?.
However, the actual partial fractions are still the same and the LHS is still (2i + (-2i) which still makes the overall integral go to 0. So how are the any different.

Does this basically back up the fact from the Cauchy-Goursat theorem that any closed contour has an integral of 0?

Or am i misinterpreting the contour conditions?

solution for (b):
--------------------------
any suggestions here? i understand the cauchy-riemann equations say $u_x=v_y$ and $v_x=-u_y$ so I have to probably show that. but dont understand how it applies to the actual question.

thanks a lot people

**ThePerfectHacker** · July 16th 2008, 12:31 PM

Originally Posted by mathfied

$ I_1 = \int\limits_{\gamma _1 (3i,2)} {\frac{1} {{2(z + 2i)}}dz} + \int\limits_{\gamma _1 (3i,2)} {\frac{1} {{2(z - 2i)}}dz} {\text{ }} = {\text{ }}\frac{1} {2}\int\limits_{\gamma _1 (3i,2)} {\frac{1} {{(z + 2i)}}} + \frac{1} {{(z - 2i)}}dz$

The first integral is zero because $-2i$ is outside this circle.
While the second integral is $(2\pi)(1/2) = \pi$ .
(You do not even need Cauchy's integral formula here).

**mathfied** · July 16th 2008, 04:11 PM

hmm, i realised on my original post for part (A) that i had included the -2i even though it was outside the circle!

i thought that since -2i is outside the circle then we simply neglect the part of the integral that has the -2i pole:

but i arrive at a differnt answer to yours. please could you see where i have misinterpreted?

for part (A),since the -2i pole is outside the contour, we only deal with (z-2i) - this corresponds to the 2i pole - hence:
the cauchy integral formula states:
$f(z_0 ) = \frac{1} {{2\pi i}}\int\limits_\gamma {\frac{{f(z)}} {{z - z_0 }}} dz$
so we let f(z)=1,

so now we get
$ (1) = \frac{1} {{2\pi i}}\frac{1} {2}\int\limits_{\gamma _1 (3i,2)} {\frac{1} {{(z - 2i)}}}dz$

$ (4\pi i) = \int\limits_{\gamma _1 (3i,2)} {\frac{1} {{(z - 2i)}}} dz$

the integral is therefore $4 \pi i$ ??
i obviously dont get Pi. are you able to see which part of my reasoning is wrong??

**mathfied** · July 16th 2008, 04:14 PM

for part (B): ${\gamma _2 = \{ z \in C:\left| z \right| = 5\} }$ , I interpret the contour being centred at 0 (or 0i) and having radius 5.

However, the actual partial fractions are still the same but the conditions mean the contour effectively cover poles from 5i to -5i. Which also means that 2i and -2i are both inside the contour.

So now we have:
$I_2 = \int\limits_{\gamma _1 (0i,5)} {\frac{1} {{2(z + 2i)}}dz} + \int\limits_{\gamma _1 (0i,5)} {\frac{1} {{2(z - 2i)}}dz} {\text{ }} = {\text{ }}\frac{1} {2}\int\limits_{\gamma _1 (0i,5)} {\frac{1} {{(z + 2i)}}} + \frac{1} {{(z - 2i)}}dz$

the cauchy integral formula states:
$f(z_0 ) = \frac{1} {{2\pi i}}\int\limits_\gamma {\frac{{f(z)}} {{z - z_0 }}} dz$

so we have:
$ (1+1) = \frac{1} {{2\pi i}}\frac{1} {2}\int\limits_{\gamma _1 (0i,5)} {\frac{1} {{(z + 2i)}}} + \frac{1} {{(z - 2i)}}dz$

$ (2)(4\pi i) = \int\limits_{\gamma _1 (0i,5)} {\frac{1} {{(z + 2i)}}} + \frac{1} {{(z - 2i)}}dz$

= $8\pi i$ ??

this is different to your answer aswell mate. are there any points where my reasoning has gone in the wrong direction???

**ThePerfectHacker** · July 16th 2008, 04:15 PM

Originally Posted by mathfied

hmm, i realised on my original post for part (A) that i had included the -2i even though it was outside the circle!

i thought that since -2i is outside the circle then we simply neglect the part of the integral that has the -2i pole:

but i arrive at a differnt answer to yours. please could you see where i have misinterpreted?

for part (A),since the -2i pole is outside the contour, we only deal with (z-2i) - this corresponds to the 2i pole - hence:
the cauchy integral formula states:
$f(z_0 ) = \frac{1} {{2\pi i}}\int\limits_\gamma {\frac{{f(z)}} {{z - z_0 }}} dz$
so we let f(z)=1,

so now we get
$ (1) = \frac{1} {{2\pi i}}\frac{1} {2}\int\limits_{\gamma _1 (3i,2)} {\frac{1} {{(z - 2i)}}}dz$

$ (4\pi i) = \int\limits_{\gamma _1 (3i,2)} {\frac{1} {{(z - 2i)}}} dz$

the integral is therefore $4 \pi i$ ??
i obviously don't get 0. are you able to see which part of my reasoning is wrong??

You are not supposed to get zero. But your answer is still wrong.
$\frac{1}{2\pi i}\oint_{\gamma} \frac{dz}{z-2i} = 1$

Thus, $\frac{1}{2}\oint_{\gamma} \frac{dz}{z-2i} = \pi i$ .

**ThePerfectHacker** · July 16th 2008, 04:18 PM

Originally Posted by mathfied

So now we have:
$I_2 = \int\limits_{\gamma _1 (0i,5)} {\frac{1} {{2(z + 2i)}}dz} + \int\limits_{\gamma _1 (0i,5)} {\frac{1} {{2(z - 2i)}}dz} {\text{ }} = {\text{ }}\frac{1} {2}\int\limits_{\gamma _1 (0i,5)} {\frac{1} {{(z + 2i)}}} + \frac{1} {{(z - 2i)}}dz$

$\oint_{\gamma} \frac{dz}{z+2i} = \oint_{\gamma} \frac{dz}{z-2i} = 2\pi i$ .

Thus, $\frac{1}{2} \left( \oint_{\gamma} \frac{dz}{z-2i} + \oint_{\gamma} \frac{dz}{z+2i} \right) = \tfrac{1}{2}(2\pi i + 2\pi i) = 2\pi i$ .

**mathfied** · July 16th 2008, 04:30 PM

yes ofcourse, both make sense perfectly now.
it makes sense the way you say it.
but do you know why it doesnt work when i do it the cauchy integral way?
do you see where the mistake is, cus as far as i can see it should give the same answer as you have given, but i get different answers!

**mathfied** · July 16th 2008, 04:34 PM

any suggestions also for the final part of the original question??

(b)
Using Cauchy Riemann prove that the function h(z)=sin(Imz) is not differentiable at any point of the strip $\{ z:\frac{{ - \pi }} {2} < {Im} z < \frac{\pi } {2}\}$ ??

thanks a great deal bro!

**PaulRS** · July 16th 2008, 06:00 PM

Originally Posted by mathfied

for part (B): ${\gamma _2 = \{ z \in C:\left| z \right| = 5\} }$ , I interpret the contour being centred at 0 (or 0i) and having radius 5.

However, the actual partial fractions are still the same but the conditions mean the contour effectively cover poles from 5i to -5i. Which also means that 2i and -2i are both inside the contour.

So now we have:
$I_2 = \int\limits_{\gamma _2 (0i,5)} {\frac{1} {{2(z + 2i)}}dz} + \int\limits_{\gamma _2 (0i,5)} {\frac{1} {{2(z - 2i)}}dz} {\text{ }} = {\text{ }}\frac{1} {2}\int\limits_{\gamma _2 (0i,5)} {\frac{1} {{(z + 2i)}}} + \frac{1} {{(z - 2i)}}dz$

the cauchy integral formula states:
$f(z_0 ) = \frac{1} {{2\pi i}}\int\limits_\gamma {\frac{{f(z)}} {{z - z_0 }}} dz$

so we have:
$ (1+1) = \frac{1} {{2\pi i}}\frac{1} {2}\int\limits_{\gamma _2 (0i,5)} {\frac{1} {{(z + 2i)}}} + \frac{1} {{(z - 2i)}}dz$

$ (2)(4\pi i) = \int\limits_{\gamma _2 (0i,5)} {\frac{1} {{(z + 2i)}}} + \frac{1} {{(z - 2i)}}dz$

= $8\pi i$ ??

this is different to your answer aswell mate. are there any points where my reasoning has gone in the wrong direction???

We have: $ \oint_{\gamma _2 } {\tfrac{{dz}} {{z - 2i}}} = 2\pi \cdot i \cdot {\text{res}}_{z = 2i} \tfrac{1} {{z - 2i}} $

$2i$ is a simple pole, indeed $ \mathop {\lim }\limits_{z \to 2i} \left( {z - 2i} \right) \cdot \tfrac{1} {{z - 2i}} = 1 $ thus we have ${\text{res}}_{z = 2i} \tfrac{1} {{z - 2i}} =1$ and $ \oint_{\gamma _2 } {\tfrac{{dz}} {{z - 2i}}} = 2\pi \cdot i $

Similarly: $ \oint_{\gamma _2 } {\tfrac{{dz}} {{z + 2i}}} = 2\pi \cdot i $

The mistake in your post is that there's an extra 2 in the denominator that came out of nowhere, in the part where you said "so we have:..."

You should get $ \tfrac{1} {2} \cdot \oint_{\gamma _2 } {\left( {\tfrac{1} {{z - 2i}} + \tfrac{1} {{z + 2i}}} \right)} dz = 2\pi \cdot i $

**ThePerfectHacker** · July 16th 2008, 06:11 PM

Originally Posted by PaulRS

We have: $ \oint_{\gamma _2 } {\tfrac{{dz}} {{z - 2i}}} = 2\pi \cdot i \cdot {\text{res}}_{z = 2i} \tfrac{1} {{z - 2i}} $

$2i$ is a simple pole, indeed $ \mathop {\lim }\limits_{z \to 2i} \left( {z - 2i} \right) \cdot \tfrac{1} {{z - 2i}} = 1 $ thus we have ${\text{res}}_{z = 2i} \tfrac{1} {{z - 2i}} =1$ and $ \oint_{\gamma _2 } {\tfrac{{dz}} {{z - 2i}}} = 2\pi \cdot i $

Similarly: $ \oint_{\gamma _2 } {\tfrac{{dz}} {{z + 2i}}} = 2\pi \cdot i $

The mistake in your post is that there's an extra 2 in the denominator that came out of nowhere, in the part where you said "so we have:..."

You should get $ \tfrac{1} {2} \cdot \oint_{\gamma _2 } {\left( {\tfrac{1} {{z - 2i}} + \tfrac{1} {{z + 2i}}} \right)} dz = 2\pi \cdot i $

There is nothing wrong with what you said Paul, but this is not the best way to procede. I know you probably learned to do integration like that from me, however, in this case the original poster did not yet reach the residue theorem. They are only doing Cauchy's Integral Formula (which is later used to derive residue theorem more rigorous than what I did). Send me a PM if you want to know more about this.

---
In fact, we do not even need to know any of those theorems. It should be a know fact that $\oint_{\gamma}\frac{dz}{z-\alpha} = 2\pi i$ if $\gamma$ is a circle (or more generally a contour) and $\alpha$ lies inside the circle.

**mathfied** · July 17th 2008, 02:38 AM

yes ofcourse guys. i was adding the extra 2 at the denominator when there was no need to!

gr8 gr8!

i realised that i was always trying to make the actual integral look like the cauchy integral formula when i should have instead adapted the cauchy integral formula to look like the form of the integral!

hope that made sense lol, but if it doesnt then just wanted to let you know that i fully understood what went wrong from what you both explained! thank you very much!

any suggestions for that last part (b)???

**kalagota** · July 17th 2008, 05:37 AM

let $z=x+yi$ .. $h(z)=\sin (Im \,\, z) = \sin y$ , $\left\{z: \frac{-\pi}{2} < y < \frac{\pi}{2} \right\}$

$u(x,y)= \sin y$ and $v(x,y)=0$

show that at least one of the C-R equations is not satisfied in the domain.

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