# Thread: Evaluate the Integral. HELP!!

1. ## Evaluate the Integral. HELP!!

Evaluate the following integral. The contour is a square centered at the origin with corners at .

20.

2. This one is different there is a z in the numerator.

3. Originally Posted by lacy1104
This one is different there is a z in the numerator.
My mistake at not loking closer - sorry.

The answer is $2 \pi i (i) = - 2 \pi$.

You can get this answer using the following formula (which is a special case of Cauchy's Integral Formula):

$\oint_{\Gamma} \frac{f(z)}{z-\alpha} dz = 2\pi i f(\alpha)$ where $\Gamma$ is any countor containing $\alpha$.

Alternatively, you can integrate along each part of the curve in the way I outlined in your other question .....

4. Originally Posted by mr fantastic
My mistake at not loking closer - sorry.

The answer is $2 \pi i (i) = - 2 \pi$.

You can get this answer using the following formula (which is a special case of Cauchy's Integral Formula):

$\oint_{\Gamma} \frac{f(z)}{z-\alpha} dz = 2\pi i f(\alpha)$ where $\Gamma$ is any countor containing $\alpha$.

Alternatively, you can integrate along each part of the curve in the way I outlined in your other question .....
Alternatively, you can integrate directly:

I'll do the integral along the line segment Im(z) = 2 from z = 2 + 2i to z = -2 + 2i:

Note that on this line segment, z = x + 2i => dz = dx:

$\int_{x = 2}^{x = -2} \frac{x + 2i}{{\color{red} (x + 2i) - i}} \, dx = \int_{x = 2}^{x = -2} \frac{x + 2i}{x + i} \, dx$

$= \int_{x = 2}^{x = -2} 1 + \frac{i}{x + i} \, dx = \int_{x = 2}^{x = -2} 1 \, dx + i \int_{x = 2}^{x = -2}\frac{1}{x + i} \, dx$

$= -4 + i \left( \ln(x + i) \bigg{|}_{2}^{-2} \right) = -4 + i (\ln (-2 + i) - \ln (2 + i) )$

$= -4 + i \left( \left[ \ln(\sqrt{5}) + \left( \pi - \arctan \left( \frac{1}{2} \right)\right) \, i \right] - \left[ \ln(\sqrt{5}) + \left(\arctan \left( \frac{1}{2} \right)\right)\, i \right] \right)$

$= -4 + i (\pi \, i) = -4 - \pi$.

The integrals along the other line segments are done similarly. Then you add together all the bits from each integral.

5. Or you can do completely avoid Cauchy theorem: $\frac{z}{z-i} = \frac{(z-i)+i}{z-i} = 1 + i\cdot \frac1{z-i}$.

6. Originally Posted by ThePerfectHacker
Or you can do completely avoid Cauchy theorem: $\frac{z}{z-i} = \frac{(z-i)+i}{z-i} = 1 + i\cdot \frac1{z-i}$.
Indeed. And because f(z) = 1 is analytic on and inside the contour, its integral around the contour is zero ..... (Of course, it's a simple matter to integrate it directly: 4 - 4 + 4 - 4 = 0)

7. Originally Posted by mr fantastic
And because f(z) = 1 is analytic on and inside the contour, its integral around the contour is zero ..... (Of course, it's a simple matter to integrate it directly: 4 - 4 + 4 - 4 = 0)
It is even easier than that. $(z)' = 1$. (So it has a primitive. And integrals over piecewise smooth closed curves of a function possesing a primitive is zero. )

And interesting question to ask on an exam is why then $\oint z^{-1} dz \not = 0$. How about $\log z$?

8. Originally Posted by ThePerfectHacker
It is even easier than that. $(z)' = 1$. (So it has a primitive. And integrals over piecewise smooth closed curves of a function possesing a primitive is zero. )

And interesting question to ask on an exam is why then $\oint z^{-1} dz \not = 0$. How about $\log z$?
Just to avoid potential confusion, the contour encloses the special point z = 0. And therein lies the answer .....