# Feynman’s trick applied to Contour Integration

A friend of mine was the TA for a graduate level  Math course for Physicists. And an exercise in that course was to solve  integrals using Contour Integration. Just for fun, I decided to mess with him by trying to solve all the contour integral problems in the prescribed textbook for the course [Arfken and Weber’s  ‘Mathematical methods for Physicists,7th edition”  exercise (11.8)] using anything BUT contour integration.

You can solve a lot of them them exclusively by using Feynman’s trick. ( If you would like to know about what the trick is – here is an introductory post) The following are my solutions:

All solutions in one pdf

Arfken-11.8.1

Arfken-11.8.2

Arfken-11.8.3

Arfken-11.8.4*

Arfken-11.8.5

Arfken-11.8.6 & 7 – not applicable

Arfken-11.8.8

Arfken-11.8.9

Arfken-11.8.10

Arfken-11.8.11

Arfken-11.8.12

Arfken-11.8.13

Arfken-11.8.14

Arfken-11.8.15

Arfken-11.8.16

Arfken-11.8.17

Arfken-11.8.18

Arfken-11.8.19

Arfken-11.8.20

Arfken-11.8.21 & Arfken-11.8.23* (Hint: Use 11.8.3)

Arfken-11.8.22

Arfken-11.8.24

Arfken-11.8.25*

Arfken-11.8.26

Arfken-11.8.27

Arfken-11.8.28

*I forgot how to solve these 4 problems without using Contour Integration. But I will update them when I remember how to do them. If you would like, you can take these to be challenge problems and if you solve them before I do send an email to 153armstrong(at)gmail.com and I will link the solution to your page. Cheers!

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# Using Complex numbers in Classical Mechanics

When one is solving problems on the two dimensional plane and you are using polar coordinates, it is always a challenge to remember what the velocity/acceleration in the radial and angular directions ($v_r , v_{\theta}, a_r, a_{\theta}$) are. Here’s one failsafe way using complex numbers that made things really easy :

$z = re^{i \theta}$

$\dot{z} = \dot{r}e^{i \theta} + ir\dot{\theta}e^{i \theta} = (\dot{r} + ir\dot{\theta} ) e^{i \theta}$

From the above expression, we can obtain $v_r = \dot{r}$ and $v_{\theta} = r\dot{\theta}$

$\ddot{z} = (\ddot{r} + ir\ddot{\theta} + i\dot{r}\dot{\theta} ) e^{i \theta} + (\dot{r} + ir\dot{\theta} )i \dot{\theta} e^{i \theta}$

$\ddot{z} = (\ddot{r} + ir\ddot{\theta} + i\dot{r}\dot{\theta} + i \dot{r} \dot{\theta} - r\dot{\theta}\dot{\theta} )e^{i \theta}$

$\ddot{z} = (\ddot{r} - r(\dot{\theta})^2+ i(r\ddot{\theta} + 2\dot{r}\dot{\theta} ) )e^{i \theta}$

From this we can obtain $a_r = \ddot{r} - r(\dot{\theta})^2$ and $a_{\theta} = (r\ddot{\theta} + 2\dot{r}\dot{\theta})$ with absolute ease.

Something that I realized only after a mechanics course in college was done and dusted but nevertheless a really cool and interesting place where complex numbers come in handy!

# nth roots of unity : A geometric approach

When one is dealing with complex numbers, it is many a times useful to think of them as transformations. The problem at hand is to find the nth roots of unity. i.e

$z^n = 1$

## Multiplication as a Transformation

Multiplication in the complex plane is mere rotation and scaling. i.e

$z_{1} = r_{1}e^{i\theta_{1}}, z_{2} = r_{2}e^{i\theta_{2}}$

$z_{1}z_{2} = \underbrace{r_{1} r_{2}}_{scaling} \underbrace{e^{i(\theta_{1} + \theta_{2})}}_{rotation}$

Now what does finding the n roots of unity mean?

If you start at 1 and perform n equal rotations( because multiplication is nothing but rotation + scaling ), you should again end up at 1.

We just need to find the complex numbers that do this.i.e

$z^n = 1$

$\underbrace{zz \hdots z}_{n} = 1$

$z = re^{i\theta}$

$r^{n}e^{i(\theta + \theta + \hdots \theta)} = 1e^{2\pi k i}$

$r^{n}e^{in\theta} =1e^{2\pi k i}$

This implies that :

$\theta = \frac{2\pi k}{n}, r = 1$

And therefore :

$z = e^{\frac{2\pi k i}{n}}$

Take a circle, slice it into n equal parts and voila you have your n roots of unity.

## Okay, but what does this imply ?

Multiplication by 1 is a $360^o/0^o$ rotation.

When you say that you are multiplying a positive real number(say 1) with 1 , we get a number(1) that is on the same positive real axis.

Multiplication by (-1) is a $180^o$ rotation.

When you multiply a positive real number (say 1) with -1, then we get a number (-1) that is on the negative real axis

The act of multiplying 1 by (-1) has resulted in a 180o transformation. And doing it again gets us back to 1.

Multiplication by $i$ is a $90^o$ rotation.

Similarly multiplying by i takes 1 from real axis to the imaginary axis, which is a 90o rotation.

This applies to -i as well.

That’s about it! – That’s what the nth roots of unity mean geometrically. Have a good one!