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-11.8.6 & 7 – not applicable














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







*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!

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!