July 20, 2018June 24, 2019 / 153armstrong / Leave a comment

If unit vectors always scared you for some reason, this neat little trick from The story of i by Paul Nahin involving complex numbers is bound to be a solace.

It allows you find the tangential and radial components of acceleration through simple differentiation. How about that!

Have a good one!

** r = r(t), θ = θ(t)

Using Complex numbers in Classical Mechanics

April 18, 2018April 18, 2018 / 153armstrong / Leave a comment

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

June 6, 2017June 6, 2017 / 153armstrong / Leave a comment

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.

drawing

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.

drawing-1

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 180^o transformation. And doing it again gets us back to 1.

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

drawing-2

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

Tricks that I wish I knew in High School : Trigonometry (#1)

June 4, 2017June 6, 2017 / 153armstrong / Leave a comment

I really wish that in High School the math curriculum would dig a little deeper into Complex Numbers because frankly Algebra in the Real Domain is not that elegant as it is in the Complex Domain.

To illustrate this let’s consider this dreaded formula that is often asked to be proved/ used in some other problems:

$cos(nx)cos(mx) =$ ?

Now in the complex domain:

$cos(x) = \frac{e^{ix} + e^{-ix}}{2}$

And therefore:

$cos(mx) = \frac{e^{imx} + e^{-imx}}{2}$

$cos(nx) = \frac{e^{inx} + e^{-inx}}{2}$

$cos(mx)cos(nx) = \left( \frac{e^{imx} + e^{-imx}}{2} \right) \left( \frac{e^{inx} + e^{-inx}}{2} \right)$

$cos(mx)cos(nx) = \frac{1}{4} \left( e^{i(m+n)x} + e^{-i(m+n)x} + e^{i(m-n)x} + e^{-i(m-n)x} \right)$

$cos(mx)cos(nx) = \frac{1}{2} \left( \left( \frac{e^{i(m+n)x} + e^{-i(m+n)x}}{2} \right) + \left( \frac{e^{i(m-n)x} + e^{-i(m-n)x}}{2} \right) \right)$

$cos(mx)cos(nx) = \frac{1}{2} \left( cos(m+n)x + cos(m-n)x \right)$
And similarly for its variants like $cos(mx)sin(nx)$ and $sin(mx)sin(nx)$ as well.

****

Now if you are in High School, that’s probably all that you will see. But if you have college friends and you took a peak what they rambled about in their notebooks, then you might this expression (for $m \neq n$ ):

$I = \int\limits_{-\pi}^{\pi} cos(mx)cos(nx) dx \\$

But you as a high schooler already know a formula for this expression:

$I = \int\limits_{-\pi}^{\pi} \left( cos(m+n)x + cos(m-n)x \right)dx \\$

$I = \int\limits_{-\pi}^{\pi} cos(\lambda_1 x) dx + \int\limits_{-\pi}^{\pi} cos(\lambda_2 x) dx \\$

where $\lambda_1$ , $\lambda_2$ are merely some numbers. Now you plot some of these values for lambda i.e ( $\lambda = 1,2, \hdots$ ) and notice that since integration is the area under the curve, the areas cancel out for any real number.

and so on….. Therefore:

$I = \int\limits_{-\pi}^{\pi} cos(mx)cos(nx)dx = 0$

This is an important result from the view point of Fourier Series!

n roots of unity

May 28, 2017 / 153armstrong / Leave a comment

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 n
roots of unity. i.e

As is common knowledge z = 1 is always a solution.

Multiplication as a transformation

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

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

This implies that :

And therefore :

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^orotation.

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 180^o 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 90^o rotation.

This applies to -i as well.

so on and so forth,

Have a great day!

December 11, 2016June 24, 2019 / 153armstrong / Leave a comment

fuckyeahphysica:

In mathematics there is a concept known as ‘Conformal Mapping’ which allows you convert a given shape to a completely different one by making a transformation.

In the joukowski transform you take all the points on a circle and apply the following transform:

And the resulting transformed points resemble an aerofoil shape. Pretty cool huh ?

** Conformal mappings are a really cool topic in complex analysis but also equally extensive. If you want to know more about them click here

EDIT:

I forgot to mention that the transformation of a circle into an aerofoil is a very special case and the transformation of other shapes would yield distinct shapes. ( very different from an aerofoil )

Here’‘s what i mean:

Different circle