# 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!

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

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.

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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!