Complex Exponentials

Complex exponentials are used immensely in math and as a result, in many fields of science. It is also used in abundance throughout this site so it is important to understand what they are for future reference. They show the relationship between exponentials and trigonometry on a fundamental level. The following is the relationship.

$e^{ix}=\textup{cos}(x)+i\textup{sin}(x)$

This relationship may seem very obscure but it can be shown to be true in many ways. Assume the derivative of both sides are taken twice as shown below.

$\left ( \frac{\mathrm{d} }{\mathrm{d} x} \right )^2e^{ix}=\left ( \frac{\mathrm{d} }{\mathrm{d} x} \right )^2(\textup{cos}(x)+i\textup{sin}(x))$
$i^2e^{ix}=-\textup{cos}(x)-i\textup{sin}(x)$
$-e^{ix}=-(\textup{cos}(x)+i\textup{sin}(x))$

The second derivatives of both functions are equal to negative of the original function. However, this may seem, in some sense, like a coincidence. There exists an alternative proof of this equivalence. Consider the Taylor series of $e^x$ , $\textup{sin}(x)$ , and $\textup{cos}(x)$ as shown below.

$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}...$
$\textup{sin}(x)=\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!}=\frac{x^1}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}...$
$\textup{cos}(x)=\sum_{n=0}^\infty \frac{x^{2n}}{(2n)!}=\frac{x^0}{0!}-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}...$

Assume $x$ is substituted with $ix$ for the definition of $e^{ix}$ . The following is the result.

$e^{ix}=\frac{(ix)^0}{0!}+\frac{(ix)^1}{1!}+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\frac{(ix)^4}{4!}+\frac{(ix)^5}{5!}+\frac{(ix)^6}{6!}...$
$=\frac{x^0}{0!}+i\frac{x^1}{1!}-\frac{x^2}{2!}-i\frac{x^3}{3!}+\frac{x^4}{4!}+i\frac{x^5}{5!}-\frac{x^6}{6!}...$

By regrouping and factoring out, one starts to see the formation of the identity.

$=\left ( \frac{x^0}{0!}-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}... \right )+i\left ( \frac{x^1}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}... \right )$

The terms grouped in parentheses are simply the Taylor series of sine and cosine as presented above. This was the desired result. The identity can be proven through Taylor series.

One can also redefine sine and cosine in terms of the exponentials as shown below.

$\textup{sin}(x)=\frac{e^{ix}-e^{-ix}}{2i}$

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

This can be tested by substituting in the values for $e^{ix}$ and $e^{-ix}$ . Note: $e^{-ix}=\textup{cos}(x)-i\textup{sin}(x)$ because $\textup{cos}(-x)=\textup{cos}(x)$ but $\textup{sin}(-x)=-\textup{sin}(x)$ . $e^{-ix}$ is basically the complex conjugate of $e^{ix}$ .
$\textup{sin}(x)=\frac{e^{ix}-e^{-ix}}{2i}$
$=\frac{(\textup{cos}(x)+i\textup{sin}(x))-(\textup{cos}(x)-i\textup{sin}(x)))}{2i}$
$=\frac{2i\textup{sin}(x)}{2i}$
$=\textup{sin}(x)$

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

$=\frac{(\textup{cos}(x)+i\textup{sin}(x))+(\textup{cos}(x)-i\textup{sin}(x))}{2}$

$\frac{2\textup{cos}(x)}{2}$

$\textup{cos}(x)$

Note their resemblance to the definition of hyperbolic sine and cosine ( $\textup{sinh}(x)=\frac{e^x-e^{-x}}{2}$ and $\textup{cosh}(x)=\frac{e^x+e^{-x}}{2}$ ). In fact, $\textup{sinh}(ix)=i\textup{sin}(x)$ and $\textup{cosh}(ix)=\textup{cos}(x)$ .

In fact, there exists an interesting connection between this identity and the plotting of complex numbers. Complex numbers are plotted in two dimensional space where their imaginary component is the $y$ coordinate and their real component is the $x$ coordinate. This means any given coordinate $(x,y)$ has the value $x+yi$ . Now consider a Cartesian space with a unit circle in it. Any point in the unit circle has the coordinates $(\textup{cos}(\theta),\textup{sin}(\theta))$ where $\theta$ is the angle the line from the point to the origin makes with the positive x-axis. If one thinks of this Cartesian space as a complex space, the value of any point on the unit circle is $e^{i\theta}$ . In fact, any point on complex space can be represented using complex exponentials by simply making the coefficient of the exponential the distance $A$ from the origin ( $Ae^{i\theta}$ ).

It’s amazing how to see how two totally different concepts are actually intertwined with each other. A complex exponential is just a wave in complex space.

If all this math is still confusing and doesn’t quite capture the relationship between the two concepts for you, watch 3Blue1Brown‘s amazing video on this concept that explains why it works geometrically.

3 responses to “Complex Exponentials”

zimbles

May 5, 2016 at 5:32 pm

Even though I understand most of the math, the relation between exponentials and trigonometry still feels mysterious and amazing to me. I think I still didn’t built a full intuitive sense of it.

The fact that it is essential in one of the most fundamental physical laws, the Schrödinger equation, makes it even better. It’s like physics is built on the most beautiful mathematical equation.

Thanks for your great post.

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1. aakashl
  
  May 5, 2016 at 5:58 pm
  
  You should watch the video at the bottom of the post! It explains the connection between exponentiation, complex numbers, and trigonometry very elegantly through visuals without using any of the complex math in this post at all. It might make things a little more clear.
  
  LikeLiked by 1 person
  
The Lagrangian | phys-x

July 16, 2016 at 5:13 pm

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Complex Exponentials

3 responses to “Complex Exponentials”

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