Euler's Formula

A convenient notation for complex numbers involves complex exponentials. It is based on Euler's formula: \begin{equation} e^{i\theta}=\cos\theta + i\sin\theta \label{Euler} \end{equation}

Euler's formula can be “proved” in two ways:

  1. Expand the left-hand and right-hand sides of Euler's equation (\ref{Euler}) in terms of known power series expansions. Compare equal powers.
  2. Show that both the left-hand and right-hand sides of Euler's equation (\ref{Euler}) are solutions of the same second order linear differential equation with constant coefficients. Since only two solutions of a second order linear equation are linearly independent, write \begin{equation} e^{i\theta}=A\cos\theta + B\sin\theta \label{Euler2} \end{equation} and choose boundary conditions and known properties of the exponential function to show that $A=1$ and $B=i$.

We can rearrange Euler's formula and its complex conjugate \begin{equation} e^{-i\theta}=\cos\theta - i\sin\theta \label{EulerConj} \end{equation} to find expressions for $\sin{\theta}$ and $\cos{\theta}$ in terms of complex exponentials: \begin{eqnarray} \cos{\theta}&=&\frac{1}{2}(e^{i\theta}+e^{-i\theta})\nonumber\\ \sin{\theta}&=&\frac{1}{2i}(e^{i\theta}-e^{-i\theta}) \label{defsincos} \end{eqnarray}

It turns out that these equations (\ref{defsincos}) can be extended to complex values of $\theta$. Rewriting (\ref{defsincos}) for a general complex-valued argument $z$, we find: \begin{eqnarray*} \cos{z}&=&\frac{1}{2}(e^{iz}+e^{-iz})\\ \sin{z}&=&\frac{1}{2i}(e^{iz}-e^{-iz}) \end{eqnarray*}

Note: If you only know properties of the exponential function for real numbers, then Euler's formula and the “proofs” above are not really proofs, rather they are an extension of the definition of exponentials to include pure imaginary coefficients. Technically, this happens through a process called analytic continuation.

The good news is that any formulas that you memorized for real exponentials also apply to complex exponentials. In particular: \begin{equation} e^{z_1 + z_2}=e^{z_1}+e^{z_2} \end{equation}

In practice, for a specific complex number $z$, these last expressions are easiest to evaluate if you express $z$ in rectangular form, $z=x+iy$. Otherwise you end with exponentials of exponentials.


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