# Griffiths' Intro to QM — Complex exponential solution

I have a question with Griffiths' approach to the infinite square well. (Please forgive me for my poor use of MathJax)

I was trying to find a way to solve the infinite square well using complex exponentials, but I found that no matter what I did, I would reach some kind of odd condition that couldn't be solved. Then I found this:

The answer to this question uses Euler's formula to decompose the following and solving it.

$$e^{2ikL} = 1.$$

I am wondering if there is a way to solve this equation (or solve for the initial conditions in the infinite square well problem) without reference to Euler's formula.

• You might well have a hard time in that $e^{ix}$ when drawn on an Argand diagram is a vector of unit length rotated by an amount dictated by $x$ and any expansion of $e^{ix}$ as a series leads to a splitting of the series into real and imaginary parts which relate to the cosine and sine function. I would have thought that Maths SE is the place to ask? math.stackexchange.com – Farcher Oct 6 '17 at 8:52

Hint: \begin{align} e^{i \theta} = \cos \theta + i \sin \theta \end{align} where $\theta$ is a real number.
Consider $e^{i \theta} = a + i b$ where $a$ and $b$ are real numbers. We have \begin{align} \cos \theta + i \sin \theta = a + i b \end{align} We match the real part and imaginary part separately and obtain \begin{align} \cos \theta = a, \qquad \sin \theta = b \end{align}