> ## Content Index
> Fetch the complete content index at: https://a-math-instructors-random-notes.ghost.io/llms.txt
> Use this file to discover other available public pages before exploring further.

# Three and a half Ways to Prove Euler's formula
- URL: https://a-math-instructors-random-notes.ghost.io/three-and-a-half-ways-to-prove-eulers-formula/
- Published: 2026-07-12T20:59:39.000Z
- Updated: 2026-07-14T02:52:06.000Z
- Author: Takashi Nakamura

Euler's formula $e^{i\\theta} = \\cos\\theta + i\\sin\\theta$ is one of the most important formulae in engineering and science. See, for example, a beautifully done video about Euler's formula with $\\theta=\\pi$:

### 1\. Maclaurin Series

This is the most common proof you see on textbooks. The Maclaurin series of $e^{i\\theta}$ reads   
\\begin{align\*}   
e^{i\\theta} &= \\sum\_{k = 0}^{\\infty} \\frac{(i\\theta)^k}{k!} = 1+i\\theta -\\frac{1}{2}\\theta^2 - \\frac{i}{3!}\\theta^3 +\\frac{1}{4!}\\theta^4+\\frac{i}{5!}\\theta^5 - \\cdots\\\\   
&= (1-\\frac{1}{2}\\theta^2+\\frac{1}{4!}\\theta^4 - \\cdots) + i(\\theta- \\frac{1}{3!}\\theta^3+\\frac{1}{5!}\\theta^5- \\cdots)   
\\end{align\*}   
We identify the Maclaurin series of $\\cos\\theta$ and $\\sin\\theta$, respectively:  
\\begin{align\*}   
1-\\frac{1}{2}\\theta^2+\\frac{1}{4!}\\theta^4 - \\cdots &= \\cos\\theta\\\\   
\\theta- \\frac{1}{3!}\\theta^3+\\frac{1}{5!}\\theta^5- \\cdots &= \\sin\\theta   
\\end{align\*}   
Thus,   
\\\[   
e^{i\\theta} = \\cos\\theta + i\\sin\\theta \\hspace{5em} \\text{Q.E.D.}   
\\\]

### 2\. First Order Differential Equation

Using a first-order ODE, we will prove that the left hand side and the right hand side of Euler's formula are identical:   
\\\[   
e^{i\\theta} = \\cos\\theta + i\\sin\\theta   
\\\]   
Let $y\_L = e^{i\\theta}$ and $y\_R = \\cos\\theta + i\\sin\\theta$. Then, the derivatives are   
\\begin{align\*}   
\\frac{dy\_L}{d\\theta} &= ie^{i\\theta} = iy\_L\\\\   
\\frac{dy\_R}{d\\theta} &= -\\sin\\theta +i\\cos\\theta = i(\\cos\\theta+i\\sin\\theta) = iy\_R.   
\\end{align\*}   
They satisfy the identical equation. Also, the initial conditions at $\\theta = 0$ are   
\\begin{align\*}   
y\_L(0) &= e^{i0} = 1\\\\ y\_R(0) &= \\cos(0) + i\\sin(0) = 1   
\\end{align\*}   
Since $y\_L$ and $y\_R$ satisfy the same first order differential equation and the same initial condition, we conclude $y\_L(\\theta) = y\_R(\\theta)$ for all values of $\\theta$. Q.E.D.

### 3\. Second Order Differential Equation

Let $y=e^{i\\theta}$. Then, we have  
\\begin{equation}\\label{first-order-eq}   
\\frac{dy}{d\\theta} = ie^{i\\theta} = iy.   
\\end{equation}   
Differentiating one more time, we obtain   
\\begin{equation}\\label{second-order-eq}   
\\frac{d^2y}{d\\theta^2} = i\\frac{dy}{d\\theta} = i^2 y = -y.   
\\end{equation}   
We identify this as the equation for a simple harmonic oscillator with $\\omega = 1$, whose general solution is   
\\begin{equation}\\label{y\_sol}   
y(\\theta) = a\\cos\\theta + b\\sin\\theta   
\\end{equation}   
and so   
\\begin{equation}\\label{Dy\_sol}   
y'(\\theta) = -a\\sin\\theta + b\\cos\\theta.   
\\end{equation}   
Substituting \\eqref{y\_sol} and \\eqref{Dy\_sol} in Eq.\\eqref{first-order-eq}, we have, for all values of $\\theta$,   
\\begin{align\*}  
 \-a\\sin\\theta + b\\cos\\theta = i(a\\cos\\theta + b\\sin\\theta),   
\\end{align\*}   
hence $-a = ib$ and $b=ia$, which are identical. Thus,   
\\\[   
y(\\theta) = a\\cos\\theta + b\\sin\\theta = a\\cos\\theta + ia\\sin\\theta   
\\\]   
But, since we know $y(0) = e^{i0} = 1$, $a$ must be 1\. Therefore,   
\\\[   
e^{i\\theta} = \\cos\\theta + i\\sin\\theta \\hspace{3em} \\text{Q.E.D.}   
\\\]

### 3'. Second Order Differential Equations of real functions

In case you felt a bit uneasy about going from Eq.\\eqref{second-order-eq} to the general solution \\eqref{y\_sol} since $y$ is complex, we can slightly modify what we did in the previous section so that we will be dealing with equations for real functions. Let $z(\\theta)=e^{i\\theta}$, and $x(\\theta) = \\operatorname{Re}\\{z(\\theta)\\}$ and $y(\\theta) = \\operatorname{Im}\\{z(\\theta)\\}$. Then, we have   
\\begin{equation\*}  
z' = ie^{i\\theta} = iz \~\~\\Leftrightarrow x'+iy' = i(x+iy).   
\\end{equation\*}   
This is a system of two coupled first order differential equations   
\\begin{align}   
x' &= -y\\label{coupled\_xprime}\\\\   
y' &= x\\label{coupled\_yprime}   
\\end{align}   
Differentiating one more time, we obtain decoupled second order equations;   
\\begin{align\*}   
x'' &= -y' = -x\\\\   
y'' &= x' = -y   
\\end{align\*}   
We identify each of these as the equation for a simple harmonic oscillator with $\\omega = 1$. Thus,   
\\begin{align\*}   
x(\\theta) &= A\\cos\\theta + B\\sin\\theta\\\\   
y(\\theta) &= C\\cos\\theta + D\\sin\\theta   
\\end{align\*}   
and so   
\\begin{align}   
x'(\\theta) &= -A\\sin\\theta + B\\cos\\theta\\label{xprime}\\\\   
y'(\\theta) &= -C\\sin\\theta + D\\cos\\theta\\label{yprime}   
\\end{align}   
Substituting \\eqref{xprime} and \\eqref{yprime} in Eqs.\\eqref{coupled\_xprime} and \\eqref{coupled\_yprime} respectively, we have, for all values of $\\theta$,   
\\begin{align\*}   
\-A\\sin\\theta + B\\cos\\theta &= -C\\cos\\theta - D\\sin\\theta\\\\   
\-C\\sin\\theta + D\\cos\\theta &= A\\cos\\theta + B\\sin\\theta   
\\end{align\*}   
hence $C = -B$ and $D = A$. Thus,   
\\begin{align\*}   
x(\\theta) &= A\\cos\\theta + B\\sin\\theta\\\\   
y(\\theta) &= -B\\cos\\theta + A\\sin\\theta   
\\end{align\*}   
But, since we know $z(0) = e^{i0} = 1$, or $x(0)=1$ and $y(0)=0$, which forces $A=1$ and $B=0$. Therefore,   
\\\[   
x(\\theta) = \\cos\\theta, \~\~\~ y(\\theta) = \\sin\\theta \\\] or \\\[ e^{i\\theta} = \\cos\\theta +i\\sin\\theta \\hspace{3em} \\text{Q.E.D.}   
\\\]