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By Euler's formula, $z=e^{i\theta }$. Factoring out the half-angle givesSince $0\le \theta <\pi$, the cosine is positive. ThereforeOn the Argand diagram, the points $0,1,z,1+z$ form a rhombus. Its diagonal from $0$ to $1+z$ bisects the angle between the unit vectors $1$ and $z$, so its direction is $\theta /2$; resolving either side along the diagonal gives its length $2\cos (\theta /2)$.

Using Euler's formula, the complex exponential function satisfiesHence

A subgroup $N\le G$ is normal whenIts left and right cosets then agree, and multiplicationis well defined on the cosets. The identity is $N$, the inverse of $gN$ is $g^{-1}N$, and associativity descends from $G$, so these cosets form the quotient group $G/N$.

For a group homomorphism $\theta :G\to H$, its kernel and image areThe kernel is normal becausewhenever $k\in \ker \theta$. Conversely, if $K⊴G$, the quotient mapis a homomorphism with kernel $K$. The image of any homomorphism is closed under products and inverses, so it is a subgroup of $H$. Finally, the mapis well defined and bijective and preserves multiplication. This is the first isomorphism theorem.

DefineEuler's formula shows that $\Phi$ is a homomorphism, its image is the complex unit circle, and its kernel is $\mathbb{Z}$. The first isomorphism theorem therefore givesThe image of $\mathbb{Q}/\mathbb{Z}$ consists exactly of the roots of unity: if $t=p/q$, then $\Phi (t{)}^q=1$, while every element of finite order on the unit circle has an argument that is a rational multiple of $2\pi$.

Euler's formula for a connected plane graph is $|V|-|E|+|F|=2$. If the graph is triangle-free, every face has boundary length at least four, soEuler's formula then gives $|E|\le 2|V|-4$, so the average degree is less than four. There is a vertex of degree at most three. Delete it, color the remaining graph inductively with four colors, and restore it using a color absent from its at most three neighbors. Thus

Using Euler's formula and the Inverse Fourier transforms of phase factors,andThese are distributional identities, with $\delta$ the Dirac delta function.

Using Euler's formula,ThusOrdering the modes as $(-1,0,1)$, let $a=1/2+i/4$. ThenConsequentlyThis matrix is triangular, so its eigenvalues are its diagonal entries: