Euler’s Formula

\begin{eqnarray} \mathop{exp}\nolimits (iθ) =\mathop{ cos}\nolimits (θ) + i\mathop{sin}\nolimits (θ)& & %&(1.44) \\ \end{eqnarray}

\begin{eqnarray} z = x + iy = r[\mathop{cos}\nolimits (θ) + i\mathop{sin}\nolimits (θ)] = r\mathop{exp}\nolimits (iθ).& & %&(1.45) \\ \end{eqnarray}

Let us have a look at the polar diagram: z is represented in the complex plane z–plane, i.e. the x–y–plane, where

\begin{eqnarray} x = Re(z),\quad y = Im(z).& & %&(1.46) \\ \end{eqnarray}

r = \sqrt{{x}^{2 } + {y}^{2}} is the modulus of z, i.e. geometrically the length of the vector (x,y). The argument arg(z) = θ is the angle between the vector (x,y) and the x–axis.

Motion along a line: we keep θ fixed and change r from small to larger values. The corresponding complex numbers move along a straight line that has a fixed angle θ with the x–axis.

Motion along a circle: we keep r fixed and change θ from 0 to larger values. We recognize that the corresponding complex numbers move along a circle with constant radius r.

1.4.1 The Unit Circle

\begin{eqnarray} z(θ) =\mathop{ cos}\nolimits (θ) + i\mathop{sin}\nolimits (θ) =\mathop{ exp}\nolimits (iθ)& & %&(1.47) \\ \end{eqnarray}

are all situated on a circle with radius r = 1 around the origin in the complex z–plane.

\begin{eqnarray} \mathop{sin}\nolimits (θ) = Im({e}^{iθ}),\quad \mathop{cos}\nolimits (θ) = Re({e}^{iθ}),& & %&(1.48) \\ \end{eqnarray}

where we wrote {e}^{z} for \mathop{exp}\nolimits (z) (both are the same, this is only a different notation). We obtain sin and cos from the exponential of imaginary argument. What happens if we calculate the exponential for a general complex argument z = x + iy that has both a real and an imaginary part? We calculate

\begin{eqnarray}{ e}^{z} = {e}^{x+iy} = {e}^{x}{e}^{i}y = {e}^{x}[\mathop{cos}\nolimits (y) + i\mathop{sin}\nolimits (y)] = {e}^{x}\mathop{ cos}\nolimits (y) + i{e}^{x}\mathop{ sin}\nolimits (y).& & %&(1.49) \\ \end{eqnarray}

Please note that the real part x only determines the modulus r = {e}^{x} of {e}^{z},

\begin{eqnarray} |{e}^{z}| = |{e}^{x+iy}| = {e}^{x}.& & %&(1.50) \\ \end{eqnarray}

\begin{eqnarray} |{e}^{z}| = |{e}^{x+iy}| = |{e}^{x}||{e}^{iy}| = {e}^{x},& & %&(1.51) \\ \end{eqnarray}

\begin{eqnarray} |{e}^{iy}| = 1.& & %&(1.52) \\ \end{eqnarray}

The circle is useful to describe, e.g., the motion of a particle on a circle. If we increase the angle θ linearily as a function of time t,

\begin{eqnarray} θ = ωt,& & %&(1.53) \\ \end{eqnarray}

\begin{eqnarray} z(t) = r\mathop{exp}\nolimits (iωt) = r\mathop{cos}\nolimits (ωt) + ir\mathop{sin}\nolimits (ωt),& & %&(1.54) \\ \end{eqnarray}

where we allowed an arbitrary radius r again. The real part of z describes the x–position, the imaginary part of z describes the y–position of the particle on the circle. The complex number z itself is not a measurable quantity, but it contains useful information (x and y–position of the particle).

Another important use of the complex exponent is in describing oscillations. This is due the fact that we can calculate the harmonic (sinusoidal) oscillations in terms of a complex exponent, when we argue that the real part is to be taken. The standard place to use such tecniques is in cicuit theory. A voltage is then represented by V = {V }_{0}\mathop{ cos}\nolimits (ωt) =\mathop{ Re}\nolimits ({V }_{0}{e}^{iωt}), and the real part in the expression will be conveniently forgotten for a while. The response of a capacitor to such an applied voltage, I = {dQ\over dt} = {d\over dt}CV then leads to the complex equation (with I = {I}_{0}{e}^{iωt}) {I}_{0} = iωC{V }_{0}, i.e. we have a simple relation between applied potential and current, just like the V = IR relation for a resistor. We can add such complex resistances, and take the rela part of the currents or volatges at the end; the phase of the complex numbers gives the phase difference betwen voltage and current.

1.4.2 Roots, n–th roots of unity

A number w is called an n–th root of a complex number z if {w}^{n} = z. We write carelessly w = {z}^{1∕n}.

The solutions of the equation {z}^{n} = 1 where n is a positive integer are called n–th roots of unity. We have

\begin{eqnarray}{ z}^{n}& =& 1 ⇝|{z}^{n}| = |z{|}^{n} = 1 ⇝|z| = 1 ⇝ z = {e}^{iθ},%&(1.55) \\ \end{eqnarray}

\begin{eqnarray}{ e}^{inθ} = 1.& & %&(1.56) \\ \end{eqnarray}

\begin{eqnarray}{ e}^{2πki} = cos(2πk) + i\mathop{sin}\nolimits (2πk) = 1,\quad k = 0,1,2,...& & %&(1.57) \\ \end{eqnarray}

we have n different solutions, i.e., n solutions where the phase is between 0 and 2π,

\begin{eqnarray} z = {e}^{2kπi∕n},k = 0,1,2,..,n − 1.& & %&(1.58) \\ \end{eqnarray}

1.4 Euler’s Formula

1.4.1 The Unit Circle

1.4.2 Roots, n–th roots of unity