1.6: Euler's Formula (2023)

P1

\(e^{i t}\) differentiates as expected:

\[\dfrac{de^{it}}{dt} = ie^{it}.\nonumber \]

Proof

This follows directly from the definition in Equation \ref{1.6.1}:

\[ \begin{align*} \dfrac{de^{it}}{dt} &= \dfrac{d}{dt} (\cos (t) + i \sin (t)) \\[4pt] &= -\sin (t) + i \cos (t) \\[4pt] &= i (\cos (t) + i \sin (t)) \\[4pt] &= ie^{it}. \end{align*} \]

P2

\[e^{i \cdot 0} = 1 . \nonumber \]

Proof

This follows directly from the definition in Equation \ref{1.6.1}:

\(e^{i \cdot 0} = \cos (0) + i \sin (0) = 1\).

P3

The usual rules of exponents hold:

\[e^{ia} e^{ib} = e^{i(a + b)}.\nonumber \]

Proof

This relies on the cosine and sine addition formulas and the definition in Equation \ref{1.6.1}:

\[\begin{align*} e^{ia} \cdot e^{ib} & = (\cos (a) + i \sin (a)) \cdot (\cos (b) + i \sin (b)) \\[4pt] & = \cos (a) \cos (b) - \sin (a) \sin (b) + i (\cos (a) \sin (b) + \sin (a) \cos (b)) \\[4pt] & = \cos (a + b) + i \sin (a + b) = e^{i (a + b)}. \end{align*} \]

P4

The definition of \(e^{i \theta}\) is consistent with the power series for \(e^x\).

Proof

To see this we have to recall the power series for \(e^x\), \(\cos (x)\) and \(\sin (x)\). They are

\[\begin{align*} e^x & = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!} + ... \\[4pt] \cos (x) & = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dfrac{x^6}{6!} + \ldots \\[4pt] \sin (x) & = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} + ... \end{align*} \]

Now we can write the power series for \(e^{i \theta}\) and then split it into the power series for sine and cosine:

\[\begin{align*} e^{i \theta} & = \sum_{0}^{\infty} \dfrac{(i\theta)^n}{n!} \\[4pt] & = \sum_{0}^{\infty} (-1)^k \dfrac{\theta ^{2k}}{(2k)!} + i \sum_{0}^{\infty} (-1)^k \dfrac{\theta ^{2k + 1}}{(2k + 1)!} \\[4pt] & = \cos (\theta) + i \sin (\theta). \end{align*} \]

So the Euler formula definition is consistent with the usual power series for \(e^x\).

Magnitude

\(|e^{i \theta}| = 1\).

Proof

\[ \begin{align*} |e^{i \theta}| &= |\cos (\theta) + i \sin (\theta)| \\[4pt] &= \sqrt{\cos ^2 (\theta) + \sin ^2 (\theta)} \\[4pt] &= 1 . \end{align*} \]

In words, this says that \(e^{i \theta}\) is always on the unit circle - this is useful to remember!

Likewise, if \(z = r e^{i \theta}\) then \(|z| = r\). You can calculate this, but it should be clear from the definitions: \(|z|\) is the distance from \(z\) to the origin, which is exactly the same definition as for \(r\).

Argument

If \(z = r e^{i \theta}\) then \(\text{arg} (z) = \theta\).

Proof

This is again the definition: the argument is the polar angle \(\theta\).

Conjugate

\(\overline{(z = r e^{i \theta})} = r e^{-i \theta}\).

Proof

\[ \begin{align*} \overline{(z = r e^{i \theta})} &= \overline{r (\cos (\theta) + i \sin (\theta))} \\[4pt] &= r (\cos (\theta) - i \sin (\theta)) \\[4pt] &= r(\cos (-\theta) + i \sin (-\theta)) \\[4pt] &= r e^{-i \theta}. \end{align*} \]

In words: complex conjugation changes the sign of the argument.

Multiplication

If \(z_1 = r_1 e^{i \theta_1}\) and \(z_2 = r_2 e^{i \theta_2}\) then

\[z_1 z_2 = r_1 r_2 e^{i (\theta_1 + \theta_2)}. \nonumber \]

This is what mathematicians call trivial to see, just write the multiplication down. In words, the formula says the for \(z_1 z_2\) the magnitudes multiply and the arguments add.

Division

Again it's trivial that

\(\dfrac{r_1 e^{i \theta_1}}{r_2 e^{i \theta_2}} = \dfrac{r_1}{r_2} e^{i (\theta_1 - \theta_2)}.\)

Example \(\PageIndex{1}\): Multiplication by 2\(i\)

Here’s a simple but important example. By looking at the graph we see that the number \(2i\) has magnitude 2 and argument \(\pi/2\). So in polar coordinates it equals \(2e^{i \pi /2}\). This means that multiplication by \(2i\) multiplies lengths by 2 and add \(\pi/2\) to arguments, i.e. rotates by \(90^{\circ}\). The effect is shown in the figures below

Example \(\PageIndex{2}\): Rasing to a power

Let's compute \((1 + i)^6\) and \((\dfrac{1 + i \sqrt{3}}{2})^3\)

Example \(\PageIndex{3}\)

Use complex replacement to compute

\[I = \int e^x \cos (2x)\ dx. \nonumber \]

Example \(\PageIndex{4}\)

Find all 5 fifth roots of 2.

Example \(\PageIndex{5}\)

Find the 4 forth roots of 1.

Example \(\PageIndex{6}\)

Find the 3 cube roots of -1.

Example \(\PageIndex{7}\)

Find the 5 fifth roots of \(1 + i\).

Example \(\PageIndex{8}\)

We should check that our technique works as expected for a simple problem. Find the 2 square roots of 4.