1.6: Euler's Formula

Euler’s (pronounced ‘oilers’) formula connects complex exponentials, polar coordinates, and sines and cosines. It turns messy trig identities into tidy rules for exponentials. We will use it a lot. The formula is the following:

$$e^{i\theta} = \cos (\theta) + i \sin (\theta). \label{1.6.1}$$

There are many ways to approach Euler’s formula. Our approach is to simply take Equation $\ref{1.6.1}$ as the definition of complex exponentials. This is legal, but does not show that it’s a good definition. To do that we need to show the $e^{i \theta}$ obeys all the rules we expect of an exponential. To do that we go systematically through the properties of exponentials and check that they hold for complex exponentials.

$e^{i \theta}$ behaves like a true exponential

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 $e^x$.

Properties P1-P4 should convince you that $e^{i \theta}$ behaves like an exponential.

Complex Exponentials and Polar Form

Now let’s turn to the relation between polar coordinates and complex exponentials.

Suppose $z = x + iy$ has polar coordinates $r$ and $\theta$. That is, we have $x = r \cos (\theta)$ and $y = r \sin (\theta)$. Thus, we get the important relationship

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

This is so important you shouldn’t proceed without understanding. We also record it without the intermediate equation.

$$z = x + iy = r e^{i \theta}. \nonumber$$

Because $r$ and $\theta$ are the polar coordinates of $(x, y)$ we call $z = r e^{i \theta}$ the polar form of $z$.

Let’s now verify that magnitude, argument, conjugate, multiplication and division are easy in polar form.

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 $(\dfrac{1 + i \sqrt{3}}{2})^3$

Solution

$1 + i$ has magnitude = $\sqrt{2}$ and $\text{arg} = \pi /4$, so $1 + i = \sqrt{2} e^{i \pi /4}$. Rasing to a power is now easy:

$(1 + i)^6 = (\sqrt{2} e^{i \pi /4})^6 = 8 e^{6i \pi /4} = 8 e^{3i \pi /2} = -8i$.

Similarly, $\dfrac{1 + i\sqrt{3}}{2} = e^{i \pi / 3}$, so $(\dfrac{1 + i\sqrt{3}}{2})^3 = (1 \cdot e^{i \pi / 3})^3 = e^{i \pi} = -1$

Complexification or Complex Replacement

In the next example we will illustrate the technique of complexification or complex replacement. This can be used to simplify a trigonometric integral. It will come in handy when we need to compute certain integrals.

Example $\PageIndex{3}$

Use complex replacement to compute

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

Solution

We have Euler's formula

$$e^{2ix} = \cos (2x) + i \sin (2x), \nonumber$$

so $\cos (2x) = \text{Re} (e^{2ix})$. The complex replacement trick is to replace $\cos (2x)$ by $e^{2ix}$. We get (justification below)

$$I_c = \int e^x \cos 2x + ie^x \sin 2x \ dx \nonumber$$

with

$$I = \text{Re} (I_c) \nonumber$$

Computing $I_c$ is straightforward:

$$I_c = \int e^x e^{i2x}\ dx = \int e^{x(1 + 2i)}\ dx = \dfrac{e^{x(1 + 2i)}}{1 + 2i}. \nonumber$$

Here we will do the computation first in rectangular coordinates. In applications, for example throughout 18.03, polar form is often preferred because it is easier and gives the answer in a more useable form.

$$\begin{array} {rcl} {I_c} & = & {\dfrac{e^{x(1 + 2i)}}{1 + 2i} \cdot \dfrac{1 - 2i}{1 - 2i}} \\ {} & = & {\dfrac{e^x (\cos (2x) + i \sin (2x)) (1 - 2i)}{5}} \\ {} & = & {\dfrac{1}{5} e^x (\cos (2x) + 2 \sin (2x) + i (-2 \cos (2x) + \sin (2x)))} \end{array} \nonumber$$

So,

$$I = \text{Re} (I_c) = \dfrac{1}{5} e^x (\cos (2x) + 2\sin (2x)). \nonumber$$

Justification of complex replacement. The trick comes by cleverly adding a new integral to $I$ as follows, Let $J = \int e^x \sin (2x)\ dx$. Then we let

$$I_c = I + iJ = \int e^x (\cos (2x) + i \sin (2x)) \ dx = \int e^x 2^{2ix}\ dx. \nonumber$$

Clearly, by construction, $\text{Re} (I_c) = I$ as claimed above.

Alternative using polar coordinates to simplify the expression for $I_c$:

In polar form, we have $1 + 2i = re^{i \phi}$, where $r = \sqrt{5}$ and $\phi = \text{arg} (1 + 2i) = \text{tan}^{-1} (2)$ in the first quadrant. Then:

$I_c = \dfrac{e^{x(1 + 2i)}}{\sqrt{5} e^{i \phi}} = \dfrac{e^x}{\sqrt{5}} e^{i(2x - \phi)} = \dfrac{e^x}{\sqrt{5}} (\cos (2x - \phi) + i \sin (2x - \phi))$.

Thus,

$$I = \text{Re} (I_c) = \dfrac{e^x}{\sqrt{5}} \cos (2x - \phi). \nonumber$$

$N$th roots

We are going to need to be able to find the $n$th roots of complex numbers, i.e., solve equations of the form

$$z^N = c, \nonumber$$

where $c$ is a given complex number. This can be done most conveniently by expressing $c$ and $z$ in polar form, $c = Re^{i \phi}$ and $z = re^{i \theta}$. Then, upon substituting, we have to solve

$$r^N e^{iN \theta} = Re^{i \phi} \nonumber$$

For the complex numbers on the left and right to be equal, their magnitudes must be the same and their arguments can only differ by an integer multiple of $2\pi$. This gives

$$r = R^{1/N} \; (N \theta = \phi + 2\pi n), \; where\;n = 0, \pm 1, \pm 2, ... \nonumber$$

Solving for $\theta$, we have

$$\theta = \dfrac{\phi}{N} + \dfrac{2\pi n}{N}. \nonumber$$

Example $\PageIndex{4}$

Find all 5 fifth roots of 2.

Solution

For $c = 2$, we have $R = 2$ and $\phi = 0$, so the fifth roots of 2 are

$z_n = 2^{1/5} e^{2n \pi i/5}$, where $n = 0, \pm 1, \pm 2, ...$

Looking at the right hand side we see that for $n = 5$ we have $2^{1/5} e^{2\pi i}$ which is exactly the same as the root when $n = 0$, i.e. $2^{1/5} e^{0i}$. Likewise $n = 6$ gives exactly the same root as $n = 1$, and so on. This means, we have 5 different roots corresponding to $n = 0, 1, 2, 3, 4$.

$z_n = 2^{1/5}, e^{1/5} e^{2\pi i/5}, e^{1/5} e^{4\pi i/5}, e^{1/5} e^{6\pi i/5}, e^{1/5} e^{8\pi i/5}$

Similarly we can say that in general $c = Re^{i \phi}$ has $N$ distinct $N$ th roots:

$z_n = r^{1/N} e^{i \phi / N + i 2\pi (n/N)}$ for $n = 0, 1, 2, ..., N - 1$.

Example $\PageIndex{5}$

Find the 4 forth roots of 1.

Solution

We need to solve $z^4 = 1$, so $\phi = 0$. So the 4 distinct fourth roots are in polar form

$$z_n = 1, e^{i \pi /2}, e^{i \pi}, e^{i 3 \pi /2} \nonumber$$

and in Cartesian representation

$$z_n = 1, i, -1, -i. \nonumber$$

Example $\PageIndex{6}$

Find the 3 cube roots of -1.

Solution

$z^2 = -1 = e^{i \pi + i 2 \pi n}$. So, $z_n = e^{i \pi + i 2 \pi (n/3)}$ and the 3 cube roots are $e^{i \pi /3}$, $e^{i \pi}$, $e^{i 5 \pi /3}$. Since $\pi /3$ radians is $60^{\circ}$ we can simplify:

$e^{i \pi /3} = \cos (\pi / 3) + i \sin (\pi /3) = \dfrac{1}{2} + i \dfrac{\sqrt{3}}{2} \Rightarrow z_n = -1, \dfrac{1}{2} \pm i \dfrac{\sqrt{3}}{2}$

Example $\PageIndex{7}$

Find the 5 fifth roots of $1 + i$.

Solution

$$z^5 = 1 + i = \sqrt{2} e^{i (\pi / 4 + 2n\pi)} \nonumber$$

for $n = 0, 1, 2, ...$. So, the 5 fifth roots are

$2^{1/10} e^{i\pi /20}$, $2^{1/10} e^{i9\pi /20}$, $2^{1/10} e^{i17\pi /20}$, $2^{1/10} e^{i25\pi /20}$, $2^{1/10} e^{i33\pi /20}$.

Using a calculator we could write these numerically as $a + bi$, but there is no easy simplification.

Example $\PageIndex{8}$

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

Solution

$z^2 = 4 e^{i2 \pi n}$. So, $z_n = 2e^{i \pi n}$, with $n = 0, 1$. So the two roots are $2e^0 = 2$ and $2e^{i\pi} = -2$ as expected!

geometry of $N$th roots

Looking at the examples above we see that roots are always spaced evenly around a circle centered at the origin. For example, the fifth roots of $1 + i$ are spaced at increments of $2\pi / 5$ radians around the circle of radius $2^{1/5}$.

Note also that the roots of real numbers always come in conjugate pairs.