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# 1.6: Euler's Formula

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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.

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

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

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

Solution

We have Euler's formula

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

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

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

with

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

Computing $$I_c$$ is straightforward:

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

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.

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

So,

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

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

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

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,

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

## $$N$$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

$z^N = c,$

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

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

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, ...$ Solving for \(\theta$$, we have

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

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

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

and in Cartesian representation

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

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

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

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!

## The 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. • Was this article helpful?