Euler’s formula

Euler's formula

The polar form of a complex number $z = r(\cos\theta + i\sin\theta)$ can be expressed in exponential form using the Maclaurin series for $\cos\theta$, $\sin\theta$, and $e^{i\theta}$.

$\cos\theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \dots \quad (6.7)$

$\sin\theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + \dots \quad (6.8)$

The Maclaurin series for $e^x$ is:

$e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \dots \quad (6.9)$

Let $x = i\theta$. Substituting $x = i\theta$ into equation (6.9):

$e^{i\theta} = 1 + \frac{i\theta}{1!} + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \frac{(i\theta)^4}{4!} + \frac{(i\theta)^5}{5!} + \dots$

$= 1 + i\theta - \frac{\theta^2}{2!} - i\frac{\theta^3}{3!} + \frac{\theta^4}{4!} + i\frac{\theta^5}{5!} + \dots$

$= \left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \dots\right) + i\left(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \dots\right) \quad (6.10)$

Substituting equations (6.7) and (6.8) into equation (6.10):

$e^{i\theta} = \cos\theta + i\sin\theta \quad (6.11)$

Multiplying both sides of equation (6.11) by $r$:

$re^{i\theta} = r(\cos\theta + i\sin\theta)$

Since $z = r(\cos\theta + i\sin\theta)$:

$z = re^{i\theta} \quad (6.12)$

Equation (6.12) is Euler's formula, which represents the complex number $z$ in exponential form. The complex conjugate of $z$ can be expressed as $z^* = re^{-i\theta}$ (6.13)

If $z_1 = r_1e^{i\theta_1}$ and $z_2 = r_2e^{i\theta_2}$, then:

1. $z_1z_2 = r_1r_2e^{i(\theta_1 + \theta_2)}$
2. $\frac{z_1}{z_2} = \frac{r_1}{r_2}e^{i(\theta_1 - \theta_2)}$
3. $\arg(z_1z_2) = \arg(z_1) + \arg(z_2)$
4. $\arg\left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2)$

Relationship between trigonometric and hyperbolic functions

Euler's formula enables us to determine the relationship between trigonometric and hyperbolic functions.

Considering the equations:

$e^{i\theta} = \cos\theta + i\sin\theta$

$e^{-i\theta} = \cos\theta - i\sin\theta$

1. Adding the two equations:

$e^{i\theta} + e^{-i\theta} = 2\cos\theta$

$\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}$

Thus, $\cos\theta = \cosh(i\theta)$.

2. Subtracting the second equation from the first:

$e^{i\theta} - e^{-i\theta} = 2i\sin\theta$

$\sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}$

Thus, $\sin\theta = -i\sinh(i\theta)$.

3. From the identities in parts 1 and 2:

$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{-i\sinh(i\theta)}{\cosh(i\theta)} = -i\tanh(i\theta)$

Thus, $\tan\theta = -i\tanh(i\theta)$.

Example 1

Express $\frac{1}{\sqrt{3}} + i$ in exponential form.

Solution:

Let $z = \frac{1}{\sqrt{3}} + i$.

$r = \sqrt{\left(\frac{1}{\sqrt{3}}\right)^2 + 1^2} = \sqrt{\frac{1}{3} + 1} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$

$\theta = \arctan\left(\frac{1}{\frac{1}{\sqrt{3}}}\right) = \arctan(\sqrt{3}) = \frac{\pi}{3}$

Thus, $z = re^{i\theta} = \frac{2\sqrt{3}}{3}e^{i\frac{\pi}{3}}$.

Example 2

Express the complex number $z = e^{\frac{13\pi}{6}i}$ in the form $z = a + ib$, where $a, b \in \mathbb{R}$.

Solution:

$z = e^{\frac{13\pi}{6}i} = \cos\frac{13\pi}{6} + i\sin\frac{13\pi}{6}$

Since $\frac{13\pi}{6} = 2\pi + \frac{\pi}{6}$,

$z = \cos\frac{\pi}{6} + i\sin\frac{\pi}{6}$

$z = \frac{\sqrt{3}}{2} + \frac{1}{2}i$

Example 3

Using Euler's formula, show that $\sin^3\theta = \frac{3\sin\theta - \sin 3\theta}{4}$.

Solution:

Using the identity $\sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}$:

$\sin^3\theta = \left(\frac{e^{i\theta} - e^{-i\theta}}{2i}\right)^3$

$= \frac{e^{3i\theta} - 3e^{2i\theta}e^{-i\theta} + 3e^{i\theta}e^{-2i\theta} - e^{-3i\theta}}{-8i}$

$= \frac{e^{3i\theta} - 3e^{i\theta} + 3e^{-i\theta} - e^{-3i\theta}}{-8i}$

$\sin^3\theta = \frac{(e^{3i\theta} - e^{-3i\theta}) - 3(e^{i\theta} - e^{-i\theta})}{-8i}$

$\sin^3\theta = \frac{2i\sin3\theta - 6i\sin\theta}{-8i}$

$\sin^3\theta = \frac{\sin 3\theta - 3\sin\theta}{-4}$

$\sin^3\theta = \frac{3\sin\theta - \sin 3\theta}{4}$

Example 4

Solve the equation $z^5 + 3\sqrt{3} + 3i = 0$, expressing the roots in exponential form. Use the argument interval $-\pi < \theta \le \pi$ and the interval $-2 \le k \le 2$ for the required roots.

Solution:

$z^5 = -3\sqrt{3} - 3i$

Let $w = -3\sqrt{3} - 3i$.

$r = \sqrt{(-3\sqrt{3})^2 + (-3)^2} = \sqrt{27 + 9} = \sqrt{36} = 6$

$\theta = \arctan\left(\frac{-3}{-3\sqrt{3}}\right) = \arctan\left(\frac{1}{\sqrt{3}}\right)$. Since both real and imaginary parts are negative, $\theta$ is in the third quadrant. Thus, $\theta = -\frac{5\pi}{6}$.

$w = 6e^{-i\frac{5\pi}{6}}$

$z = w^{\frac{1}{5}} = 6^{\frac{1}{5}}e^{i\left(\frac{-\frac{5\pi}{6} + 2\pi k}{5}\right)}$

$z_k = \sqrt[5]{6}e^{i\left(\frac{-5\pi + 12\pi k}{30}\right)}$

For $k = -2, -1, 0, 1, 2$:

$z_{-2} = \sqrt[5]{6}e^{-i\frac{29\pi}{30}}$

$z_{-1} = \sqrt[5]{6}e^{-i\frac{17\pi}{30}}$

$z_{0} = \sqrt[5]{6}e^{-i\frac{5\pi}{30}} = \sqrt[5]{6}e^{-i\frac{\pi}{6}}$

$z_{1} = \sqrt[5]{6}e^{i\frac{7\pi}{30}}$

$z_{2} = \sqrt[5]{6}e^{i\frac{19\pi}{30}}$

Example 5

Solve for $z$ given that $\cos z = 4$.

Solution:

Using Euler's formula, $\cos z = \frac{e^{iz} + e^{-iz}}{2}$.

$\frac{e^{iz} + e^{-iz}}{2} = 4$

$e^{iz} + e^{-iz} = 8$

Multiply both sides by $e^{iz}$:

$(e^{iz})^2 - 8e^{iz} + 1 = 0$

Let $u = e^{iz}$. Then $u^2 - 8u + 1 = 0$.

Using the quadratic formula:

$u = \frac{8 \pm \sqrt{64 - 4}}{2} = \frac{8 \pm \sqrt{60}}{2} = 4 \pm \sqrt{15}$

$e^{iz} = 4 \pm \sqrt{15}$

$iz = \ln(4 \pm \sqrt{15})$

$z = -i\ln(4 \pm \sqrt{15})$