Euler's Identity

Since $z = x + jy$ is the algebraic expression of $z$ in terms of its rectangular coordinates, the corresponding expression in terms of its polar coordinates is

$$z = r\cos(\theta) + j\,r\sin(\theta).$$

There is another, more powerful representation of $z$ in terms of its polar coordinates. In order to define it, we must introduce Euler's identity:

$$\zbox {e^{j\theta}=\cos(\theta)+j\sin(\theta)}$$ (2.5)

A proof of Euler's identity is given in the next chapter. Before, the only algebraic representation of a complex number we had was $z = x + jy$, which fundamentally uses Cartesian (rectilinear) coordinates in the complex plane. Euler's identity gives us an alternative representation in terms of polar coordinates in the complex plane:

$$\zbox {z = re^{j\theta}}$$

We'll call $re^{j\theta}$ the polar form of the complex number $z$, in contrast with the rectangular form $z = x + jy$. Polar form often simplifies algebraic manipulations of complex numbers, especially when they are multiplied together. Simple rules of exponents can often be used in place of messier trigonometric identities. In the case of two complex numbers being multiplied, we have

$$z_1 z_2 = \left(r_1 e^{j \theta_1}\right) \left(r_2 e^{j \theta_2... ...eta_1} e^{j \theta_2}\right) = r_1 r_2 e^{j \left(\theta_1 + \theta_2\right)}.$$

A corollary of Euler's identity is obtained by setting $\theta=\pi$ to get

$$e^{j\pi} + 1 = 0.$$

This has been called the ``most beautiful formula in mathematics'' due to the extremely simple form in which the fundamental constants $e, j, \pi, 1$, and 0, together with the elementary operations of addition, multiplication, exponentiation, and equality, all appear exactly once.

For another example of manipulating the polar form of a complex number, let's again verify $z\overline{z} = \left\vert z\right\vert^2$, as we did above in Eq. (2.4), but this time using polar form:

$$z \overline{z} = r e^{j \theta} r e^{-j \theta} = r^2 e^0 = r^2 = \vert z\vert^2$$

As mentioned in §2.7, any complex expression can be conjugated by replacing $j$ by $-j$ wherever it occurs. This implies $\overline{r e^{j \theta}} = r e^{-j \theta}$, as used above. The same result can be obtained by using Euler's identity to expand $re^{j\theta}$ into $r \cos(\theta) + j r \sin(\theta)$ and negating the imaginary part to obtain $\overline{r e^{j\theta}} = r \cos(\theta) - j r \sin(\theta) = r \cos(-\theta) + j r \sin(-\theta) = r e^{-j \theta}$, where we used also the fact that cosine is an even function ( $\cos(-\theta) = \cos(\theta)$) while sine is odd ( $\sin(-\theta) = -\sin(\theta)$).

We can now easily add a fourth line to that set of examples:

$$z/\overline{z} = \frac{r e^{j \theta}}{r e^{-j \theta}} = e^{j2\theta} = e^{j2\angle{z}}$$

Thus, $\left\vert z/\overline{z}\right\vert=1$ for every $z\neq 0$.

Euler's identity can be used to derive formulas for sine and cosine in terms of $e^{j\theta }$:

$$\begin{eqnarray*} e^{j \theta} + \overline{e^{j \theta}}&=&e^{j \theta} + e^{-j ... ...+ \left[\cos(\theta) - j \sin(\theta)\right]\\ &=&2\cos(\theta) \end{eqnarray*}$$

Similarly, $e^{j \theta} - \overline{e^{j \theta}} = 2j\, \sin(\theta)$, and we obtain the following classic identities:

$\parbox{0.8\textwidth}{% \begin{displaymath} \cos(\theta) = \display... ...heta) = \displaystyle\frac{e^{j \theta} - e^{-j \theta}}{2j} \end{displaymath}}$