Euler's formula

Complex numbers. By setting the imaginary unit ``$  i $'' which satisfies `` $ {i^2 = -1}$,'' we can introduce complex numbers of the form $ z = a + bi$, where $a$ and $b$ are real numbers.

Sum, product, conjugate and absolute value. Let $ z = a + bi$ and $ w = c + di$. Then the sum ``$ z + w$'' and the product ``$ zw$'' of the two complex numbers are defined respectively by

$\displaystyle z + w = (a + c) + (b + d)i$    and $\displaystyle \quad
zw = (ac - bd) + (ad + bc)i .
$

The conjugate $ \bar{z}$ and the absolute value $ \vert z\vert$ of $ z$ are defined respectively by

$\displaystyle \bar{z} = a - bi$    and $\displaystyle \quad
\vert z\vert = \sqrt{a^2 + b^2}.
$

Complex plane. Each complex number  $ z = a + bi$ uniquely determines the point $ P$ at $(a,b)$ in the coordinate plane, which is referred as complex plane. In the complex plain, the $x$-axis and the $y$-axis are repectively called the real axis and the imaginary axis.

\includegraphics{lec16a.ps}

Trigonometric forms for complex numbers. Let $ r = \sqrt{a^2 + b^2}$ be the length of $ OP$, and let $ \theta$ be the angle from the real axis to $ OP$. (Recall Lecture summary No.13; $ \theta = \tan^{-1}(b/a)$ if $ a \ge 0$; $ \theta = \pi + \tan^{-1}(b/a)$ if $a < 0$.) Then we have $ (a, b) = (r\cos\theta, r\sin\theta)$, and therefore,

$\displaystyle a + bi = r\cos\theta + (r\sin\theta)i = r(\cos\theta + i\sin\theta).$ (3-1)

\includegraphics{lec16b.ps}

Euler's formula. In order to relate the trigonometric form  $ \cos\theta + i\sin\theta$ to the exponential form  $ e^{i\theta}$, we can (at least superficially) introduce Euler's formula

$\displaystyle e^{i\theta} = \cos\theta + i\sin\theta$ (3-2)

where $ e$ denotes the base for natural logarithm. Combining (3-1) and (3-2) together, we obtain $ z = r e^{i\theta}$.

Calculations of complex numbers. Let $ v = \cos\alpha + i\sin\alpha$ and $ w = \cos\beta + i\sin\beta$. Then we can compute

$\displaystyle v w$ $\displaystyle = (\cos\alpha + i\sin\alpha) (\cos\beta + i\sin\beta)$    
  $\displaystyle = (\cos\alpha\cos\beta - \sin\alpha\sin\beta) + i(\cos\alpha\sin\beta + \sin\alpha\cos\beta)$    
  $\displaystyle = \cos(\alpha + \beta) + i\sin(\alpha + \beta) ,$    

which coincides with the ``formal'' calculation $ v w = e^{i\alpha} e^{i\beta} = e^{i(\alpha + \beta)}$ of exponents via (3-2). Since $ z^n = (r e^{i\theta})^n = r^n e^{i n \theta}$, we obtain the following result (De Moivre's Theorem):

$\displaystyle [r (\cos\theta + i\sin\theta)]^n = r^n (\cos n\theta + i\sin n\theta) .
$

Roots of $ x^n = z$. Recalling that the cosine and sine functions are periodic with period $ 2 \pi$, we obtain

$\displaystyle e^{i(\theta + 2\pi k)} = \cos(\theta + 2\pi k) + i\sin(\theta + 2\pi k) = \cos\theta + i\sin\theta = e^{i\theta}$   $\displaystyle \mbox{ for every integer $k$. }$ (3-3)

Observing $ e^{2\pi k i} = e^0 = 1$ when $ \theta = 0$ in (3-3), we can find all the solutions to $ x^n = 1$, known as ``$n$th roots of unity'' as follows:

$\displaystyle x = \left(e^{2\pi k i}\right)^{\frac{1}{n}}
= e^{\frac{2\pi k}{n} i}
= \cos\left(\frac{2\pi k}{n}\right)
+ i\sin\left(\frac{2\pi k}{n}\right)$    for $ k = 0, 1, \ldots, n-1$.

When $ z = r (\cos\theta + i\sin\theta) = r e^{i\theta}$, all the solutions to $ x^n = z$ (that is, $n$th roots of $ z$) are given by

$\displaystyle x = \sqrt[n]{r} \left[
\cos\left(\frac{\theta + 2\pi k}{n}\right)
+ i\sin\left(\frac{\theta + 2\pi k}{n}\right) \right]$    for $ k = 0, 1, \ldots, n-1$.



Department of Mathematics
Last modified: 2005-09-29