Revision of the exponential form of a complex number

2 Revision of the exponential form of a complex number

Recall that a complex number in Cartesian form which is written as

$z = a + i b,$

where $a$ and $b$ are real numbers and $i^{2} = - 1,$ can be written in polar form as

$z = r (cos θ + i sin θ)$

where $r = |z| = \sqrt{a^{2} + b^{2}}$ and $θ$ , the argument or phase of $z$ , is such that

$a = r cos θ b = r sin θ .$

A more concise version of the polar form of $z$ can be obtained by defining a complex exponential quantity $e^{i θ}$ by Euler’s relation

$e^{i θ} \equiv cos θ + i sin θ$

The polar angle $θ$ is normally expressed in radians . Replacing $i$ by $- i$ we obtain the alternative form

$e^{- i θ} \equiv cos θ - i sin θ$

Task!

Write down in $cos θ \pm i sin θ$ form and also in Cartesian form

$e^{i π ∕ 6}$
$e^{- i π ∕ 6}$ .

Use Euler’s relation:

We have, by definition,

$e^{i π ∕ 6} = cos (\frac{π}{6}) + i sin (\frac{π}{6}) = \frac{\sqrt{3}}{2} + \frac{1}{2} i$
$e^{- i π ∕ 6} = cos (\frac{π}{6}) - i sin (\frac{π}{6}) = \frac{\sqrt{3}}{2} - \frac{1}{2} i$

Task!

Write down

$cos (\frac{π}{6})$
$sin (\frac{π}{6})$ in terms of $e^{i π ∕ 6}$ and $e^{- i π ∕ 6}$ .

We have, adding the two results from the previous task

$e^{i π ∕ 6} + e^{- i π ∕ 6} = 2 cos (\frac{π}{6}) or cos (\frac{π}{6}) = \frac{1}{2} (e^{i π ∕ 6} + e^{- i π ∕ 6})$

Similarly, subtracting the two results,

$e^{i π ∕ 6} - e^{- i π ∕ 6} = 2 i sin (\frac{π}{6}) or sin (\frac{π}{6}) = \frac{1}{2 i} (e^{i π ∕ 6} - e^{- i π ∕ 6})$

(Don’t forget the factor $i$ in this latter case.)

Clearly, similar calculations could be carried out for any angle $θ$ . The general results are summarised in the following Key Point.

Key Point 8

Euler’s Relations

\begin{array}{rcl} e^{i θ} & \equiv & cos θ + i sin θ, e^{- i θ} \equiv cos θ - i sin θ \\ cos θ & \equiv & \frac{1}{2} (e^{i θ} + e^{- i θ}) sin θ \equiv \frac{1}{2 i} (e^{i θ} - e^{- i θ}) \end{array}

Using these results we can redraft an expression of the form

$a_{n} cos n θ + b_{n} sin n θ$

in terms of complex exponentials.

(This expression, with $θ = ω_{0} t$ , is of course the $n^{th}$ harmonic of a trigonometric Fourier series.)

Task!

Using the results from the Key Point 8 (with $n θ$ instead of $θ$ ) rewrite

$a_{n} cos n θ + b_{n} sin n θ$

in complex exponential form.

First substitute for $cos n θ$ and $sin n θ$ with exponential expressions using Key Point 8:

We have

$a_{n} cos n θ = \frac{a_{n}}{2} (e^{i n θ} + e^{- i n θ}) b_{n} sin n θ = \frac{b_{n}}{2 i} (e^{i n θ} - e^{- i n θ})$

$a_{n} cos n θ + b_{n} sin n θ = \frac{a_{n}}{2} (e^{i n θ} + e^{- i n θ}) + \frac{b_{n}}{2 i} (e^{i n θ} - e^{- i n θ})$

Now collect the terms in $e^{i n θ}$ and in $e^{- i n θ}$ and use the fact that $\frac{1}{i} = - i$ :

We get

$\frac{1}{2} (a_{n} + \frac{b_{n}}{i}) e^{i n θ} + \frac{1}{2} (a_{n} - \frac{b_{n}}{i}) e^{- i n θ}$

or, since $\frac{1}{i} = \frac{i}{i^{2}} = - i$ $\frac{1}{2} (a_{n} - i b_{n}) e^{i n θ} + \frac{1}{2} (a_{n} + i b_{n}) e^{- i n θ} .$

Now write this expression in more concise form by defining

$c_{n} = \frac{1}{2} (a_{n} - i b_{n})$ which has complex conjugate $c_{n}^{*} = \frac{1}{2} (a_{n} + i b_{n}) .$

Write the concise complex exponential expression for $a_{n} cos n θ + b_{n} sin n θ$ :

$a_{n} cos n θ + b_{n} sin n θ = c_{n} e^{i n θ} + c_{n}^{*} e^{- i n θ}$

Clearly, we can now rewrite the trigonometric Fourier series

$\frac{a_{0}}{2} + \sum_{n = 1}^{\infty} (a_{n} cos n ω_{0} t + b_{n} sin n ω_{0} t)$ as $\frac{a_{0}}{2} + \sum_{n = 1}^{\infty} (c_{n} e^{i n ω_{0} t} + c_{n}^{*} e^{- i n ω_{0} t})$ (3)

A neater, and particularly concise, form of this expression can be obtained as follows:

Firstly write $\frac{a_{0}}{2} = c_{0}$ (which is consistent with the general definition of $c_{n}$ since $b_{0} = 0$ ).

The second term in the summation

$\sum_{n = 1}^{\infty} c_{n}^{*} e^{- i n ω_{0} t} = c_{1}^{*} e^{- i ω_{0} t} + c_{2}^{*} e^{- 2 i ω_{0} t} + \dots$

can be written, if we define $c_{- n} = c_{n}^{*} = \frac{1}{2} (a_{n} + i b_{n}),$ as

$c_{- 1} e^{- i ω_{0} t} + c_{- 2} e^{- 2 i ω_{0} t} + c_{- 3} e^{- 3 i ω_{0} t} + \dots = \sum_{n = - 1}^{- \infty} c_{n} e^{i n ω_{0} t}$

Hence (3) can be written $c_{0} + \sum_{n = 1}^{\infty} c_{n} e^{i n ω_{0} t} + \sum_{n = - 1}^{- \infty} c_{n} e^{i n ω_{0} t}$ or in the very concise form $\sum_{n = - \infty}^{\infty} c_{n} e^{i n ω_{0} t} .$

The complex Fourier coefficients $c_{n}$ can be readily obtained as follows using (1) and (2) for $a_{n}, b_{n}$ .

Firstly

$c_{0} = \frac{a_{0}}{2} = \frac{1}{T} \int_{- \frac{T}{2}}^{\frac{T}{2}} f (t) d t$ (4)

For $n = 1, 2, 3, \dots$ we have

$c_{n} = \frac{1}{2} (a_{n} - i b_{n}) = \frac{1}{T} \int_{- \frac{T}{2}}^{\frac{T}{2}} f (t) (cos n ω_{0} t - i sin n ω_{0} t) d t i.e. c_{n} = \frac{1}{T} \int_{- \frac{T}{2}}^{\frac{T}{2}} f (t) e^{- i n ω_{0} t} d t$ (5)

Also for $n = 1, 2, 3, \dots$ we have

$c_{- n} = c_{n}^{*} = \frac{1}{2} (a_{n} + i b_{n}) = \frac{1}{T} \int_{- \frac{T}{2}}^{\frac{T}{2}} f (t) e^{i n ω_{0} t} d t$

This last expression is equivalent to stating that for $n = - 1, - 2, - 3, \dots$

$c_{n} = \frac{1}{T} \int_{- \frac{T}{2}}^{\frac{T}{2}} f (t) e^{- i n ω_{0} t} d t$ (6)

The three equations (4), (5), (6) can thus all be contained in the one expression

$c_{n} = \frac{1}{T} \int_{- \frac{T}{2}}^{\frac{T}{2}} f (t) e^{- i n ω_{0} t} d t for n = 0, \pm 1, \pm 2, \pm 3, \dots$

The results of this discussion are summarised in the following Key Point.

Key Point 9

Fourier Series in Complex Form

A function $f (t)$ of period $T$ has a complex Fourier series

$f (t) = \sum_{n = - \infty}^{\infty} c_{n} e^{i n ω_{0} t} where c_{n} = \frac{1}{T} \int_{- \frac{T}{2}}^{\frac{T}{2}} f (t) e^{- i n ω_{0} t} d t$

For the special case $T = 2 π$ , so that $ω_{0} = 1$ , these formulae become particularly simple:

\begin{array}{rcl} f (t) = \sum_{n = - \infty}^{\infty} c_{n} e^{i n t} c_{n} = \frac{1}{2 π} \int_{- π}^{π} f (t) e^{- i n t} d t . \end{array}

2 Revision of the exponential form of a complex number

Task!

Answer

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Key Point 8

Task!

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Answer

Key Point 9