# Complex Form of Fourier Series

Let the function $$f\left( x \right)$$ be defined on the interval $$\left[ { – \pi ,\pi } \right].$$ Using the well-known Euler’s formulas

${\cos \varphi = \frac{{{e^{i\varphi }} + {e^{ – i\varphi }}}}{2},\;\;\;}\kern0pt {\sin \varphi = \frac{{{e^{i\varphi }} – {e^{ – i\varphi }}}}{{2i}},}$

we can write the Fourier series of the function in complex form:

${f\left( x \right) = \frac{{{a_0}}}{2} }+{ \sum\limits_{n = 1}^\infty {\left( {{a_n}\cos nx + {b_n}\sin nx} \right)} } = {\frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {\left( {{a_n}\frac{{{e^{inx}} + {e^{ – inx}}}}{2} }\right.}}+{{\left.{ {b_n}\frac{{{e^{inx}} – {e^{ – inx}}}}{{2i}}} \right)} } = {\frac{{{a_0}}}{2} }+{ \sum\limits_{n = 1}^\infty {\frac{{{a_n} – i{b_n}}}{2}{e^{inx}}} }+{ \sum\limits_{n = 1}^\infty {\frac{{{a_n} + i{b_n}}}{2}{e^{ – inx}}} } = {\sum\limits_{n = – \infty }^\infty {{c_n}{e^{inx}}} .}$

Here we have used the following notations:

${{c_0} = \frac{{{a_0}}}{2},\;\;\;}\kern0pt {{c_n} = \frac{{{a_n} – i{b_n}}}{2},\;\;\;}\kern0pt {{c_{ – n}} = \frac{{{a_n} + i{b_n}}}{2}.}$

The coefficients $${c_n}$$ are called complex Fourier coefficients. They are defined by the formulas

${{c_n} = \frac{1}{{2\pi }}\int\limits_{ – \pi }^\pi {f\left( x \right){e^{ – inx}}dx} ,\;\;}\kern0pt{n = 0, \pm 1, \pm 2, \ldots }$

If necessary to expand a function $$f\left( x \right)$$ of period $$2L,$$ we can use the following expressions:

$f\left( x \right) = \sum\limits_{n = – \infty }^\infty {{c_n}{e^{\frac{{in\pi x}}{L}}}} ,$

where

${{c_n} = \frac{1}{{2L}}\int\limits_{ – L}^L {f\left( x \right){e^{ – \frac{{in\pi x}}{L}}}dx} ,\;\;}\kern0pt{n = 0, \pm 1, \pm 2, \ldots }$

The complex form of Fourier series is algebraically simpler and more symmetric. Therefore, it is often used in physics and other sciences.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Using complex form, find the Fourier series of the function
${f\left( x \right) = \text{sign}\,x }= {\begin{cases} -1, & -\pi \le x \le 0 \\ 1, & 0 \lt x \le \pi \end{cases}.}$

### Example 2

Using complex form find the Fourier series of the function $$f\left( x \right) = {x^2},$$ defined on the interval $$\left[ { – 1,1} \right].$$

### Example 3

Using complex form find the Fourier series of the function
${f\left( x \right) = \frac{{a\sin x}}{{1 – 2a\cos x + {a^2}}},\;\;}\kern-0.3pt{\left| a \right| \lt 1.}$

### Example 1.

Using complex form, find the Fourier series of the function
${f\left( x \right) = \text{sign}\,x }= {\begin{cases} -1, & -\pi \le x \le 0 \\ 1, & 0 \lt x \le \pi \end{cases}.}$

Solution.

We calculate the coefficients $${c_0}$$ and $${c_n}$$ for $$n \ne 0:$$

$\require{cancel} {{c_0} = \frac{1}{{2\pi }}\int\limits_{ – \pi }^\pi {f\left( x \right)dx} } = {\frac{1}{{2\pi }}\left[ {\int\limits_{ – \pi }^0 {\left( { – 1} \right)dx} + \int\limits_0^\pi {dx} } \right] } = {\frac{1}{{2\pi }}\left[ {\left. {\left( { – x} \right)} \right|_{ – \pi }^0 + \left. x \right|_0^\pi } \right] } = {\frac{1}{{2\pi }}\left( { – \cancel{\pi} + \cancel{\pi }} \right) }={ 0,}$

${{c_n} = \frac{1}{{2\pi }}\int\limits_{ – \pi }^\pi {f\left( x \right){e^{ – inx}}dx} } = {{\frac{1}{{2\pi }}\left[ {\int\limits_{ – \pi }^0 {\left( { – 1} \right){e^{ – inx}}dx} }\right.}+{\left.{ \int\limits_0^\pi {{e^{ – inx}}dx} } \right] }} = {{\frac{1}{{2\pi }}\left[ { – \frac{{\left. {\left( {{e^{ – inx}}} \right)} \right|_{ – \pi }^0}}{{ – in}} }\right.}+{\left.{ \frac{{\left. {\left( {{e^{ – inx}}} \right)} \right|_0^\pi }}{{ – in}}} \right] }} = {{\frac{i}{{2\pi n}}\left[ { – \left( {1 – {e^{in\pi }}} \right) }\right.}+{\left.{ {e^{ – in\pi }} – 1} \right] }} = {\frac{i}{{2\pi n}}\left[ {{e^{in\pi }} + {e^{ – in\pi }} – 2} \right] } = {\frac{i}{{\pi n}}\left[ {\frac{{{e^{in\pi }} + {e^{ – in\pi }}}}{2} – 1} \right] } = {\frac{i}{{\pi n}}\left[ {\cos n\pi – 1} \right] } = {\frac{i}{{\pi n}}\left[ {{{\left( { – 1} \right)}^n} – 1} \right].}$

If $$n = 2k,$$ then $${c_{2k}} = 0.$$ If $$n = 2k – 1,$$ then $${c_{2k – 1}}$$ $$= – {\large\frac{{2i}}{{\left( {2k – 1} \right)\pi }}\normalsize}.$$

Hence, the Fourier series of the function in complex form is

${f\left( x \right) = \text{sign}\,x } = { – \frac{{2i}}{\pi }\sum\limits_{k = – \infty }^\infty {\frac{1}{{2k – 1}}{e^{i\left( {2k – 1} \right)x}}} .}$

We can transform the series and write it in the real form. Rename: $$n = 2k – 1,$$ $$n = \pm 1, \pm 2, \pm 3, \ldots$$ Then

${f\left( x \right) = \text{sign}\,x } = { – \frac{{2i}}{\pi }\sum\limits_{k = – \infty }^\infty {\frac{1}{{2k – 1}}{e^{i\left( {2k – 1} \right)x}}} } = { – \frac{{2i}}{\pi }\sum\limits_{n = – \infty }^\infty {\frac{{{e^{inx}}}}{n}} } = { – \frac{{2i}}{\pi }\sum\limits_{n = 1}^\infty {\left( {\frac{{{e^{ – inx}}}}{{ – n}} + \frac{{{e^{inx}}}}{n}} \right)} } = {\frac{4}{\pi }\sum\limits_{n = 1}^\infty {\frac{{{e^{inx}} – {e^{ – inx}}}}{{2in}}} } = {\frac{4}{\pi }\sum\limits_{n = 1}^\infty {\frac{{\sin nx}}{n}} } = {\frac{4}{\pi }\sum\limits_{k = 1}^\infty {\frac{{\sin \left( {2k – 1} \right)x}}{{2k – 1}}} .}$

Graph of the function and its Fourier approximation for $$n = 5$$ and $$n = 50$$ are shown in Figure $$1.$$

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Problem 1
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Problems 2-3