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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 ${\cos \varphi = \frac{{{e^{i\varphi }} + {e^{ - i\varphi }}}}{2},}\;\; {\sin \varphi = \frac{{{e^{i\varphi }} - {e^{ - i\varphi }}}}{{2i}},}$ we can write the Fourier series of the function in complex form: ${f\left( x \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {\left( {{a_n}\cos nx + {b_n}\sin nx} \right)} } = {\frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {\left( {{a_n}\frac{{{e^{inx}} + {e^{ - inx}}}}{2} + {b_n}\frac{{{e^{inx}} - {e^{ - inx}}}}{{2i}}} \right)} } = {\frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {\frac{{{a_n} - i{b_n}}}{2}{e^{inx}}} + \sum\limits_{n = 1}^\infty {\frac{{{a_n} + i{b_n}}}{2}{e^{ - inx}}} } = {\sum\limits_{n = - \infty }^\infty {{c_n}{e^{inx}}} .}$ Here we have used the following notations: ${{c_0} = \frac{{{a_0}}}{2},}\;\; {{c_n} = \frac{{{a_n} - i{b_n}}}{2},}\;\; {{c_{ - n}} = \frac{{{a_n} + i{b_n}}}{2}.}$ The coefficients $${c_n}$$ are called complex Fourier coefficients. They are defined by the formulas ${c_n} = \frac{1}{{2\pi }}\int\limits_{ - \pi }^\pi {f\left( x \right){e^{ - inx}}dx} ,\;\;n = 0, \pm 1, \pm 2, \ldots$ 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 ${c_n} = \frac{1}{{2L}}\int\limits_{ - L}^L {f\left( x \right){e^{ - \frac{{in\pi x}}{L}}}dx} ,\;\;n = 0, \pm 1, \pm 2, \ldots$ The complex form of Fourier series is algebraically simpler and more symmetric. Therefore, it is often used in physics and other sciences.

Example 1
Using complex form, find the Fourier series of the function $f\left( x \right) = \text{sign}\,x = \begin{cases} -1, & -\pi \le x \le 0 \\ 1, & 0 < x \le \pi \end{cases}.$
Solution.
Calculate the coefficients $${c_0}$$ and $${c_n}$$ for $$n \ne 0:$$ $\require{cancel} {{c_0} = \frac{1}{{2\pi }}\int\limits_{ - \pi }^\pi {f\left( x \right)dx} } = {\frac{1}{{2\pi }}\left[ {\int\limits_{ - \pi }^0 {\left( { - 1} \right)dx} + \int\limits_0^\pi {dx} } \right] } = {\frac{1}{{2\pi }}\left[ {\left. {\left( { - x} \right)} \right|_{ - \pi }^0 + \left. x \right|_0^\pi } \right] } = {\frac{1}{{2\pi }}\left( { - \cancel{\pi} + \cancel{\pi }} \right) = 0,}$ ${{c_n} = \frac{1}{{2\pi }}\int\limits_{ - \pi }^\pi {f\left( x \right){e^{ - inx}}dx} } = {\frac{1}{{2\pi }}\left[ {\int\limits_{ - \pi }^0 {\left( { - 1} \right){e^{ - inx}}dx} + \int\limits_0^\pi {{e^{ - inx}}dx} } \right] } = {\frac{1}{{2\pi }}\left[ { - \frac{{\left. {\left( {{e^{ - inx}}} \right)} \right|_{ - \pi }^0}}{{ - in}} + \frac{{\left. {\left( {{e^{ - inx}}} \right)} \right|_0^\pi }}{{ - in}}} \right] } = {\frac{i}{{2\pi n}}\left[ { - \left( {1 - {e^{in\pi }}} \right) + {e^{ - in\pi }} - 1} \right] } = {\frac{i}{{2\pi n}}\left[ {{e^{in\pi }} + {e^{ - in\pi }} - 2} \right] } = {\frac{i}{{\pi n}}\left[ {\frac{{{e^{in\pi }} + {e^{ - in\pi }}}}{2} - 1} \right] } = {\frac{i}{{\pi n}}\left[ {\cos n\pi - 1} \right] } = {\frac{i}{{\pi n}}\left[ {{{\left( { - 1} \right)}^n} - 1} \right].}$ 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 $$1.$$
 Fig.1, n = 5, n = 50 Fig.2, n = 2, n = 5
Example 2
Using complex form find the Fourier series of the function $$f\left( x \right) = {x^2},$$ defined on the interval $$\left[ { - 1,1} \right].$$

Solution.
Here the half-period is $$L = 1.$$ Therefore, the coefficient $${c_0}$$ is ${{c_0} = \frac{1}{{2L}}\int\limits_{ - L}^L {f\left( x \right)dx} } = {\frac{1}{2}\int\limits_{ - 1}^1 {{x^2}dx} } = {\frac{1}{2}\left[ {\left. {\left( {\frac{{{x^3}}}{3}} \right)} \right|_{ - 1}^1} \right] } = {\frac{1}{6}\left[ {{1^3} - {{\left( { - 1} \right)}^3}} \right] } = {\frac{1}{3}.}$ For $$n \ne 0$$, ${{c_n} = \frac{1}{{2L}}\int\limits_{ - L}^L {f\left( x \right){e^{ - \frac{{in\pi x}}{L}}}dx} } = {\frac{1}{2}\int\limits_{ - 1}^1 {{x^2}{e^{ - {in\pi x}}}dx} .}$ Integrating by parts twice, we obtain ${{c_n} = \frac{1}{2}\left[ {\left. {\left( {\frac{{{x^2}{e^{ - in\pi x}}}}{{ - in\pi }}} \right)} \right|_{ - 1}^1 - \int\limits_{ - 1}^1 {\frac{{2x{e^{ - in\pi x}}}}{{ - in\pi }}dx} } \right] } = {\frac{1}{2}\left[ {\left. {\left( {\frac{{{x^2}{e^{ - in\pi x}}}}{{ - in\pi }}} \right)} \right|_{ - 1}^1 + \frac{2}{{in\pi }}\int\limits_{ - 1}^1 {x{e^{ - in\pi x}}dx} } \right] } = {\frac{1}{2}\left\{ {\left. {\left( {\frac{{{x^2}{e^{ - in\pi x}}}}{{ - in\pi }}} \right)} \right|_{ - 1}^1 + \frac{2}{{in\pi }}\left[ {\left. {\left( {\frac{{x{e^{ - in\pi x}}}}{{ - in\pi }}} \right)} \right|_{ - 1}^1 - \int\limits_{ - 1}^1 {\frac{{{e^{ - in\pi x}}}}{{ - in\pi }}dx} } \right]} \right\} } = { - \frac{1}{{2in\pi }}\left[ {\left. {\left( {{x^2}{e^{ - in\pi x}} + \frac{2}{{in\pi }}x{e^{ - in\pi x}} + \frac{2}{{{{\left( {in\pi } \right)}^2}}}{e^{ - in\pi x}}} \right)} \right|_{ - 1}^1} \right] } = {- \frac{1}{{2in\pi }}\left[ {{e^{ - in\pi }} + \frac{2}{{in\pi }}{e^{ - in\pi }} + \frac{2}{{{{\left( {in\pi } \right)}^2}}}{e^{ - in\pi }}} \right. } + {\left. {\frac{2}{{in\pi }}{e^{in\pi }} - \frac{2}{{{{\left( {in\pi } \right)}^2}}}{e^{in\pi }}} \right] } = {\frac{1}{{2in\pi }}\left[ {{e^{in\pi }} - {e^{ - in\pi }} - \frac{2}{{in\pi }}\left( {{e^{in\pi }} + {e^{ - in\pi }}} \right)} \right. } + {\left. {\frac{2}{{{{\left( {in\pi } \right)}^2}}}\left( {{e^{in\pi }} - {e^{ - in\pi }}} \right)} \right] } = {\frac{1}{{n\pi }} \cdot \frac{{{e^{in\pi }} - {e^{ - in\pi }}}}{{2i}} } + {\frac{2}{{{n^2}{\pi ^2}}} \cdot \frac{{{e^{in\pi }} + {e^{ - in\pi }}}}{2} } - {\frac{2}{{{n^3}{\pi ^3}}} \cdot \frac{{{e^{in\pi }} - {e^{ - in\pi }}}}{{2i}} } = {\frac{1}{{n\pi }} \cdot \sin n\pi + \frac{2}{{{n^2}{\pi ^2}}}\cos n\pi - \frac{2}{{{n^3}{\pi ^3}}}\sin n\pi .}$ Substituting $$\sin n\pi = 0$$ and $$\cos n\pi = {\left( { - 1} \right)^n},$$ we get the compact expression for the coefficients $${c_n}:$$ ${c_n} = \frac{2}{{{n^2}{\pi ^2}}}{\left( { - 1} \right)^n}.$ Thus, the Fourier extension in complex form is given by ${f\left( x \right) = {x^2} } = {\frac{1}{3} + \sum\limits_{n = 1}^\infty {\frac{2}{{{n^2}{\pi ^2}}}{{\left( { - 1} \right)}^n}{e^{in\pi x}}} } + {\sum\limits_{n = 1}^\infty {\frac{2}{{{{\left( { - n} \right)}^2}{\pi ^2}}}{{\left( { - 1} \right)}^{ - n}}{e^{ - in\pi x}}} .}$ Taking into account that $${\left( { - 1} \right)^{ - n}} = {\left( { - 1} \right)^n},$$ we can finally write: ${f\left( x \right) = {x^2} } = {\frac{1}{3} + \frac{4}{{{\pi ^2}}}\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{n^2}}}\frac{{{e^{in\pi x}} + {e^{ - in\pi x}}}}{2}} } = {\frac{1}{3} + \frac{4}{{{\pi ^2}}}\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{n^2}}}\cos n\pi x} .}$ Graphs of the function and the Fourier series expansions are shown in Figure $$2$$ above.

Example 3
Using complex form find the Fourier series of the function $f\left( x \right) = \frac{{a\sin x}}{{1 - 2a\cos x + {a^2}}},\;\;\left| a \right| \lt 1.$
Solution.
We apply the formulas ${\cos x = \frac{{{e^{ix}} + {e^{ - ix}}}}{2},}\;\; {\sin x = \frac{{{e^{ix}} - {e^{ - ix}}}}{{2i}}.}$ This results in the following expression: ${f\left( x \right) = \frac{{a \cdot \frac{{{e^{ix}} - {e^{ - ix}}}}{{2i}}}}{{1 - 2a \cdot \frac{{{e^{ix}} + {e^{ - ix}}}}{2} + {a^2}}} } = {\frac{1}{{2i}} \cdot \frac{{a\left( {{e^{ix}} - {e^{ - ix}}} \right)}}{{1 - a\left( {{e^{ix}} + {e^{ - ix}}} \right) + {a^2}}} } = {\frac{1}{{2i}} \cdot \frac{{a\left( {{e^{ix}} - {e^{ - ix}}} \right)}}{{1 - a{e^{ix}} - a{e^{ - ix}} + {a^2}{e^{ix}}{e^{ - ix}}}} } = {\frac{1}{{2i}} \cdot \frac{{a\left( {{e^{ix}} - {e^{ - ix}}} \right)}}{{\left( {1 - a{e^{ix}}} \right) - a{e^{ - ix}}\left( {1 - a{e^{ix}}} \right)}} } = {\frac{1}{{2i}} \cdot \frac{{a\left( {{e^{ix}} - {e^{ - ix}}} \right)}}{{\left( {1 - a{e^{ix}}} \right)\left( {1 - a{e^{ - ix}}} \right)}}.}$ Using partial decomposition, we can write: ${f\left( x \right) = \frac{1}{{2i}} \cdot \frac{{a\left( {{e^{ix}} - {e^{ - ix}}} \right)}}{{\left( {1 - a{e^{ix}}} \right)\left( {1 - a{e^{ - ix}}} \right)}} } = {\frac{1}{{2i}}\left( {\frac{A}{{1 - a{e^{ix}}}} + \frac{B}{{1 - a{e^{ - ix}}}}} \right).}$ Calculate the coefficients $$A, B:$$ ${A\left( {1 - a{e^{ - ix}}} \right) + B\left( {1 - a{e^{ix}}} \right) = a{e^{ix}} - a{e^{ - ix}},}\;\; {\Rightarrow A - aA{e^{ - ix}} + B - aB{e^{ix}} = a{e^{ix}} - a{e^{ - ix}},}\;\; {\Rightarrow A = 1,\;B = - 1.}$ As a result, the function $$f\left( x \right)$$ can be written in the form $f\left( x \right) = \frac{1}{{2i}}\left( {\frac{1}{{1 - a{e^{ix}}}} - \frac{1}{{1 - a{e^{ - ix}}}}} \right).$ We see that ${\left| {a{e^{ix}}} \right| = \left| a \right|\left| {{e^{ix}}} \right| } = {\left| a \right|\sqrt {{{\cos }^2}x + {{\sin }^2}x} } = {\left| a \right| \lt 1.}$ For conjugate, we have the same result: $\left| {a{e^{ - ix}}} \right| = \left| {a{e^{ix}}} \right| = \left| a \right| \lt 1.$ Expanding the fractions into power series, we get $\frac{1}{{1 - a{e^{ix}}}} = {\left( {1 - a{e^{ix}}} \right)^{ - 1}} = \sum\limits_{n = 0}^\infty {{a^n}{e^{inx}}} ,$ $\frac{1}{{1 - a{e^{ - ix}}}} = {\left( {1 - a{e^{ - ix}}} \right)^{ - 1}} = \sum\limits_{n = 0}^\infty {{a^n}{e^{ - inx}}} .$ Thus, the Fourier series of the function $$f\left( x \right)$$ is ${f\left( x \right) = \frac{1}{{2i}}\sum\limits_{n = 0}^\infty {{a^n}\left( {{e^{inx}} - {e^{ - inx}}} \right)} } = {\sum\limits_{n = 0}^\infty {{a^n}\sin nx} .}$ Since $$\sin nx = 0,$$ the final answer is $f\left( x \right) = \sum\limits_{n = 1}^\infty {{a^n}\sin nx} .$