Show that the Fourier series (1.7.1) can be expressed in the complex form $$ f(\theta)=\sum_{n=-\infty}^{+\infty} A_{n} e^{i n \theta} $$ where the complex Fourier coefficients are given by $$ \breve{A}_{\mathrm{n}}=\frac{1}{2 \mathrm{x}} \int_{-\pi}^{\pi} f(\theta) e^{-i \text { in } \theta} d \theta $$ Here, the entire series \((1.7 .11)\), not just its real part, represents \(f(\theta)\). If \(f(\theta)\) is real, this implies that the imaginary part of the series sums to zero. Relate the \(A_{n}\) to the coefficients \(a_{0}, a_{n}, b_{n}\) given by (1.7.2). We make considerable use of Fourier series in the complex form in later chapters.

Question: Given the complex form of the Fourier series: $$ f(\theta) = \sum_{n=-\infty}^{\infty} A_n e^{in\theta} $$ Where \(A_n = \frac{a_n - ib_n}{2}\) for \(n<0\), \(A_0 = \frac{a_0}{2}\), and \(A_n = \frac{a_n + ib_n}{2}\) for \(n>0\). Find the relationship between the complex coefficients \(A_n\) and the real coefficients \(a_0, a_n, b_n\). Answer: The relationship between the complex coefficients \(A_n\) and the real coefficients \(a_0, a_n, b_n\) is: $$ A_n = \begin{cases} \frac{a_0}{2}, & n=0 \\ \frac{a_n + ib_n}{2}, & n>0 \\ \frac{a_n - ib_n}{2}, & n<0 \end{cases} $$ And to solve for the real coefficients, we can use the relationships: $$ a_0 = 2A_0 \\ a_n = 2 (A_n + A_{-n}) \quad \text{for} \;\; n > 0 \\ b_n = 2i (A_n - A_{-n}) \quad \text{for} \;\; n > 0 $$