The function \(f(x) = \exp(2x)\sin(2x)\) can be expanded as the Taylor series:
\(\displaystyle\sum_{n=0}^{\infty} 2^{3n/2}\,\sin\!\left(\dfrac{n\pi}{4}\right)\dfrac{x^{n}}{n!}\)
Write \(f(x) = \operatorname{Im}\!\left(e^{2x} e^{2ix}\right) = \operatorname{Im}\!\left(e^{2(1+i)x}\right)\).
Since \(1 + i = \sqrt{2}\,e^{i\pi/4}\), we get \(2(1+i) = 2\sqrt{2}\,e^{i\pi/4}\), so \(e^{2(1+i)x} = \sum_{n=0}^{\infty} \dfrac{(2\sqrt{2})^{n} e^{i n\pi/4}}{n!}\,x^{n} = \sum_{n=0}^{\infty} \dfrac{2^{3n/2} e^{i n\pi/4}}{n!}\,x^{n}\).
Taking the imaginary part gives \(f(x) = \sum_{n=0}^{\infty} 2^{3n/2}\,\sin\!\left(\dfrac{n\pi}{4}\right)\dfrac{x^{n}}{n!}\).
Hence, the Taylor series is \(\sum_{n=0}^{\infty} 2^{3n/2}\sin(n\pi/4)\,x^{n}/n!\).
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