Differentiating & Integrating Trigonometric Functions Using Complex Numbers

Since a complex number can be expressed using $\cos$ and $\sin$, the inverse is also true. $\sin (\theta )$ can be represented as $Im(e^{i\theta })$ and similarly, $cos(\theta )=Re(e^{i\theta })$ . This can aid integration and derivation.

Examples §

1: Finding the 40th Derivative

Solve the following derivative: $\frac{d^{40}}{d{x}^{40}}(e^{-x}\cos (x))$
$\cos (x)$ can be represented as a complex number: $e^{ix}=\cos (x)+i\sin (x)$ $\text{Re}(e^{ix})=\cos (x)$ So: $\frac{d^{40}}{d{x}^{40}}(e^{-x}\times \text{Re}(e^{ix}))$ We can ‘pull’ out the Re() function, because real numbers do not affect the Re(%20function.). $\text{Re}\left[\frac{d^{40}}{d{x}^{40}}(e^{-x}\times e^{ix})\right]=\text{Re}\left[\frac{d^{40}}{d{x}^{40}}(e^{x(-1+i)})\right]$ $\text{Re}((-1+i{)}^{40}\times e^{x(-1+i)})]$ We can use De Moivre’s Theorem to simplify the first part: $|-1+i|=\sqrt{2}$ and $Arg(-1+i)=\frac{3\pi }{4}$ $(\sqrt{2}{e}^{3\pi i/4}{)}^{40}=2^{20}\times e^{30\pi i}=2^{20}\times e^0=2^{20}$ $\text{Re}(2^{20}\times e^{x(-1+i)})]$ We can pull out the real $2^{20}$ and $e^{-x}$ $2^{20}\times \text{Re}(e^{-x}\times e^{ix})$ $2^{20}\times e^{-x}\times \text{Re}(e^{ix})$
$=2^{20}{e}^{-x}\cos (x)$