Calculus – Complex numbers – Applications

The most important application is at the finding of solutions of homogeneous linear differential equations with constant coefficients of order two and higher. Such a differential equation has the form

\[c_ny^{(n)}(x)+c_{n-1}y^{(n-1)}(x)+\cdots+c_1y'(x)+c_0y(x)=0,\quad n\geq2.\]

If we \(y(x)=e^{rx}\) try as a solution, then we have that \(y'(x)=re^{rx},\;y''(x)=r^2e^{rx},\;\ldots,\;y^{(n)}(x)=r^ne^{rx}\). Substitution then leads to:

\[c_nr^ne^{rx}+c_{n-1}r^{n-1}e^{rx}+\cdots+c_1re^{rx}+c_0e^{rx}=0\quad\Longleftrightarrow\quad e^{rx}\left(c_nr^n+c_{n-1}r^{n-1}+\cdots+c_1r+c_0\right)=0.\]

Since \(e^{rx}\neq0\) this implies the characteristic equation

\[c_nr^n+c_{n-1}r^{n-1}+\cdots+c_1r+c_0=0.\]

The fundamental theorem of algebra then implies that this equation has \(n\) solutions for \(r\) in \(\mathbb{C}\). In the case of real differential equations (all coefficients are real) nonreal solutions can only occur in complex conjugate pairs. In the case of these nonreal solutions linear combinations of the complex solutions give rise to real solutions. If \(r=\alpha\pm i\beta\) with \(\beta\neq 0\), then we have:

\[e^{(\alpha+i\beta)x}=e^{\alpha x}\cdot e^{i\beta x}=e^{\alpha x}\left(\cos(\beta x)+i\sin(\beta x)\right)\quad\text{and}\quad e^{(\alpha-i\beta)x}=e^{\alpha x}\cdot e^{-\beta x}=e^{\alpha x}\left(\cos(\beta x)-i\sin(\beta x)\right)\]

are complex solutions. Linear combinations then give rise to the linear independent real solutions

\[e^{\alpha x}\cos(\beta x)\quad\text{and}\quad e^{\alpha x}\sin(\beta x).\]

More information and details can be found on the pages concerning homogeneous linear differential equations with constant coefficients of order two and higher.

Integrals of the form \(\displaystyle\int e^{\alpha x}\cos(\beta x)\,dx\) and \(\displaystyle\int e^{\alpha x}\sin(\beta x)\,dx\) can be evaluated by means of integration by parts, however using the observation that \(e^{\alpha x}\cos(\beta x)=\text{Re}\left(e^{(\alpha+i\beta)x}\right)\) and \(e^{\alpha x}\sin(\beta x)=\text{Im}\left(e^{(\alpha+i\beta)x}\right)\) and using

\[\int e^{(\alpha+i\beta)x}\,dx=\frac{1}{\alpha+i\beta}e^{(\alpha+i\beta)x}=\frac{\alpha-i\beta}{\alpha^2+\beta^2}e^{\alpha x}\left(\cos(\beta x)+i\sin(\beta x)\right)\]

it easily follows that

\[\int e^{\alpha x}\cos(\beta x)\,dx=\frac{\alpha}{\alpha^2+\beta^2}e^{\alpha x}\cos(\beta x)+\frac{\beta}{\alpha^2+\beta^2}e^{\alpha x}\sin(\beta x)\]

and

\[\int e^{\alpha x}\sin(\beta x)\,dx=\frac{\alpha}{\alpha^2+\beta^2}e^{\alpha x}\sin(\beta x)-\frac{\beta}{\alpha^2+\beta^2}e^{\alpha x}\cos(\beta x).\]

Another application is in fact a byproduct of the theory of complex numbers. Although De Moivre's theorem is proved by using trigonometric identities, conversely this can be used to deduce these identities. So one does not need to memorize these identities (anymore), because they are easily deduced by using De Moivre's theorem. Some examples:

The "double angle" formulas

\[\left\{\begin{array}{l}\sin(2\theta)=2\sin(\theta)\cos(\theta)\\[2.5mm]\cos(2\theta)=\cos^2(\theta)-\sin^2(\theta)\end{array}\right.\]

can be deduced as follows:

\begin{align*} \cos(2\theta)+i\sin(2\theta)&=e^{2\theta i}=\left(e^{\theta i}\right)^2=\left(\cos(\theta)+i\sin(\theta)\right)^2\\[2.5mm] &=\cos^2(\theta)+2i\sin(\theta)\cos(\theta)+i^2\sin^2(\theta)\\[2.5mm] &=\cos^2(\theta)-\sin^2(\theta)+2i\sin(\theta)\cos(\theta). \end{align*}

Comparing both the real and the imaginary part leads to the wanted result. More general we have:

\begin{align*} \cos(\theta+\phi)+i\sin(\theta+\phi)&=e^{(\theta+\phi)i}=e^{\theta i}\cdot e^{\phi i}=\left(\cos(\theta)+i\sin(\theta)\right)\left(\cos(\phi)+i\sin(\phi)\right)\\[2.5mm] &=\cos(\theta)\cos(\phi)+i\sin(\theta)\cos(\phi)+i\cos(\theta)\sin(\phi)+i^2\sin(\theta)\sin(\phi)\\[2.5mm] &=\cos(\theta)\cos(\phi)-\sin(\theta)\sin(\phi)+i\left(\sin(\theta)\cos(\phi)+\cos(\theta)\sin(\phi)\right). \end{align*}

Comparing both the real and the imaginary part leads to the addition formulas:

\[\left\{\begin{array}{l}\sin(\theta+\phi)=\sin(\theta)\cos(\phi)+\cos(\theta)\sin(\phi)\\[2.5mm] \cos(\theta+\phi)=\cos(\theta)\sin(\phi)-\sin(\theta)\sin(\phi).\end{array}\right.\]

Stewart Appendix H, Exercise 47
Use De Moivre's theorem to find formulas for \(\sin(3\theta)\) en \(\cos(3\theta)\).

Solution:
\begin{align*} \cos(3\theta)+i\sin(3\theta)&=e^{3\theta i}=\left(e^{\theta i}\right)^3=\left(\cos(\theta)+i\sin(\theta)\right)^3\\[2.5mm] &=\cos^3(\theta)+3\cos^2(\theta)\cdot i\sin(\theta)+3\cos(\theta)\cdot i^2\sin^2(\theta)+i^3\sin^3(\theta)\\[2.5mm] &=\cos^3(\theta)-3\cos(\theta)\sin^2(\theta)+i\left(3\cos^2(\theta)\sin(\theta)-\sin^3(\theta)\right). \end{align*}

Using \(\sin^2(\theta)=1-\cos^2(\theta)\) and \(\cos^2(\theta)=1-\sin^2(\theta)\) we conclude that:

\[\left\{\begin{array}{l}\cos(3\theta)=\cos^3(\theta)-3\cos(\theta)\sin^2(\theta)=4\cos^3(\theta)-3\cos(\theta)\\[2.5mm] \sin(3\theta)=3\cos^2(\theta)\sin(\theta)-\sin^3(\theta)=3\sin(\theta)-4\sin^3(\theta).\end{array}\right.\]

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Last modified on March 1, 2021