Prerequisite Math: Review of Complex Numbers

Quick Reference

Item	Statement
Imaginary unit	$i=\sqrt{-1}$, so $i^2=-1$, $i^3=-i$, $i^4=1$
Euler's formula	$e^{x+iy}=e^x(\cos y+i\sin y)$
Pure imaginary case	$e^{i\theta }=\cos \theta +i\sin \theta$
Shorthand	$\cos y+i\sin y$ is sometimes written $\text{cis }y$
Cosine from exponentials	$\cos x={\displaystyle \frac{e^{ix}+e^{-ix}}{2}}$
Sine from exponentials	$\sin x={\displaystyle \frac{e^{ix}-e^{-ix}}{2i}}$
Conjugate	$\overline{a+bi}=a-bi$
Modulus	$|a+bi|=\sqrt{a^2+b^2}$, and $z\bar{z}={\left|z\right|}^2$
Real polynomial roots	Complex roots occur in conjugate pairs

Deriving Euler's Formula

To motivate the definition of $e^{x+iy}$, where $x$ and $y$ are real numbers and $i=\sqrt{-1}$, we may proceed as follows. Using the exponential series $e^w=1+w+{\displaystyle \frac{w^2}{2!}}+\cdots$ with $w=iy$,

\begin{aligned} e^{z} = e^{x+iy} &= e^{x}e^{iy} \\ &= e^{x}\left[1 + iy + \frac{(iy)^{2}}{2!} + \frac{(iy)^{3}}{3!} + \frac{(iy)^{4}}{4!} + \cdots\right] \\ &= e^{x}\left[\left(1 - \frac{y^{2}}{2!} + \frac{y^{4}}{4!} - \cdots\right) + i\left(y - \frac{y^{3}}{3!} + \frac{y^{5}}{5!} - \cdots\right)\right] \end{aligned}

The last step used the powers of $i$ to separate real from imaginary terms: $(iy{)}^2=-y^2$, $(iy{)}^3=-i{y}^3$, $(iy{)}^4=y^4$, and the pattern repeats with period four. The even powers are real and alternate in sign; the odd powers are imaginary and alternate in sign.

Now recognize the Taylor series

\begin{aligned} \sin x &= x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \cdots \\ \cos x &= 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \cdots \end{aligned}

sitting inside the two groups. We obtain

• This equation, known as Euler's formula, does not merely relate quantities already defined: it defines $e^{x+iy}$. Raising $e$ to a complex power has no prior meaning, and the series computation shows this is the only definition consistent with the usual exponential.
• For brevity, the expression in parentheses is sometimes written $\text{cis }y$.

Sine and Cosine as Complex Exponentials

It follows from

\begin{aligned} e^{ix} &= \cos x + i\sin x \\ e^{-ix} &= \cos x - i\sin x \end{aligned}

where the second line replaces $x$ by $-x$ and uses $\cos (-x)=\cos x$ and $\sin (-x)=-\sin x$, that adding and subtracting give

Adding the two lines cancels the sines and gives $2\cos x$; subtracting cancels the cosines and gives $2i\sin x$.

Where This Is Used in the Book

• Section 4.2. Complex characteristic roots $\alpha \pm i\beta$ give the complex solutions $e^{(\alpha \pm i\beta )x}$, which Euler's formula converts into the real pair $e^{\alpha x}\cos \beta x$ and $e^{\alpha x}\sin \beta x$.
• Section 4.5. Complex exponents in a Cauchy-Euler equation give $x^{\alpha \pm i\beta }$, which becomes $x^{\alpha }\cos (\beta \ln x)$ and $x^{\alpha }\sin (\beta \ln x)$ after writing $x^{i\beta }=e^{i\beta \ln x}$.
• Section 5.1. Complex eigenvalues produce complex solution vectors, whose real and imaginary parts are the two real solutions.
• Section 7.1. The transform of $e^{ikt}$ yields the transforms of $\sin kt$ and $\cos kt$ at once, by taking real and imaginary parts.

The pattern is the same in every case: do the computation with complex exponentials, which are easy to differentiate and integrate, then extract real and imaginary parts at the end.

Conjugates and Real Answers

Two facts about conjugates are used repeatedly.

This is why complex characteristic roots and complex eigenvalues always appear in pairs, and why the second member of a pair never requires separate computation.

This is the fact behind the substitution $A=\frac{1}{2}(c_1-i{c}_2)$, $B=\frac{1}{2}(c_1+i{c}_2)$ used in Section 4.2 to turn a complex general solution into a real one.

Worked Examples

Exercises

Frequently Asked Questions

What is Euler's formula?

The identity $e^{x+iy}=e^x(\cos y+i\sin y)$, and in particular $e^{i\theta }=\cos \theta +i\sin \theta$. It is what gives meaning to a complex exponent, and it is the bridge between exponential and trigonometric behavior.

Is Euler's formula proved or assumed?

Both, depending on the starting point. If $e^w$ for complex $w$ is defined by its power series, the formula is a theorem, proved by the rearrangement shown above. If it is not yet defined, the formula is taken as the definition, and the series computation is what shows the definition is the only reasonable one.

Why do complex roots always come in pairs?

Because the characteristic polynomial has real coefficients. Conjugating the equation $p(z)=0$ leaves the coefficients unchanged and turns $z$ into $\bar{z}$, so $p(\bar{z})=0$ as well. The same argument applies to the eigenvalues of a real matrix.

How do complex solutions become real ones?

Take real and imaginary parts. If $z(x)$ solves a linear equation with real coefficients, then so do $\mathfrak{Re}z$ and $\mathfrak{Im}z$, since they are the combinations $\frac{1}{2}(z+\bar{z})$ and $\frac{1}{2i}(z-\bar{z})$ of the two conjugate solutions. This device recurs in Sections 4.2, 4.5, 5.1, and 7.1.

Why divide by $2i$ in the formula for $\sin x$?

Because subtracting $e^{-ix}$ from $e^{ix}$ leaves $2i\sin x$, so recovering $\sin x$ requires dividing by $2i$. The asymmetry with the cosine formula is real, and forgetting the $i$ is a common slip.

What is the relationship to the hyperbolic functions?

They are the same construction without the $i$: $\cosh x={\displaystyle \frac{e^x+e^{-x}}{2}}$ against $\cos x={\displaystyle \frac{e^{ix}+e^{-ix}}{2}}$. Formally, $\cos (ix)=\cosh x$ and $\sin (ix)=i\sinh x$, which is why every trigonometric identity has a hyperbolic counterpart differing only in signs, and why their Laplace transforms differ only in the sign of $k^2$.