Every complex Fourier series ahead in this course is written entirely in terms of e^{ikx}. There are no sines and no cosines. It's just a single family of complex exponentials, indexed by an integer k. Euler's formula is the only fact needed to move freely between that compact form and the real sines and cosines it's built from. It's worth having automatic, before the notation leans on it that hard.
e^{i\theta} is not exponential growth, in the usual sense. Multiplying by it never changes a number's distance from the origin. It only rotates that number by angle \theta. Euler's formula is the key between that turning picture and the ordinary cosine/sine picture. The real part of e^{i\theta} is \cos\theta. The imaginary part is \sin\theta. That's the entire content of the formula.
Here's why that link has to be true, not just a naming rule. Track how fast the point \cos\theta+i\sin\theta moves, as \theta grows. The answer is exactly i times the point's own current position. That's the same "rate equals a constant times the current value" pattern an ordinary exponential e^{kx} obeys. Here the constant k just gets replaced by i. Two things that start in the same place, and always grow at that same rate, have to stay equal forever. \cos\theta+i\sin\theta and e^{i\theta} both start at 1. They both grow that way too. So they're one and the same function.
Euler's formula states e^{i\theta}=\cos\theta+i\sin\theta, for any real \theta. So a complex exponential e^{i\theta} is exactly the point at angle \theta on the unit circle in the complex plane.
The formula also follows from a one-line derivative check. It's not just a definition to accept on faith. Let f(\theta)=\cos\theta+i\sin\theta. Then f'(\theta)=-\sin\theta+i\cos\theta=i(\cos\theta+i\sin\theta)=if(\theta), and f(0)=1. That is exactly the initial value problem y'=iy,\ y(0)=1. A first-order linear equation y'=ky with y(0)=y_0 has exactly one solution, y=y_0e^{kx}. So f(\theta) solves the identical problem e^{i\theta} is defined to solve. The two must be the same function.
Complex conjugate exponents \alpha\pm i\beta show up constantly. For instance, they're the two roots of a real quadratic equation with a negative discriminant. The complex expressions e^{(\alpha\pm i\beta)x}=e^{\alpha x}e^{\pm i\beta x} combine, using Euler's formula, into the real functions e^{\alpha x}\cos\beta x and e^{\alpha x}\sin\beta x. That's the pair a constant-coefficient equation with such roots always produces.
A complex pair, and the real oscillation it produces
Drag the root's imaginary part -- watch the oscillation frequency change. Drag the real part -- watch the envelope grow or decay around it.
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Worked example
Euler's formula at a specific angle
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1
Evaluate e^{i\pi/3} using Euler's formula.
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e^{i\pi/3} = \cos(\pi/3)+i\sin(\pi/3).
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3\boxed{e^{i\pi/3} = \tfrac12 + \tfrac{\sqrt3}{2}i}
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Worked example
Complex roots to a real solution
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Write the real general solution for a second-order linear ODE whose two roots are -\tfrac12\pm3i.
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By Euler's formula, e^{(-1/2\pm3i)x}=e^{-x/2}(\cos3x\pm i\sin3x), whose real and imaginary parts are each real solutions on their own.
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3\boxed{y = e^{-x/2}\left(C_1\cos3x+C_2\sin3x\right)}
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Where this goes next
Complex Fourier series
Writing a Fourier series entirely in complex exponentials e^{ikx} does two things. It turns every coefficient formula into a single integral, instead of two separate sine and cosine cases. And it turns differentiating a series term by term into just multiplying by ik. None of that shortcut works without one thing. You need to move freely between e^{i\theta} and \cos\theta,\sin\theta. That's exactly the formula reviewed here.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Ordinary Differential Equations
For roots r=1+-2i, what is beta (the coefficient controlling oscillation frequency)?
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Why
\beta=2, the imaginary part's magnitude.
Correct answer
Also accepted:
Why
r=alpha+-beta*i; read off beta directly.
From Ordinary Differential Equations
Why does the real coefficients of the ODE let us split a complex solution into two real ones?
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The complex solution is genuinely used -- its real and imaginary parts are extracted, not discarded.
This applies specifically to the case where the imaginary part is nonzero (genuine complex roots).
Real coefficients are exactly what makes the real and imaginary parts separately satisfy the equation.
Why
Because the equation's coefficients are real, substituting a complex solution and separating real/imaginary parts gives two independent real equations, each automatically satisfied.
Think about substituting a complex-valued y into a real-coefficient equation and separating real/imaginary parts.
From Ordinary Differential Equations
For roots r=-2+-5i, write the exponential envelope factor e^(alpha*x) as an expression in x.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
\alpha=-2, so the envelope is e^{-2x}.
Correct answer
Also accepted:
Why
alpha is the real part of the roots.