Review / Part I — Explicit & Qualitative Methods / Ch 5

Complex Numbers Refresher

Ch 5 — Part I — Explicit & Qualitative Methods

Complex numbers \(z = a + bi\) with \(i^2 = -1\) are essential for second-order linear ODEs and systems. Two views coexist: rectangular \((a, b)\) and polar \((r, \theta)\) with \(r = |z|,\; \theta = \arg z\).

Euler's formula

$$e^{i\theta} = \cos\theta + i\sin\theta.$$ Multiplication in polar form becomes modulus multiplication + argument addition: \((r_1 e^{i\theta_1})(r_2 e^{i\theta_2}) = r_1 r_2 e^{i(\theta_1+\theta_2)}\).

The complex exponential

\(e^{(a+bi)t} = e^{at}(\cos bt + i \sin bt)\). This is why real solutions of linear ODEs with complex roots involve \(e^{at}\cos bt\) and \(e^{at}\sin bt\).

Conjugates and magnitude

\(\bar z = a - bi\), \(|z|^2 = z\bar z = a^2 + b^2\). Real-valued ODEs with complex roots always produce roots in conjugate pairs.

Practice Problems

Problem 1medium
Write \((1 + i)^8\) in rectangular form.
Hint
Convert to polar first: \(1+i = \sqrt 2 e^{i\pi/4}\).
Solution

\((1+i)^8 = (\sqrt 2)^8 e^{i 2\pi} = 16 \cdot 1 = 16\).

Answer: 16.

Problem 2easy
Simplify \(e^{i\pi} + 1\).
Hint
Euler's identity.
Solution

\(e^{i\pi} = \cos\pi + i\sin\pi = -1\), so \(e^{i\pi} + 1 = 0\).

Answer: 0.

Problem 3medium
Find all roots of \(z^3 = 8\).
Hint
Polar form: \(8 = 8 e^{i2\pi k}\), so \(z = 2 e^{i 2\pi k/3}\) for \(k = 0,1,2\).
Solution

\(z_0 = 2,\; z_1 = 2(\cos 120°+i\sin 120°) = -1 + i\sqrt 3,\; z_2 = -1 - i\sqrt 3\).

Answer: \(2,\; -1\pm i\sqrt 3\).

Problem 4easy
If \(\lambda = -2 + 3i\), what is \(\operatorname{Re}(e^{\lambda t})\)?
Hint
\(e^{\lambda t} = e^{-2t}(\cos 3t + i \sin 3t)\).
Solution

Taking real part: \(e^{-2t}\cos 3t\).

Answer: \(e^{-2t}\cos(3t)\).