Complex Numbers and Oscillatory Modes

Use complex arithmetic to interpret conjugate eigenvalues of a real matrix.

Builds on Diagonalization and Discrete Dynamics

The bigger question: Which directions keep their identity as a system evolves?

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Extend the number system

Define i2=−1i^2=-1. A complex number is z=a+ibz=a+ib, with conjugate zˉ=a−ib\bar z=a-ib and magnitude ∣z∣=a2+b2|z|=\sqrt{a^2+b^2}. Add components and multiply using i2=−1i^2=-1. To divide by nonzero zz, multiply numerator and denominator by its conjugate.

Polar form is z=r(cos⁡θ+isin⁡θ)=reiθz=r(\cos\theta+i\sin\theta)=re^{i\theta}. Multiplication multiplies magnitudes and adds angles, so zk=rkeikθz^k=r^ke^{ik\theta}. This identity follows from the angle-addition formulas and explains rotation combined with scaling.

Visual guide

VISUAL GUIDEComplex eigenvalues can describe real rotation
The real matrix [[0, −1], [1, 0]] rotates vectors by a quarter turn. The four arrows show repeated application to (1, 0). No nonzero real direction is preserved, consistent with eigenvalues ±i.-1.7-1.7-0.85-0.85000.850.851.71.7xy
  • Unit circle
The real matrix [[0, −1], [1, 0]] rotates vectors by a quarter turn. The four arrows show repeated application to (1, 0). No nonzero real direction is preserved, consistent with eigenvalues ±i.

Worked example: arithmetic and roots

(1+2i)(3−i)=5+5i(1+2i)(3-i)=5+5i. Also 1/(1+i)=(1−i)/21/(1+i)=(1-i)/2. The equation λ2+1=0\lambda^2+1=0 has roots ii and −i-i; there is no real solution because a real square cannot be negative.

For real matrices, nonreal eigenvalues occur in conjugate pairs. Conjugating Av=λvAv=\lambda v gives Avˉ=λˉvˉA\bar v=\bar\lambda\bar v since AA itself is real.

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

What is i²?

Hint 1 · Find a starting point

The imaginary unit is defined as a square root of −1.

Hint 2 · Take the next step

Squaring i returns that defining value.

Show the reasoning

Answer: −1

i²=−1, which lets real two-dimensional rotations be represented with complex numbers.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: a real rotation

Let R=(0−110)R=\begin{pmatrix}0&-1\\1&0\end{pmatrix}. Its eigenvalues are ±i\pm i. For λ=i\lambda=i, a complex eigenvector is (1,−i)T(1,-i)^T. This does not mean the original real plane contains a fixed real direction: a quarter-turn changes every nonzero real direction.

Nevertheless real motion is easy to interpret. Identifying (x,y)(x,y) with z=x+iyz=x+iy, applying RR multiplies zz by ii, a 90∘90^\circ counterclockwise rotation. For A=ρRA=\rho R, repeated application multiplies magnitudes by ρk\rho^k. With 0<ρ<10<\rho<1, states spiral toward zero; with ρ>1\rho>1, they spiral outward.

Complex vector inner products use conjugate transpose u∗vu^*v, rather than ordinary transpose, to ensure v∗vv^*v is real and nonnegative. Our earlier projection formulas were stated for real vectors.

Practice

  1. Find ∣3−4i∣|3-4i| and its conjugate.
  2. Compute i4i^4.
  3. Describe repeated action of 0.8R0.8R on a nonzero real vector.
Show worked solutions
  1. Magnitude 55; conjugate 3+4i3+4i.
  2. i2=−1i^2=-1, so i4=1i^4=1.
  3. Rotate by 90∘90^\circ each step and shrink magnitude by 0.80.8; states tend to zero.
MAKE IT YOURS

Pause before the next idea.

Can you explain how the visualization connects to this lesson’s goal? If a step still feels uncertain, put this lesson on your review list and try the check again another day.

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