Complex roots and roots of unity

A connected application in three stages

Use zⱼ(t)=2eᵗ cis(t+2jπ/3), j=0,1,2, t≥0, with angles in radians. Interpret each complex number as a planar position in length units. First study t=0, then prove general features, then vary the radius and angular rate. The geometry and vector calculus connect separate areas of study; the extension lesson continues the same context

Stage 1: investigate the initial triangle, ungraded

For zⱼ(t)=2eᵗ cis(t+2jπ/3), j=0,1,2, find all three positions at t=0, identify their common polynomial, calculate side length and area, and draw a labelled Argand diagram. Explain why the vertices are distinct. Give exact answers and a geometric justification, not just a software picture

Model: compute and justify the specific case

At t=0 the vertices are 2, −1+i√3 and −1−i√3. Each cubes to 8, and their arguments differ by 2π/3, so these are all three distinct roots of z³−8. The side length is √(3²+(√3)²)=2√3. Taking the vertical side 2√3 and horizontal altitude 3 gives area 3√3. The equal-scale diagram checks the arrangement; the calculation establishes the exact area

The triangle has vertices 2, −1+i√3 and −1−i√3. What is its squared side length?

Which statement is justified by the displayed initial triangle?

Stage 2: generalise the configuration, ungraded

Let zⱼ(t)=2eᵗ cis(t+2jπ/3), j=0,1,2, for t≥0. Prove that the three vertices are precisely the roots of z³=8e³ᵗ cis(3t). Find their sum, side length and area as functions of t, and obtain the area rate. Explain why rotation does not change distances

Model: use a common complex multiplier

At t=0 the vertices are 2, −1+i√3 and −1−i√3, with area 3√3. The three initial roots are multiplied by eᵗ cis t. Multiplication by cis t preserves modulus and rotates every difference by t; multiplication by eᵗ scales every length by eᵗ. Cubing gives 8e³ᵗ cis(3t), with three distinct arguments modulo 2π. Their sum remains zero. Thus side length is 2√3eᵗ, area A(t)=3√3e²ᵗ and A′(t)=6√3e²ᵗ. A common nonzero multiplier preserves the number of distinct vertices

Given A(t)=3√3e²ᵗ, what is the constant ratio A′(t)/A(t)?

Stage 3: vary the assumptions, ungraded

Replace 2eᵗ cis(t+2jπ/3) by R eᵃᵗ cis(ωt+2jπ/3), where R>0, a and ω are real and t≥0. Derive area and area rate. Describe what happens for a=0 and for negative ω. Then replace the equally spaced arguments by 0, π/2 and π at t=0: identify which conclusions fail

Model: separate radial growth from rotation

Let zⱼ(t)=R eᵃᵗ cis(ωt+2jπ/3), j=0,1,2, where R>0 and a and ω are real. With equal spacing, A(t)=(3√3/4)R²e²ᵃᵗ and A′=2aA, independent of ω. If a=0, area is constant; negative ω reverses rotation. R>0 keeps the vertices distinct. The altered initial angles give a right triangle with vertices (R,0),(0,R),(−R,0), area R² and sum iR, so the equilateral area formula and zero-sum claim fail. A common multiplier still scales any triangle area by e²ᵃᵗ. State the changed geometry rather than silently retaining the old conclusions

For zⱼ(t)=R cis(ωt+2jπ/3), j=0,1,2, with R>0, increasing |ω| changes which quantity?

Written response

The three complete stages are ungraded