Complex Numbers: Solving Equations (with Example) [dark version]

TL;DR

Convert the right-hand side a into exponential form a = R·e^{iθ} to make root extraction manageable.

Briefing Cornell Notes

Briefing

Solving equations like z^n = a in the complex numbers comes down to taking roots in polar (exponential) form—and the exponent n dictates both how many solutions exist and how they’re spaced. Unlike the real line, where x^2 = 4 has just two ordered solutions (±2), the complex plane turns “distance from zero” into circles and “sign” into geometry. For any nonzero complex number a, every solution z must lie on the circle centered at the origin with radius |a|^(1/n), and the n solutions are evenly distributed around that circle.

The transcript builds the method using a concrete example: z^3 = 2 + 2i. First, 2 + 2i is converted to polar/exponential form. Its modulus is computed as R = √(2^2 + 2^2) = √8, and its argument is found from the right-triangle geometry: the point (2,2) lies at a 45° angle, i.e., π/4. In exponential form, 2 + 2i becomes √8 · e^{iπ/4}. The key algebra move is rewriting roots using the fact that angles differing by full turns represent the same complex number: e^{i(θ + 2πk)} for any integer k. That freedom produces multiple distinct cube roots.

Next comes the counting and spacing rule. For z^3 = a (with a ≠ 0), the solutions must sit on the circle of radius |a|^(1/3) and split that circle into three equal arcs. Geometrically, that means the arguments differ by 2π/3 (120°). Starting from one cube root, rotating by 120° yields the next solution, and rotating again yields the third; after three such rotations, the process returns to the starting point.

The transcript then performs the formal calculation using the exponent 1/3. The general cube-root solutions take the form z_k = (|a|)^{1/3} · e^{i( (arg(a) + 2πk)/3 )}, with k = 0,1,2. For the example, this produces three angles: π/12, π/12 + 2π/3, and π/12 + 4π/3, all sharing the same modulus (√8)^{1/3} = √2. The final result is three points on the complex plane that are uniformly spaced around the circle, matching the 120° separation predicted by the exponent.

The same logic generalizes. For z^4 = a (with a ≠ 0), the solutions again lie on a circle, but now the exponent 4 splits the circle into four equal parts, giving four solutions separated by 90°. The method also naturally recovers real solutions when they occur—when one of the evenly spaced points lands on the real axis. The takeaway is practical: convert the right-hand side to exponential form, use the “add 2πk” angle rule, divide the argument by n, and generate exactly n roots arranged evenly on a circle.

Cornell Notes

For equations of the form z^n = a (with a a fixed complex number), every solution z lies on a circle centered at 0 with radius |a|^(1/n). Writing a in exponential (polar) form, a = R·e^{iθ}, makes taking roots straightforward: the n roots come from dividing the angle by n while allowing θ to shift by full turns θ + 2πk. The exponent n determines both the number of solutions and their spacing: the solutions are evenly distributed around the circle, separated by 2π/n radians. For the example z^3 = 2 + 2i, the modulus is √8 and the argument is π/4, giving three cube roots with arguments π/12, π/12 + 2π/3, and π/12 + 4π/3 and the same modulus √2.

Why does solving z^n = a in the complex plane require polar/exponential form instead of Cartesian coordinates?

Because powers and roots act cleanly on the argument (angle) and modulus (radius) when the number is written as a = R·e^{iθ}. In exponential form, raising to a power multiplies angles: (R·e^{iθ})^n = R^n·e^{inθ}. That structure makes extracting roots a matter of dividing the angle by n, while the modulus becomes R^(1/n). In Cartesian form, the same operation would mix real and imaginary parts in a harder-to-control way.

How does the “add 2πk” rule generate multiple distinct roots?

Angles that differ by full turns represent the same complex number: e^{i(θ+2πk)} = e^{iθ} for any integer k. When taking n-th roots, dividing (θ+2πk) by n produces different arguments for different k values. For z^3 = a, choosing k = 0, 1, 2 yields three distinct cube roots; larger k values repeat the same set because the circle closes after n steps.

What determines the number of solutions to z^n = a (when a ≠ 0)?

The exponent n. All solutions must lie on the circle |z| = |a|^(1/n), and the arguments split that circle into n equal arcs. That means there are exactly n distinct solutions, evenly spaced by an angle of 2π/n radians.

In the example z^3 = 2 + 2i, how are the modulus and argument found?

The modulus is R = √(2^2 + 2^2) = √8. The argument comes from the point (2,2) in the complex plane: the right triangle has equal legs, so the angle is 45°, i.e., θ = π/4. Thus 2 + 2i = √8·e^{iπ/4}.

How are the three cube roots of z^3 = 2 + 2i constructed from one solution?

Once one cube root is known, the others come from rotating by 120° (2π/3). The transcript’s formal result matches this geometry: the arguments are π/12, π/12 + 2π/3, and π/12 + 4π/3, all with the same modulus (√8)^(1/3) = √2. Each rotation lands on the next solution point on the circle.

What changes when moving from z^3 = a to z^4 = a?

The circle still sets the modulus, but the exponent changes the spacing and count. For z^4 = a (with a ≠ 0), there are exactly four solutions, evenly spaced by 2π/4 = π/2 = 90° around the circle. The same exponential-root procedure yields those four arguments.

Review Questions

For z^n = a with a ≠ 0, what are the modulus and argument rules for generating all solutions?
Why do only k = 0, 1, …, n−1 produce distinct solutions, while other k values repeat them?
In z^4 = 3 + 2i, what angular separation between solutions should be expected, and why?

Key Points

1
Convert the right-hand side a into exponential form a = R·e^{iθ} to make root extraction manageable.
2
For z^n = a (a ≠ 0), every solution satisfies |z| = |a|^(1/n), so all roots lie on a circle centered at 0.
3
Use the angle equivalence θ + 2πk to generate multiple roots; dividing by n turns those angle shifts into distinct solutions.
4
The exponent n determines the number of solutions: exactly n distinct roots for a ≠ 0.
5
The solutions are evenly spaced around the circle with argument separation 2π/n radians.
6
For z^3 = 2 + 2i, the modulus is √8 and the argument is π/4, giving three roots with modulus √2 and arguments π/12 + 2πk/3 for k = 0,1,2.
7
Real solutions appear automatically when one of the evenly spaced complex roots lands on the real axis.

Highlights

In the complex plane, x^2 = 4 becomes a circle of radius 2, and the “two solutions” correspond to two points on that circle rather than ± on a line.

Writing a = R·e^{iθ} turns taking n-th roots into dividing the argument by n and adjusting by 2πk.

For z^3 = 2 + 2i, the cube roots are three points equally spaced by 120° with the same modulus √2.

For z^4 = a (a ≠ 0), the four roots are separated by 90° on the circle |z| = |a|^(1/4).

Topics

Complex Roots
Polar Form
De Moivre’s Theorem
Geometric Spacing
Solving Equations