Roots of complex numbers

Textbook reading

The problem that opens the lesson

Find all cube roots of $8i$ .

Solution

Begin by identifying the mathematical object and the information that fixes it. Write the target in polar form, apply the root formula, list exactly $n$ distinct roots, and verify by raising one or all roots to the nth power. The relevant conditions are not optional bookkeeping: The zero number has one nth root, zero, rather than $n$ distinct roots. Following that structure gives $8i=8cis(\frac{pi}{2})$ ; roots have modulus $2$ and arguments $\frac{pi}{6}+\frac{2kpi}{3}$ for $k=0,1,2$ .

Why this works

The $2pi k$ term is essential because the same target number has infinitely many arguments before division by $n$ . The calculation and the representation should agree, so a graph, diagram, table, or substitution check should support the same conclusion.

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What this lesson is really about

The nth roots of a nonzero complex number are equally spaced around a circle.

If $w=R$ cis phi, then each root has modulus $R^{\frac{1}{n}}$ and argument $\frac{phi+2pi k}{n}$ for $k=0,...,n-1$ . Adding $n$ to $k$ repeats the first root.

The point is not merely to reproduce a formula. A learner should be able to identify the quantities or geometric objects involved, explain why the relationship has its stated form, and recognize when the same idea appears in a graph, table, diagram, or model.

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Why the relationship works

The $2pi k$ term is essential because the same target number has infinitely many arguments before division by $n$ .

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A reliable way to work

Write the target in polar form, apply the root formula, list exactly $n$ distinct roots, and verify by raising one or all roots to the nth power.

The zero number has one nth root, zero, rather than $n$ distinct roots.

After the symbolic work is complete, check the result. Depending on the lesson, this may mean substituting into an original equation, comparing coordinates, examining a graph, checking units, testing an interval, or confirming that every branch of a periodic solution has been included.

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What commonly goes wrong

A common error is using only the principal argument and finding one root.

The repair is to return to the definition and identify the first step where the invalid solution stops describing the original mathematical object. Later algebra cannot rescue a first step that changed the domain, orientation, branch, or meaning of the problem.

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Worked examples

Worked example 1

Find all cube roots of $8i$

Solution

Begin by identifying the mathematical object and the information that fixes it. Write the target in polar form, apply the root formula, list exactly $n$ distinct roots, and verify by raising one or all roots to the nth power. The relevant conditions are not optional bookkeeping: The zero number has one nth root, zero, rather than $n$ distinct roots. Following that structure gives $8i=8cis(\frac{pi}{2})$ ; roots have modulus $2$ and arguments $\frac{pi}{6}+\frac{2kpi}{3}$ for $k=0,1,2$ .

Why this works

The $2pi k$ term is essential because the same target number has infinitely many arguments before division by $n$ . The calculation and the representation should agree, so a graph, diagram, table, or substitution check should support the same conclusion.

Transfer example

Problem

Find fourth roots of $16$

Worked development

Write the target in polar form, apply the root formula, list exactly $n$ distinct roots, and verify by raising one or all roots to the nth power. In this example, the first useful move is to make the defining structure visible rather than to search for a memorized answer. If $w=R$ cis phi, then each root has modulus $R^{\frac{1}{n}}$ and argument $\frac{phi+2pi k}{n}$ for $k=0,...,n-1$ . Adding $n$ to $k$ repeats the first root. Then apply the conditions explicitly: The zero number has one nth root, zero, rather than $n$ distinct roots. Finish by checking the result in a second representation and explaining what the result means.

Interpretation

Complex roots complete polynomial solution sets and create regular polygons in the complex plane.

Reasoning example

Problem

Solve $z^5=-1$

Worked development

Write the target in polar form, apply the root formula, list exactly $n$ distinct roots, and verify by raising one or all roots to the nth power. In this example, the first useful move is to make the defining structure visible rather than to search for a memorized answer. If $w=R$ cis phi, then each root has modulus $R^{\frac{1}{n}}$ and argument $\frac{phi+2pi k}{n}$ for $k=0,...,n-1$ . Adding $n$ to $k$ repeats the first root. Then apply the conditions explicitly: The zero number has one nth root, zero, rather than $n$ distinct roots. Finish by checking the result in a second representation and explaining what the result means.

Interpretation

Complex roots complete polynomial solution sets and create regular polygons in the complex plane.

Worked example 4: quick check

How many distinct seventh roots does a nonzero complex number have?

Solution

Begin by identifying the mathematical object and the information that fixes it. Write the target in polar form, apply the root formula, list exactly $n$ distinct roots, and verify by raising one or all roots to the nth power. The relevant conditions are not optional bookkeeping: The zero number has one nth root, zero, rather than $n$ distinct roots. Following that structure gives Seven.

Why this works

The $2pi k$ term is essential because the same target number has infinitely many arguments before division by $n$ . The calculation and the representation should agree, so a graph, diagram, table, or substitution check should support the same conclusion.

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Application and interpretation

Complex roots complete polynomial solution sets and create regular polygons in the complex plane.

A contextual answer must include units, a meaningful domain, and the assumptions that make the model plausible. An exact mathematical relationship should not be diluted into a decimal unless a measurement or comparison requires it.

Practice