3.5.11 · D5Complex Numbers
Question bank — nth roots of complex numbers — finding all n roots
Before you start, one anchor to keep in your head:
Recall The whole engine in one sentence
One number has many equivalent angles () because a full turn returns you to the same spot; dividing those "same" angles by spreads them apart into exactly evenly-spaced roots. Every trap on this page is a failure to remember this.
The figure above shows the geometry every item below relies on: same radius , angles stepping by , vertices of a regular -gon.
True or false — justify
A nonzero complex number always has exactly distinct -th roots.
True — adding a full turn lands the arrow on the same point ( and have period ), so can be written as for any integer ; feeding these into for gives different angles, and repeats a full turn later, so the count stops at .
The -th roots of a real number are all real.
False — e.g. has one real root () and two complex roots (). Being real just means , but the roots still fan out by around the circle.
All of the roots have the same modulus.
True — from (equating moduli in ), every root has modulus , the unique non-negative real -th root of ; only the angle changes with , so they all sit on one circle of radius .
The roots are equally spaced around their circle.
True — consecutive angles differ by , a fixed step, so the roots are the vertices of a regular -gon.
Taking versus gives different root sets.
False — the roots repeat with period in ; gives the same point as . Any consecutive integers produce the identical set of roots.
The sum of all -th roots of unity is for every .
False — it is only for . For the single root is , and . The cancellation needs at least two symmetric vectors.
has exactly distinct -th roots.
False — forces only; the " distinct roots" theorem requires because collapses every angle to the same point.
If is one -th root of , then is another, where .
True — multiplying by (a root of unity) keeps the modulus and rotates the angle by exactly , landing on the next root. All roots are .
The square roots of a complex number are always negatives of each other.
True — with the two angles differ by , i.e. a rotation, which is exactly multiplication by , so .
The principal ("calculator") root is the only correct answer to .
False — the principal root is one chosen value by convention; the equation genuinely has solutions over . Reporting one is incomplete unless the question asks only for the principal value.
Spot the error
"For I wrote , then only, so the root is — done."
The error is dropping ; that yields just ONE of four roots. Correct angles are for : .
", so and I take for the angle."
You root the modulus, not the angle — the angle is divided, not rooted. It's , not .
" has modulus , so ."
Modulus is always non-negative: , and . A radius cannot be negative; the "minus" belongs to the argument , not to .
"I found the roots of by adding but dividing by ."
Degrees and radians are mixed. If you write and add , divide by in degrees (); never combine with a divisor.
"The roots of unity form a GP , so their sum is , which I can't evaluate."
You can: , so the numerator is , giving sum (for , where so the denominator is nonzero).
", so and I divide by ."
Using is fine because angles are mod — but be consistent: for you'd get , which just corresponds to a different (still valid) starting . The set of roots is unchanged; the error is thinking you get a "wrong" root — you don't, only a relabelled one.
"Since has roots that are negatives, once I find I don't need ."
Correct that they're negatives, but is still a distinct root you must report. "They're related" doesn't mean "there's only one."
Why questions
Why do we write in polar form before taking roots?
Because multiplication of complex numbers multiplies moduli and adds arguments; polar form makes powering (De Moivre) and its reverse — root-taking — into simple arithmetic on and .
Why does adding before dividing by matter, but adding it after would not help?
Divide-then-add would give , which is the same direction repeated (full turns) — one root. Add-then-divide gives , where the steps are fractions of a turn, producing genuinely different directions.
Why is De Moivre's Theorem the right tool here rather than expanding algebraically?
Expanding by the binomial theorem gives a messy pair of real polynomial equations; De Moivre collapses to , turning the problem into two clean equations and .
Why do the -th roots of unity sum to zero geometrically?
They are the vertices of a regular -gon centred at the origin; the position vectors point symmetrically in all directions and cancel in balanced opposite pairs (or triples etc.), leaving the centroid at .
Why must consecutive roots be exactly apart, no more, no less?
Because incrementing by adds to the angle formula ; this is forced by the single per full turn shared across roots.
Why can a real, positive number like have non-real cube roots?
The Fundamental Theorem of Algebra guarantees has exactly roots in ; realness of only fixes , but the spacing still rotates two roots off the real axis.
Why is always taken as the positive real root?
Because is a length, and is also a length; the equation with both sides non-negative has a unique non-negative solution, which is the ordinary positive -th root.
Edge cases
What are the -th roots of ?
Only , with multiplicity . The " distinct roots" result fails here because places every angle at the same origin point.
For , how many roots does have and where are they?
Exactly one: itself. A "-gon" is a single point, and there are no full turns to spread, so the trap of "always distinct spread-out roots" degenerates cleanly.
If is a negative real number and is even, is any root real?
No — a real number raised to an even power is , so it can never equal a negative . All roots must be genuinely complex, symmetric across both axes.
If is a positive real and is odd, how many real roots are there?
Exactly one (the positive real ); the other roots come in complex-conjugate pairs off the real axis, since odd leaves no room for a second real root.
What happens to the root circle's radius as ?
The radius , so all roots collapse toward the origin — the regular -gon shrinks to a point, matching the degenerate case.
If two of the listed roots ever coincided, what would that imply?
It would require two angles and to differ by a multiple of , i.e. — impossible for distinct in , which is why they're guaranteed distinct.
Does choosing a different valid value of (say instead of ) change the root set?
No — it merely shifts which integer produces which root; the complete set of points on the circle is identical, because both descriptions cover the same angles mod .