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Every non-zero number has exactly $n$ complex $n$th roots, equally spaced round a circle as the corners of a regular polygon, because dividing an argument that is fixed only up to whole turns leaves $n$ answers rather than one.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to find every complex nth root of a given number, say why there are exactly n of them, plot them as the corners of a regular polygon, recognise a primitive root of unity and find its order, and use the fact that the nth roots of unity sum to zero.
De Moivre's theorem, which says that raising to a power multiplies the argument. Running it backwards is the whole of this lesson: if a power multiplies the argument, a root divides it — and since the argument was never a single number to begin with, there is more than one way to do the dividing.
An nth root of $w$ is any $z$ with $z^{n} = w$. The nth roots of unity are the nth roots of $1$; a root of unity is primitive of order $n$ when $n$ is the smallest positive power that returns it to $1$. The roots of a non-zero number are the corners of a regular polygon inscribed in a circle about the origin.
The formula. Let $w = re^{i\theta}$ be non-zero. Its $n$th roots are
$$z_j = r^{1/n}\,e^{i(\theta + 2\pi j)/n}, \qquad j = 0, 1, \ldots, n - 1,$$
where $r^{1/n}$ is the ordinary positive real root of the positive real number $r$. There are exactly $n$ of them: $j = n$ adds a whole turn to the argument of $z_0$ and gives the same point back.
Why there are $n$ and not one. The argument of $w$ is not a number but a number modulo whole turns. Dividing by $n$ turns an ambiguity of $2\pi$ into an ambiguity of $2\pi/n$, and $n$ steps of that size fit inside one turn. This is the first place the ambiguity of the argument produces something real rather than something to be tidied away, and it is the pattern for the logarithm two lessons from now.
The picture. All $n$ roots have the same modulus, so they lie on one circle; their arguments are equally spaced; so they are the corners of a regular $n$-gon. Two consequences fall out of the picture immediately: the roots of unity sum to zero for $n \ge 2$, because a regular polygon centred at the origin is balanced; and the set of $n$th roots of unity is closed under multiplication, because adding two of the angles gives another of them.
Roots of unity generate all the others. If $z_0$ is one $n$th root of $w$, every other is $z_0\zeta$ with $\zeta$ an $n$th root of unity. So the whole problem reduces to the roots of $1$, which is why they get a name.
Another way: picture
The cube roots of $8$ are the corners of an equilateral triangle on the circle of radius $2$: one at $2$ on the real axis, the others a third of a turn each way. A real number does not have one cube root and two impostors; it has three roots, and exactly one of them happens to be real. Widening the number system did not add exceptions, it removed them.
Another way: steps
This looks like a lesson about extracting roots, and it is in fact the first appearance of a structure that returns in almost every later unit.
They are the symmetries of a polygon. Multiplying by a primitive $n$th root of unity rotates the plane by one $n$th of a turn, and doing it $n$ times is the identity. The $n$th roots of unity, under multiplication, are the rotation group of the regular $n$-gon.
They sum to zero, and that is a cancellation theorem. The same cancellation is what makes the integral of $z^{k}$ round a circle vanish for every $k$ except $-1$, which is the single computation the residue theorem rests on. When you meet that fact in unit 3, it will be this polygon again, with infinitely many corners.
They make hard sums easy. Averaging a function over the $n$th roots of unity picks out every $n$th coefficient of its power series and kills the rest, because the other coefficients are multiplied by sums of roots of unity and cancel. That trick is the finite ancestor of the Fourier transform, and it is the reason a first course in complex analysis is also, quietly, a first course in harmonic analysis.
Over the positive reals, the square root means the positive one, and that convention is so well worn that it is invisible. Over the complex numbers it cannot be kept: there is no way to choose one square root of every complex number continuously, and a choice that is not continuous is of no use in a subject about limits.
So three habits have to go.
Writing $\sqrt{z}$ as though it named a number. It names two numbers, and which one you mean has to be said. The same goes for $z^{1/n}$ and, worse, for $z^{c}$ with $c$ not a whole number at all.
Expecting $\sqrt{zw} = \sqrt{z}\sqrt{w}$. With $z = w = -1$ the left side is a square root of $1$ and the right side is $i \times i = -1$. The rule is not false so much as meaningless until a branch has been chosen, and it can fail even then.
Solving $z^{n} = w$ and stopping at one answer. There are $n$, always, for $w \ne 0$. A polynomial equation of degree $n$ over the complex numbers has $n$ roots counted properly, and this is the easiest case of that theorem.
Solve $z^{3} = -8$. In polar form $-8$ has modulus $8$ and argument $180^\circ$ plus any whole number of turns.
Write the ambiguity in before dividing.
The modulus of each root is $2$; the arguments are $(180 + 360j)/3 = 60^\circ, 180^\circ, 300^\circ$.
Divide argument and whole turns alike by three.
So the roots are $1 + i\sqrt{3}$, $-2$ and $1 - i\sqrt{3}$: an equilateral triangle, of which only the middle one is real.
Three roots, one polygon.
Among the sixth roots of unity, $e^{i\,60^\circ}$ returns to $1$ only after six steps: it is primitive of order six.
Six steps of a sixth of a turn.
But $e^{i\,120^\circ}$ is also a sixth root of unity, and it returns after three steps: it is primitive of order three, and it is a cube root of unity that happens to be a sixth root too.
Every divisor of six contributes its own roots.
$16$ has modulus $16$ and argument $0$, so every root has modulus $16^{1/4} = 2$ and argument $(0 + 360j)/4$.
Modulus and argument separately.
The arguments are $0^\circ, 90^\circ, 180^\circ, 270^\circ$: the roots are $2$, $2i$, $-2$ and $-2i$.
Plot all four complex fourth roots of $1296$, reading the first coordinate as the real part and the second as the imaginary part.
Plot your answer on the grid:
How many different complex $4$th roots does a non-zero number have?
Answer:
Consider the $4$th roots of $1$. Fill in the table, giving every angle in degrees.
| Value | |
|---|---|
| How many roots there are | |
| The modulus of each root | |
| The angle in degrees from one root to the next | |
| The argument in degrees of the second root anticlockwise from 1 | |
| The sum of all the roots |
Put the steps of finding every complex $5$th root of a given number in order.
Number the steps in order (write the number in the box):
$w = e^{i\pi/7}$. What is the smallest positive integer $n$ with $w^{n} = 1$?
Answer:
Add together all $4$ of the complex $4$th roots of $1$. What do you get?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Plot all four complex fourth roots of $81$, reading the first coordinate as the real part and the second as the imaginary part.
Plot your answer on the grid:
You can extract all n roots of a number rather than one, and you can say where the extra roots come from. Say in your own words why the roots of unity add up to zero. Next: the exponential and the logarithm, where the same ambiguity produces infinitely many values instead of n.
9. Your turn: the four fourth roots of $16$, step 2