Solving and Plotting Complex nth Roots

Hello! Last lesson, you used De Moivre’s theorem to find a power by multiplying an angle by the exponent. This lesson reverses the question: if $z^n=w$, what could $z$ have been? The challenge is that angles differing by a full revolution describe the same $w$, but can produce different answers when divided by $n$.

You will derive a method that finds every solution, use it for a non-real target, and place the solutions accurately on an Argand diagram. In an exam, finding one root is rarely enough: you need to show why your list is complete.

Why one angle gives several roots

Let $n$ be a positive integer and suppose first that $w\ne 0$. Write

where $r,\rho >0$. By De Moivre’s theorem,

Thus $z^n=w$ requires ${\rho }^n=r$. For the angles, however, we need not have $n\varphi =\theta$ exactly: angles that differ by a whole number of revolutions represent the same complex number. Consequently,

Here $r^{1/n}$ means the positive real $n$th root of $r$. Substituting gives the complete formula:

Watch the short derivation in De Moivre’s Formula | How to Find the nth Roots of A Complex Number from Let’s Get Complex. The video calls the number being rooted $z$ and a possible root $w$; our notation reverses those roles, but the equation and reasoning are the same.

Why stop there? Increasing $k$ by $n$ increases a root’s angle by $2\pi$, returning to the same point. Conversely, two values among $0,1,\dots ,n-1$ differ by less than $n$, so their angles cannot differ by a whole $2\pi$. The formula therefore gives exactly $n$ distinct roots when $w\ne 0$.

This is also why taking just $r^{1/n}\mathrm{cis}(\theta /n)$ is insufficient. It gives the $k=0$ root, not the complete solution set.

A worked equation: $z^3=8i$

The target $8i$ has modulus $8$ and argument $\pi /2$, so

Every cube root has modulus $8^{1/3}=2$. Its angle comes from

Keeping $k$ visible makes it easy to account for all three answers:

Hence

If the question requests principal arguments, replace $3\pi /2$ with $-\pi /2$; these are two arguments of the same root. Do not alter an angle merely because it exceeds $\pi$ unless the requested range calls for it.

A quick check combines algebra and geometry. For $\sqrt{3}+i=2\mathrm{cis}(\pi /6)$, cubing gives $8\mathrm{cis}(\pi /2)=8i$. The other two angles exceed $\pi /6$ by $2\pi /3$ and $4\pi /3$; cubing adds respectively $2\pi$ and $4\pi$ to the resulting angle, so they also produce $8i$. As a direct sign check, $(-2i{)}^3=8i$.

From solutions to an Argand diagram

To draw the three roots of $8i$, first sketch and label the real and imaginary axes. Draw a circle of radius $2$ centred at the origin: every root must lie on it. Then plot and label

These coordinates correspond, respectively, to angles $\pi /6$, $5\pi /6$, and $3\pi /2$. Each successive point is $2\pi /3$ around the circle. Joining them would make an equilateral triangle, though the equation asks you to represent the points, not necessarily to draw its sides.

The following image shows a related triangle: the cube roots of $1$, rather than the cube roots of $8i$. Compare its equal angular spacing with ours. Its points are on the unit circle and begin at angle $0$; ours are on a circle of radius $2$ and begin at $\pi /6$.

In general, the roots of $z^n=w$, for $w\ne 0$, all lie on the circle of radius $r^{1/n}$. Their angles differ successively by $2\pi /n$. For $n\ge 3$, they are the vertices of a regular $n$-sided polygon centred at the origin; for $n=2$, the two points are diametrically opposite.

Watch a fourth-root example to see the same method applied to a target in the second quadrant. In the video, the target is $-8+8\sqrt{3}i$. Notice why its argument is $2\pi /3$, and follow the plotting of all four roots.

Making the method reliable under exam conditions

A compact solution should expose four things: the target’s correct polar form, the general expression for $z_k$, the specified values of $k$, and a diagram whose radius and labels agree with the algebra. If Cartesian answers are requested, evaluate each cosine and sine rather than leaving only a $\mathrm{cis}$ formula.

Pay particular attention to the target’s quadrant. A negative real target has argument $\pi$, not $0$; a negative imaginary target can have argument $-\pi /2$. Either of two arguments differing by $2\pi$ will work, provided you still include $2k\pi$ before dividing by $n$. The choice may change the order in which you list the roots, but not the set of answers.

There is one boundary case. If $w=0$, the equation $z^n=0$ has only $z=0$. Zero has no argument, so the polar root formula is not used. If $n=1$, the sole solution is simply $z=w$.

Key takeaways

For nonzero $w=r\mathrm{cis}\theta$, solving $z^n=w$ means taking the positive real $n$th root of the modulus while allowing for all whole revolutions in the angle. The values $k=0,\dots ,n-1$ give every distinct solution. On an Argand diagram, check two features at a glance: a common radius $r^{1/n}$ and equal spacing $2\pi /n$.

Next, you will examine the special case $w=1$ more closely. The image’s equally spaced points are the beginning of that story.