Pj14–Pj17 - Exponential form and nth roots

Writing a complex number in exponential form

How many different numbers have fourth power $-16$, and why do they add to zero? Dividing one angle by $4$ finds only one of them. The others come from remembering that angles can make whole extra turns.

There is a shorter way to write the unit complex number $\cos\theta+i\sin\theta$. For example, the point at angle $\pi/2$ is $i$, so $e^{i\pi/2}=i$. This is a complex exponential; it need not be a positive real number, unlike $e^x$ for real $x$.

The definition gives the memorable identity $e^{i\pi}+1=0$: the same line contains $e$, $i$, $\pi$, $1$ and $0$. Its explanation here is simply the point at angle $\pi$, which is $-1$. No mysterious cancellation is needed.

For conversion, find the modulus and quadrant before writing the exponential. The coefficient outside the exponential must be positive if it is to be the modulus.

Change only the sign of the imaginary part. The modulus stays the same, while the principal argument reflects in the real axis.

Writing $re^{i\theta}$ and writing $r(\cos\theta+i\sin\theta)$ describe the same number, but are different requested forms. Read the form in the question carefully.

Choose the principal argument again, this time for a steeper third-quadrant point.

Finding every nth root

To find the $n$-th roots of a number, solve $w^n=z$. If $z=re^{i\theta}$ and $w=\rho e^{i\phi}$, then $w^n=\rho^ne^{in\phi}$. Matching moduli gives $\rho^n=r$. Matching arguments allows every whole turn: $n\phi=\theta+2k\pi$.

For instance, the fourth roots of $-16=16e^{i\pi}$ have modulus $2$, with angles $(\pi+2k\pi)/4$. The extra turns must be included before dividing by $4$.

The booklet gives roots of unity, $w^n=1$, as a special case; the general non-zero right side above must be constructed. Write the original modulus and argument first, especially under detailed reasoning.

Change only the sign of the given imaginary number. The modulus remains $8$, but the starting argument changes from $\pi/2$ to $-\pi/2$.

For a non-zero right side there are $n$ distinct roots, not one root with $n$ different names. The next part explains why the list stops exactly where it does.

The opening puzzle changes the degree to $4$. Keep the whole-turn term inside the division.

Count distinct roots after dividing the full angle below.

Why the roots form a regular polygon

The fourth roots just found all lie at distance $2$ from the origin. Their arguments increase by $\pi/2$, so they are the vertices of a square centred at the origin.

[DIAGRAM: asset_name: The four fourth roots of minus sixteen; asset_slug: h645_b010_fourth_roots; caption: The fourth roots of minus sixteen form a square on the circle of radius 2; description: An equal-scale Argand diagram shows a circle of radius 2 and the four vertices at angles pi over 4, 3 pi over 4, 5 pi over 4 and 7 pi over 4. Cartesian labels show square root 2 plus square root 2 i and its three sign variants. A central arc marks angular spacing pi over 2. A side panel states modulus 2, spacing 2 pi over 4, and fourth powers all equal minus sixteen.]

In the general formula, increasing $k$ by $1$ increases the argument by $2\pi/n$, while leaving the modulus $r^{1/n}$ unchanged. Thus for $n\ge3$ the roots form a regular $n$-gon centred at the origin. For $n=2$, they are opposite ends of a diameter; for $n=1$, there is just the original number.

Why are there exactly $n$ distinct values? If $0\le j<k\le n-1$, the angle difference $2(k-j)\pi/n$ lies strictly between $0$ and $2\pi$, so the points differ. Increasing $k$ by $n$ adds a full turn and returns to the same point. Any integer $k$ therefore reproduces one of the listed values.

The non-zero condition matters. The equation $w^n=0$ has only the value $w=0$; it is a repeated root algebraically, with no argument or polygon. The equal-spacing result is not in the formula booklet: learn it from the root formula.

Use this distinction between starting angle and spacing to diagnose a tempting mistake.

Why the roots add to zero

The four roots of $-16$ have Cartesian forms $\sqrt2+\sqrt2\,i$, $-\sqrt2+\sqrt2\,i$, $-\sqrt2-\sqrt2\,i$ and $\sqrt2-\sqrt2\,i$. Adding them cancels both parts. With an odd number of roots, opposite pairs are unavailable, so we need an argument that works for every $n\ge2$.

Let $a=r^{1/n}e^{i\theta/n}$ be one root and let $\omega=e^{2\pi i/n}$. The roots are

This is a finite geometric series, with $\omega^n=1$ and $\omega\ne1$ when $n\ge2$. Hence

For instance, the cube roots of $8i$ give $(\sqrt3+i)+(-\sqrt3+i)-2i=0$, despite having no pair of opposite roots.

There is also an algebraic explanation. In $w^n-z=0$, the coefficient of $w^{n-1}$ is zero for $n\ge2$, so the sum of roots is zero. The geometric-series proof makes the cancellation explicit without needing each root in Cartesian form.

On the diagram, the zero sum says that the position vectors balance at the origin. A regular triangle balances just as a square does; it does not need opposite vertices.

The opening puzzle is now complete: four distinct roots sit at the square's vertices, and their position vectors sum to zero. Explain why that conclusion extends to an odd number of roots as well.

Quick Recap

• $e^{i\theta}=\cos\theta+i\sin\theta$, so $re^{i\theta}$ records a modulus and an angle compactly.
• For $w^n=re^{i\theta}$, take modulus $r^{1/n}$ and arguments $(\theta+2k\pi)/n$, for $k=0,\ldots,n-1$.
• All $n$ roots are distinct and equally spaced; for $n\ge3$ they form a regular polygon centred at the origin.
• For $n\ge2$, a geometric series proves that all roots sum to zero. The number $0$ and the case $n=1$ need separate treatment.

The fourth roots of $-16$ are $\pm\sqrt2\pm\sqrt2\,i$, with the signs chosen independently, and their sum is $0$.