Complex depth: rotations, root geometry and parameter cases

In the complete ordered real field, for $b>0$ and integer $m\ge1$, let $S=\{t\ge0:t^m\le b\}$. It contains $\min\{1,b\}>0$ and is bounded above by $\max\{1,b\}$, so $r=\sup S>0$ exists. Finite factorization gives, for nonnegative $t,r\le M$,

$$t^m-r^m=(t-r)\sum_{j=0}^{m-1}t^{m-1-j}r^j,\qquad |t^m-r^m|\le mM^{m-1}|t-r|.$$

Take $M=r+1$. If $r^m<b$, choose $0<h<1$ so $mM^{m-1}h<b-r^m$; then $(r+h)^m<b$, contradicting the upper bound $r$. If $r^m>b$, choose $0<h<\min\{1,r\}$ with $mM^{m-1}h<r^m-b$. Then $(r-h)^m>b$, and monotonicity of integer powers on nonnegative numbers makes $r-h$ an upper bound of $S$, contradicting its supremum. Hence $r^m=b$. At $b=0$ the unique nonnegative root is zero. The same factorization shows $t^m>s^m$ whenever $t>s\ge0$, so this positive root is unique.

Learning goals and preparation

The expressions $a+bi$ and $r(\cos\theta+i\sin\theta)$ are two descriptions of the same point, but they emphasize different operations. This chapter builds a route from Cartesian arithmetic to rotations, roots and loci, and shows how a parameter changes an equation's geometry. Prerequisites are real algebra, radian trigonometry, the conjugate/modulus formulas and the finite geometric-sum identity.

1. Multiplication and division are geometric maps

For fixed $w=u+iv$, multiplication sends $z=x+iy$ to

$$wz=(ux-vy)+i(vx+uy).$$

In real coordinates its matrix is $\begin{pmatrix}u&-v\\v&u\end{pmatrix}$. Its determinant is $u^2+v^2=|w|^2$. If $w\ne0$, the map is invertible, with inverse multiplication by $\overline w/|w|^2$; if $w=0$, it collapses every point to the origin.

The geometric addition identities for sine and cosine are

$$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta,\qquad \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta.$$

Proof

Take as a geometric fact that a rotation about the origin preserves distances and orientation and is linear in coordinate components. Let $R_\alpha$ rotate counterclockwise through $\alpha$. On the unit circle, it sends the first and second coordinate basis vectors to

$$R_\alpha e_1=(\cos\alpha,\sin\alpha)^T,\qquad R_\alpha e_2=(-\sin\alpha,\cos\alpha)^T.$$

By linearity, its action on the point at angle $\beta$ is

$$R_\alpha(\cos\beta,\sin\beta)^T =(\cos\alpha\cos\beta-\sin\alpha\sin\beta,\ \sin\alpha\cos\beta+\cos\alpha\sin\beta)^T.$$

Geometrically, successive rotations add their angles, so this same point is $(\cos(\alpha+\beta),\sin(\alpha+\beta))^T$. Comparing coordinates proves both identities without assuming them. Setting $\beta=\alpha=\theta$ gives $\sin2\theta=2\sin\theta\cos\theta$ and $\cos2\theta=\cos^2\theta-\sin^2\theta$. An identity holds for every input in its domain; an equation asks which inputs make a particular equality true.

Write $w=\rho\operatorname{cis}\phi$, where $\rho>0$ and $\operatorname{cis}\phi=\cos\phi+i\sin\phi$. The matrix is $\rho$ times the rotation matrix through $\phi$. Thus multiplication by a nonzero complex number is exactly a uniform dilation followed by rotation. Lengths scale by $\rho$, and oriented areas scale by $\rho^2>0$. Multiplication alone does not reflect the plane. Conjugation, with matrix $\operatorname{diag}(1,-1)$, is the reflection across the real axis.

For nonzero $z,w$, division scales lengths by $1/|w|$ and subtracts an argument of $w$. It is legal only because $w$ is nonzero. The reciprocal map $z\mapsto1/z$ itself is not a single linear rotation-dilation: its scale varies with the input, so geometrical claims about fixed multiplication must not be transferred to it unchanged.

Worked map. Multiplication by $1+i$ has scale $\sqrt2$ and rotation $\pi/4$. The image of the line $z=t$ for real $t$ is $t+it$, the line $y=x$. The image of the circle $|z-2|=1$ is $|w-2(1+i)|=\sqrt2$, because $|w-(1+i)2|=|1+i||z-2|$. Both the center and radius must be transformed.

2. Arguments are equivalence classes of angles

For $z\ne0$, every argument differs from every other by $2k\pi$. With principal branch $-\pi<\operatorname{Arg}z\le\pi$, a product obeys addition modulo $2\pi$, not always literal equality of principal values.

For example $z=w=e^{3\pi i/4}$, using $e^{i\theta}$ here only as notation for $\operatorname{cis}\theta$. Their principal arguments sum to $3\pi/2$, but the product has principal argument $-\pi/2$. The difference is $2\pi$, so polar multiplication remains correct. This notation does not require an analytic construction of the complex exponential.

When taking an $n$th root of $R\operatorname{cis}\phi$ with $n$ a positive integer and $R>0$, use all angles $(\phi+2k\pi)/n$ for $k=0,\ldots,n-1$. Choosing a different argument $\phi+2m\pi$ permutes these roots rather than creating a different set. Keeping only the principal angle divided by $n$ selects one root and does not solve the whole polynomial equation.

For completeness, polar multiplication gives $(r\operatorname{cis}\theta)^n=r^n\operatorname{cis}(n\theta)$ by induction starting at $n=1$: multiply the $n$th case by $r\operatorname{cis}\theta$ and add the arguments. Thus the complete root list is

$$z_k=R^{1/n}\operatorname{cis}\left(\frac{\phi+2k\pi}{n}\right),\qquad k=0,\ldots,n-1,$$

where $R^{1/n}$ is the unique positive real $n$th root. Direct substitution gives $z_k^n=R\operatorname{cis}\phi$. Conversely, any root is nonzero and has polar form $r\operatorname{cis}\theta$; equality of moduli forces $r=R^{1/n}$, and equality of unit-circle points forces $n\theta-\phi=2m\pi$ for an integer $m$. Write $m=qn+k$ with $0\le k<n$; the extra angle $2q\pi$ leaves the point unchanged, so that root equals $z_k$. If $z_j=z_k$, their argument difference $2(j-k)\pi/n$ is a multiple of $2\pi$, hence $n$ divides $j-k$ and the chosen index range forces $j=k$. This proves existence, exhaustiveness and distinctness without the fundamental theorem of algebra.

3. Roots of unity explain the regular polygon

Let $n\ge2$ be an integer and $\zeta=\operatorname{cis}(2\pi/n)$. The $n$ solutions of $z^n=1$ are $1,\zeta,\ldots,\zeta^{n-1}$, equally spaced on the unit circle. They are distinct because $\zeta^j=\zeta^k$ forces $j-k$ to be a multiple of $n$. Their sum is zero:

$$1+\zeta+\cdots+\zeta^{n-1}=\frac{1-\zeta^n}{1-\zeta}=0,$$

where $\zeta\ne1$ makes division valid. Thus the polygon's vertex average is the origin, justified algebraically rather than only by symmetry.

More generally for an integer $m$,

$$\sum_{k=0}^{n-1}\zeta^{mk}=\begin{cases}n,&n\mid m,\\0,&n\nmid m.\end{cases}$$

If $n\mid m$, every term is one. Otherwise $\zeta^m\ne1$ but $(\zeta^m)^n=1$, and the finite geometric identity proves zero. The denominator-zero case must be separated, not canceled.

For $w\ne0$, the equation $z^n=w$ produces a rotated and dilated version of this polygon: choose one root $z_0$ and multiply it by each root of unity. This describes both the exhaustive root set and its geometric arrangement. If $w=0$, the only root of $z^n=0$ is $z=0$.

4. A parameter can move roots from a line to a circle

Consider $z^2-2tz+1=0$ for a real parameter $t$. Completing the square gives $(z-t)^2=t^2-1$. The cases are:

• If $|t|<1$, the roots are $t\pm i\sqrt{1-t^2}$. They are conjugates with modulus one, so they lie on the unit circle.
• If $t=1$ or $t=-1$, the roots merge into the repeated real root $1$ or $-1$.
• If $|t|>1$, the roots $t\pm\sqrt{t^2-1}$ are distinct and real. Their product is one, and exactly one has modulus less than one while the other has modulus greater than one.

For the last assertion, neither root is zero and neither can have modulus one: the only real numbers of modulus one are $\pm1$, which solve the equation only at $t=\pm1$. Since the positive product of their moduli is one and they are unequal in modulus away from the boundary cases, one lies inside and the other outside the unit circle.

The classification is exhaustive. A complex number $a+ib$ whose square is real satisfies $2ab=0$. If its square is positive, the roots are the two real square roots; if negative, the roots are the two purely imaginary ones. If zero, the root is zero. Thus the completed-square equation supplies every root, not merely plausible candidates.

A converse interpretation. Dividing the original equation by nonzero $z$ gives $z+1/z=2t$. For $z=re^{i\theta}$, its imaginary part is $(r-r^{-1})\sin\theta$. For the sum to be real, either $r=1$ or $\sin\theta=0$. Hence solutions lie either on the unit circle or on the nonzero real axis, exactly the two families found algebraically.

5. Distance loci must be transformed without losing equivalence

The equation $|z-a|=k|z-b|$, $k>0$, asks for a fixed distance ratio. Both sides are nonnegative, so squaring is equivalent to the original equation. For $k=1$ and $a\ne b$, the quadratic terms cancel, giving the perpendicular-bisector line. For $k\ne1$, they do not cancel. Expanding $|z-a|^2=|z|^2-2\operatorname{Re}(z\overline a)+|a|^2$ and completing the squared modulus gives center $c=(a-k^2b)/(1-k^2)$ and radius $k|a-b|/|1-k^2|$. Indeed the radius squared is $|c|^2-(|a|^2-k^2|b|^2)/(1-k^2)=k^2|a-b|^2/(1-k^2)^2$. Distinct foci make the radius positive, so the locus is genuinely a circle rather than a collapsed point.

The equation here is the undivided equality $|z-a|=k|z-b|$, defined for every $z\in\mathbb C$, not a quotient with denominator $|z-b|$. If $a=b$ and $k=1$, every complex number is a solution. If $a=b$ and $k\ne1$, it reduces to $(1-k)|z-a|=0$, so the locus is the single point $a$ (the radius-zero case of the formula). If one instead asks for the quotient equation $|z-a|/|z-b|=k$, its domain excludes $z=b$; the equality's locus must then be intersected with that smaller domain. In particular, for $a=b$ the quotient locus is $\mathbb C\setminus\{a\}$ when $k=1$ and empty when $k\ne1$.

Worked locus. For $|z-1|=2|z+1|$, write $z=x+iy$ and square:

$$(x-1)^2+y^2=4[(x+1)^2+y^2].$$

Rearrangement gives $3x^2+10x+3+3y^2=0$, hence $(x+5/3)^2+y^2=16/9$. The locus is a circle centered at $(-5/3,0)$ with radius $4/3$. Because the distance equation had nonnegative sides, every point of this circle also satisfies the unsquared relation. The squaring step did not introduce extraneous points.

Graded synthesis and diagnostics

1. Geometry. Describe the image of $|z-(1-i)|=2$ under $w=iz$, and explain why no reflection occurs.
2. Branch diagnosis. A student says $\operatorname{Arg}(z^2)=2\operatorname{Arg}z$ for every nonzero $z$. Refute the literal assertion and state the correct modular relation.
3. Root structure. For an integer $n\ge2$ and any nonzero $w$, prove that the sum of all distinct roots of $z^n=w$ is zero, using roots of unity rather than a general coefficient-root theorem.
4. Parameter analysis. Classify all complex solutions of $z+1/z=s$ as real $s$ varies. State the zero exclusion and include repeated-root cases.
5. Locus. Find all $z$ satisfying $|z-2i|=|z+2i|$, and prove both inclusions between your answer and the original locus.
6. Challenge. Let $\zeta=\operatorname{cis}(2\pi/5)$. Evaluate $\sum_{k=0}^4\zeta^{3k}$ and $\sum_{k=0}^4\zeta^{5k}$, explaining why the same fraction formula cannot be applied blindly to both.

Full solutions

1. The center maps to $i(1-i)=1+i$, and $|i|=1$ leaves the radius two unchanged. Thus the image is $|w-(1+i)|=2$. The multiplication matrix is a rotation through $\pi/2$ with determinant one, not a reflection.
2. Take $z=\operatorname{cis}(3\pi/4)$. Then $z^2=\operatorname{cis}(3\pi/2)$ has principal argument $-\pi/2$, whereas twice the principal input argument is $3\pi/2$. The correct relation is equality modulo $2\pi$; after doubling, select the representative in the chosen principal interval.
3. Choose one nonzero root $z_0$. Every root is $z_0\zeta^k$ for $k=0,\ldots,n-1$: division by $z_0$ reduces its $n$th power to one, and the root formula shows the list exhaustive and distinct. Their sum is $z_0\sum\zeta^k=0$ because $n\ge2$.
4. The equation is defined only for $z\ne0$ and becomes $z^2-sz+1=0$, whose roots are never zero. Set $t=s/2$. If $|s|<2$, the two roots are $s/2\pm i\sqrt{1-s^2/4}$ on the unit circle. At $s=2$ the repeated root is one; at $s=-2$ it is minus one. For $|s|>2$, the roots are $(s\pm\sqrt{s^2-4})/2$, distinct real reciprocals, one of modulus below one and the other above one.
5. With $z=x+iy$, squared distances give $x^2+(y-2)^2=x^2+(y+2)^2$, equivalent to $y=0$. Every original solution must have $y=0$ by subtraction, and every real $z$ has equal distances to the two conjugate foci, proving the converse. The locus is the real axis.
6. Since five does not divide three, $\zeta^3\ne1$ and its fifth power is one, so the first sum is zero by the finite geometric identity. Since five divides five, every term in the second sum is one and the sum is five. A quotient with denominator $1-\zeta^5=0$ is undefined; it cannot replace this direct case check.

Summary and terms

A rotation-dilation changes every vector by one common angle and scale. A principal argument is a chosen angle representative, not a globally additive real function. Roots of unity organize every nonzero root equation into a regular polygon. A parameter classification splits at values where root type or multiplicity changes. A distance-ratio locus turns modulus equations into exact Cartesian geometry only after preserving the equivalence of every transformation.