10. Complex numbers, polar form and roots

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.

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.

Learning goals and definitions

Calculate with complex numbers, use conjugates and moduli, multiply in polar form, and solve root equations without omitting branches. A complex number is $z=a+bi$, where $a,b\in\mathbb R$ and $i^2=-1$. Addition is componentwise; multiplication distributes with $i^2=-1$. Equality requires equal real parts and equal imaginary parts.

The conjugate is $\overline z=a-bi$ and modulus is $|z|=\sqrt{a^2+b^2}$. Direct multiplication gives $z\overline z=|z|^2$. Thus for $z\ne0$, $z^{-1}=\overline z/|z|^2$. The identity $|zw|=|z||w|$ follows by expanding squared moduli or using conjugation. In the Argand plane, $|z-w|$ is ordinary Euclidean distance.

For $z\ne0$, polar form is $z=r(\cos\theta+i\sin\theta)$ with $r=|z|>0$. An argument $\theta$ is determined only modulo $2\pi$; zero has no argument. A principal argument uses one convention, here $-\pi<\operatorname{Arg}z\le\pi$. A quotient can cross that branch cut even when the arithmetic is continuous.

Theorem: polar multiplication and De Moivre's identity

For $\operatorname{cis}\theta=\cos\theta+i\sin\theta$,

$$\operatorname{cis}\alpha\operatorname{cis}\beta=\operatorname{cis}(\alpha+\beta),\qquad (r\operatorname{cis}\theta)^n=r^n\operatorname{cis}(n\theta)$$

for $r>0$ and nonnegative integers $n$; the power identity also holds for $r=0$ when $n\ge1$.

Proof. Expanding the product and using the sine and cosine addition formulas gives the first identity. Repeated multiplication proves the second by induction, including $n=0$ when the base is nonzero. At base zero we use the power identity only for $n\ge1$, avoiding the undefined expression $0^0$.

If $w=R\operatorname{cis}\phi\ne0$ and $n\ge1$, all solutions of $z^n=w$ are

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

Indeed moduli force $|z|^n=R$, and arguments force $n\theta=\phi+2k\pi$. These $n$ values are distinct modulo $2\pi$, and every integer $k$ reduces to one of them modulo $n$. If $w=0$, the only root is zero. A single real radical does not list all complex roots.

Worked examples

Division. Multiply numerator and denominator of $(2+3i)/(1-i)$ by $1+i$. The result is $(-1+5i)/2=-1/2+(5/2)i$. A useful check is to multiply this answer by $1-i$ and recover $2+3i$.

Cube roots. Since $-8=8\operatorname{cis}\pi$, its cube roots have modulus $2$ and arguments $\pi/3,\pi,5\pi/3$. They are $1+\sqrt3i,-2,1-\sqrt3i$. The negative real root is only one member of the set.

A locus. The equation $|z-1|=|z+1|$ says that a point is equally distant from $1$ and $-1$. Writing $z=x+iy$ and squaring gives $(x-1)^2+y^2=(x+1)^2+y^2$, hence $x=0$. The locus is the imaginary axis, including zero; arguments are unnecessary.

Graded practice

1. Foundation. Compute $(1+2i)(3-i)$ and the modulus of the product.
2. Core. Find all fourth roots of $16$.
3. Core. Solve $z^2-2z+5=0$ over $\mathbb C$ and verify both roots.
4. Proof. Show that a polynomial with real coefficients has conjugate root pairs: if $p(z)=0$, then $p(\overline z)=0$. Explain why the real-coefficient hypothesis matters.

Full solutions

1. The product is $5+5i$, with modulus $\sqrt{25+25}=5\sqrt2$. Independently, the two factor moduli are $\sqrt5$ and $\sqrt{10}$, whose product is also $5\sqrt2$.
2. The roots have modulus $2$ and arguments $0,\pi/2,\pi,3\pi/2$: $2,2i,-2,-2i$. Each fourth power is $16$, and the root formula proves the list exhaustive.
3. Completing the square gives $(z-1)^2=-4$, hence $z=1\pm2i$. For either sign, $(z-1)^2+4=0$, so substitution verifies the original equation.
4. Conjugation preserves sums and products. If every coefficient $a_j$ is real, then $p(\overline z)=\sum a_j\overline z^j=\overline{\sum a_jz^j}=\overline{p(z)}=0$. For nonreal coefficients the claim can fail: $p(z)=z-i$ has root $i$ but not $-i$.

Summary and term definitions

The conjugate reflects across the real axis; the modulus is distance from zero; an argument is an angle modulo $2\pi$. A principal argument is one chosen representative, not an algebraically additive real function. Polar multiplication adds angles, and root extraction divides angles only after adding every $2k\pi$ branch. Cartesian form is often simpler for sums, quotients and distance loci.