6.1 Complex numbers, polar form, and geometry

Why complex numbers live in a plane

The equation $x^2+1=0$ has no real solution. Introducing a symbol i with $i^2=-1$ is not only an algebraic trick; it also suggests a geometric extension of the real line. Multiplication by $-1$ rotates a real number by $180^\circ$. Since $i^2=-1$, multiplication by i should behave like a $90^\circ$ anticlockwise rotation. Thus $i\cdot 1$ is naturally placed at the point (0,1) rather than somewhere on the real line.

This motivates the formal construction: a complex number is a point in the plane, but with addition and multiplication chosen so that the rule $i^2=-1$ is built into the arithmetic.

These definitions are chosen so that the informal expression $x+iy$ is not just notation but actually agrees with the ordered-pair construction. Indeed, for real x and y,

$$(x,0)+(0,1)(y,0)=(x,0)+(0,y)=(x,y).$$

So from now on we write

$$z=x+iy$$

instead of (x,y).

Equality of complex numbers is equality of ordered pairs. Thus

$$a+ib=c+id \quad\Longleftrightarrow\quad a=c\text{ and }b=d.$$

This simple point is often the reason complex equations can be solved by comparing real and imaginary parts.

Arithmetic and field structure

The ordered-pair rules make $\mathbb C$ behave like a number system. Addition and multiplication are commutative and associative, multiplication distributes over addition, $0=0+i0$ is the additive identity, and $1=1+i0$ is the multiplicative identity.

The inverse formula is worth deriving once. If $z=x+iy\ne0$, then $x^2+y^2>0$, and

$$(x+iy)\left(\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}\right) =\frac{x^2+y^2}{x^2+y^2} +i\frac{-xy+xy}{x^2+y^2} =1.$$

Because non-zero complex numbers have multiplicative inverses, there are no zero divisors in $\mathbb C$: if $zw=0$, then $z=0$ or $w=0$.

Subtraction and division

For $z=a+ib$ and $w=c+id$,

$$z-w=(a-c)+i(b-d).$$

If $w\ne0$, then division means multiplication by the inverse of w:

$$\frac{z}{w} =\frac{a+ib}{c+id} =(a+ib)\left(\frac{c}{c^2+d^2}-i\frac{d}{c^2+d^2}\right).$$

After multiplying out,

$$\frac{a+ib}{c+id} =\frac{ac+bd}{c^2+d^2} +i\frac{bc-ad}{c^2+d^2}.$$

The Argand plane

The same ordered pair (x,y) can be read geometrically as a point or vector in the coordinate plane. For $z=x+iy$, the horizontal coordinate is the real part and the vertical coordinate is the imaginary part. This plane is called the complex plane or Argand plane. The horizontal axis is the real axis, and the vertical axis is the imaginary axis.

Addition is vector addition:

$$(x_1+iy_1)+(x_2+iy_2) =(x_1+x_2)+i(y_1+y_2).$$

Multiplication is less visible in rectangular coordinates, but it becomes clear after we introduce modulus and argument.

Conjugate and modulus

The conjugate is reflection across the real axis. The modulus is the distance from the origin to the point representing z.

The basic identities are

$$\overline{\overline z}=z,\qquad z=\overline z\Longleftrightarrow z\in\mathbb R,$$

and

$$\operatorname{Re}(z)=\frac12(z+\overline z), \qquad \operatorname{Im}(z)=\frac{1}{2i}(z-\overline z).$$

Conjugation respects arithmetic:

$$\overline{z_1\pm z_2}=\overline{z_1}\pm\overline{z_2}, \qquad \overline{z_1z_2}=\overline{z_1}\,\overline{z_2}, \qquad \overline{\left(\frac{z_1}{z_2}\right)} =\frac{\overline{z_1}}{\overline{z_2}}$$

when $z_2\ne0$.

The modulus is tied to conjugation by

$$|z|^2=z\overline z.$$

Hence $z\overline z$ is always a non-negative real number. Also,

$$|z|=|-z|=|\overline z|, \qquad |z_1z_2|=|z_1||z_2|, \qquad \left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}$$

for $z_2\ne0$.

Argument and polar form

Let $z=x+iy\ne0$. An argument of z is any angle from the positive real axis to the point (x,y), measured anticlockwise. If $\\theta$ is one argument, then all arguments are

$$\theta+2k\pi,\qquad k\in\mathbb Z.$$

The principal argument is the unique argument in the interval

$$-\pi<\operatorname{Arg}(z)\le\pi.$$

If $r=|z|$, then

$$z=r(\cos\theta+i\sin\theta),$$

where $\\theta$ is any argument of z. This is the polar form of z.

Polar form also explains conjugates and inverses:

$$\overline z=r(\cos(-\theta)+i\sin(-\theta)), \qquad z^{-1}=\frac1r(\cos(-\theta)+i\sin(-\theta)).$$

In particular, if $|z|=1$, then $z^{-1}=\overline z$.

Multiplication as rotation and scaling

Suppose

$$z_1=r_1(\cos\theta_1+i\sin\theta_1), \qquad z_2=r_2(\cos\theta_2+i\sin\theta_2).$$

Using the angle-sum formulas,

$$z_1z_2 =r_1r_2\bigl(\cos(\theta_1+\theta_2) +i\sin(\theta_1+\theta_2)\bigr).$$

Thus multiplication multiplies moduli and adds arguments:

$$|z_1z_2|=|z_1||z_2|, \qquad \arg(z_1z_2)=\arg(z_1)+\arg(z_2)$$

up to multiples of $2\pi$.

Division reverses the second rotation and divides the lengths:

$$\frac{z_1}{z_2} =\frac{r_1}{r_2} \bigl(\cos(\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)\bigr).$$

De Moivre's theorem

Integral powers are defined by repeated multiplication. If $z\ne0$, then $z^0=1$ and $z^{-n}=(z^{-1})^n$ for positive integers n.

For positive n, the result follows by induction from the multiplication rule in polar form. The case $n=0$ is immediate. For negative n, write $n=-m$, use the positive case for m, and use the fact that a unit complex number has inverse equal to its conjugate.

Euler's formula and exponential form

The power series for sine, cosine, and the exponential suggest what $e^{i\\theta}$ should mean:

$$e^{i\theta} =1+i\theta-\frac{\theta^2}{2!}-i\frac{\theta^3}{3!} +\frac{\theta^4}{4!}+\cdots =\cos\theta+i\sin\theta.$$

In this course context, we use that motivation to write

$$e^{i\theta}=\cos\theta+i\sin\theta.$$

A non-zero complex number can then be written in exponential form as

$$z=re^{i\theta},$$

where $r=|z|$ and $\\theta$ is an argument of z.

This notation compresses multiplication and De Moivre's theorem:

$$r_1e^{i\theta_1}\,r_2e^{i\theta_2} =r_1r_2e^{i(\theta_1+\theta_2)}, \qquad (e^{i\theta})^n=e^{in\theta}.$$

It also gives

$$\cos\theta=\frac12(e^{i\theta}+e^{-i\theta}), \qquad \sin\theta=\frac{1}{2i}(e^{i\theta}-e^{-i\theta}).$$

Putting $\\theta=\pi$ gives Euler's identity:

$$e^{i\pi}+1=0.$$

Triangle inequality

The proof is a good example of how conjugates avoid coordinate expansion:

$$\begin{aligned} |z_1+z_2|^2 &=(z_1+z_2)(\overline{z_1}+\overline{z_2})\\ &=|z_1|^2+2\operatorname{Re}(z_1\overline{z_2})+|z_2|^2\\ &\le |z_1|^2+2|z_1\overline{z_2}|+|z_2|^2\\ &=(|z_1|+|z_2|)^2. \end{aligned}$$

Taking square roots gives the inequality because both sides are non-negative.

Polynomial roots over $\mathbb C$

The construction of complex numbers began because $x^2+1=0$ has no real solution. The deeper closure statement is that, once we allow complex coefficients and complex roots, every non-constant polynomial has a root.

The theorem is not proved in this unit, but its consequences are central. If $\alpha$ is a root, then $p(z)=(z-\alpha)q(z)$ for a polynomial q whose degree is one smaller. Repeating this factorization gives

$$p(z)=a_n(z-\alpha_1)(z-\alpha_2)\cdots(z-\alpha_n),$$

where roots are counted with multiplicity.

When the polynomial has real coefficients, non-real roots come in conjugate pairs. Indeed, if

$$p(z)=a_nz^n+a_{n-1}z^{n-1}+\cdots+a_1z+a_0,\qquad a_k\in\mathbb R,$$

and $p(\alpha)=0$, then taking conjugates of the equation gives $p(\overline\alpha)=0$.

This explains why real polynomials factor into real linear and real quadratic pieces:

$$(z-\alpha)(z-\overline\alpha) =z^2-2\operatorname{Re}(\alpha)z+|\alpha|^2.$$

Roots of unity and solving $z^n=z_0$

Polar form makes roots of equations such as $z^n=1$ visible. Let

$$\omega=e^{2\pi i/n} =\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n}.$$

Then $\omega^n=1$, and

$$1,\omega,\omega^2,\ldots,\omega^{n-1}$$

are the n distinct n-th roots of unity. Therefore

$$z^n-1=(z-1)(z-\omega)(z-\omega^2)\cdots(z-\omega^{n-1}).$$

Over the real numbers, conjugate roots are paired into quadratic factors. For example, if $n=5$, then

$$z^5-1 =(z-1) \left(z^2-2\cos\frac{2\pi}{5}\,z+1\right) \left(z^2-2\cos\frac{4\pi}{5}\,z+1\right).$$

The same idea solves $z^n=z_0$. If $z_0=r_0e^{i\theta_0}$ with $r_0>0$, then the solutions are

$$\alpha_k=r_0^{1/n} e^{i(\theta_0+2k\pi)/n}, \qquad k=0,1,\ldots,n-1.$$

Each root has the same modulus $r_0^{1/n}$; the arguments are equally spaced by $2\pi/n$.

Geometry with complex ratios

Differences of complex numbers represent directed vectors. For distinct points $z_1,z_2,z_3$, write

$$\frac{z_1-z_3}{z_2-z_3}=re^{i\theta}.$$

Then $r=|z_1-z_3|/|z_2-z_3|$, and $\\theta$ is the oriented angle from the vector $z_2-z_3$ to the vector $z_1-z_3$. Thus one complex ratio records both a side-length ratio and an angle.

This gives a compact collinearity test:

The real-ratio condition says the two directed vectors have argument difference 0 or $\pi$; that is exactly the condition that they lie on the same line.

Complex ratios also encode similarity of oriented triangles:

$$\triangle z_1z_2z_3\sim \triangle w_1w_2w_3 \quad\Longleftrightarrow\quad \frac{z_1-z_3}{z_2-z_3} = \frac{w_1-w_3}{w_2-w_3},$$

when the orientation is the same. The equality means the corresponding side-ratio and included angle both agree.

The source also introduces the cross ratio

$$(z_1,z_2;z_3,z_4) =\frac{(z_1-z_3)(z_2-z_4)} {(z_2-z_3)(z_1-z_4)}$$

for four distinct points. At this level, the important lesson is not to memorize every identity at once, but to notice the recurring strategy: translation is represented by differences, scaling and rotation by multiplication, and shape information by ratios.

Locus equations and transformations

Many plane loci can be written cleanly in complex notation. The equation

$$|z-z_0|=r$$

is the circle centered at $z_0$ with radius r. Squaring gives

$$(z-z_0)(\overline z-\overline{z_0})=r^2,$$

which is often easier to expand.

Similarly,

$$|z-z_1|=|z-z_2|$$

is the perpendicular bisector of the segment joining $z_1$ and $z_2$, because it is the set of points equally distant from the two endpoints. A line through $z_1$ and $z_2$ can be described by the real-ratio condition

$$\frac{z-z_2}{z_1-z_2}\in\mathbb R.$$

For a circle through three non-collinear points, the cross-ratio test gives

$$(z_1,z_2;z_3,z)\in\mathbb R.$$

The source also treats complex transformations, meaning functions from the complex plane to itself. The basic transformations are:

• scaling: $f(z)=rz$, with real $r\ne0$;
• translation: $f(z)=z+a$;
• rotation: $f(z)=e^{i\\theta}z$;
• inversion: $f(z)=1/z$;
• reflection in the real axis: $f(z)=\overline z$.

For example, $f(z)=z^2$ sends $z=re^{i\\theta}$ to

$$f(z)=r^2e^{2i\theta},$$

so it squares distances from the origin and doubles arguments.

Translations, rotations, and conjugation preserve distances. Translations, rotations, and scalings preserve the three-point ratio

$$\frac{z_1-z_3}{z_2-z_3}.$$

This explains why they preserve similarity. Cross ratios are even more stable: they are preserved by scalings, translations, rotations, and inversions, hence by Mobius transformations

$$f(z)=\frac{az+b}{cz+d},\qquad ad-bc\ne0.$$

This is why Mobius transformations send straight lines and circles to straight lines and circles. One algebraic way to summarize both objects is the equation

$$A z\overline z+Bz+\overline B\,\overline z+C=0,$$

where $A,C\in\mathbb R$; $A\ne0$ gives a circle, while $A=0$ gives a line.

Quick checks

Exercises

1. Prove that the embedding $\phi:\mathbb R\to\mathbb C$, $\phi(x)=x+i0$, preserves addition and multiplication.
2. Express $(3-2i)/(1+i)$ in the form $a+bi$.
3. Prove that $z\overline z=|z|^2$ and use it to derive the inverse formula for non-zero z.
4. Find the modulus, all arguments, principal argument, and polar form of $-\sqrt3+i$.
5. Prove $|1-\overline{z_1}z_2|^2-|z_1-z_2|^2=(1-|z_1|^2)(1-|z_2|^2)$.
6. Use De Moivre's theorem to express $\\cos(5\\theta)$ and $\\sin(5\\theta)$ in terms of powers of $\cos\\theta$ and $\sin\\theta$.
7. Prove by induction that $|\sum_{k=1}^n z_k|\le \sum_{k=1}^n |z_k|$.
8. For $\\theta\ne2k\pi$, use the geometric sum for $e^{i\\theta}$ to derive the formula $\sum_{k=1}^n \\cos(k\\theta)=\\sin(n\\theta/2)\cos((n+1)\\theta/2)/\\sin(\\theta/2)$.
9. Factor $z^6-1$ over $\mathbb C$ and then group the non-real conjugate roots to factor it over $\mathbb R$.
10. Solve $z^4=16$ in polar form.
11. Show that if $p(z)\in\mathbb R[z]$ and $p(\alpha)=0$, then $p(\overline\alpha)=0$.
12. Prove that $z_1,z_2,z_3$ are collinear exactly when $(z_1-z_3)/(z_2-z_3)$ is real.
13. Show directly that $|z-1|=3|z+1|$ is the circle $|z+5/4|=3/4$.
14. Let $f(z)=az+b$ with $a\ne0$. Prove that $(f(z_1)-f(z_3))/(f(z_2)-f(z_3))=(z_1-z_3)/(z_2-z_3)$ for distinct $z_1,z_2,z_3$.

Guided solutions

1. For real a,b, $\phi(a+b)=(a+b)+i0=(a+i0)+(b+i0)=\phi(a)+\phi(b)$. Also $\phi(ab)=ab+i0=(a+i0)(b+i0)=\phi(a)\phi(b)$. Thus real arithmetic is preserved inside $\mathbb C$.

2. Multiply by the conjugate of the denominator:

   $$\frac{3-2i}{1+i} =\frac{(3-2i)(1-i)}{(1+i)(1-i)} =\frac{1-5i}{2} =\frac12-\frac52 i.$$

3. If $z=x+iy$, then $z\overline z=(x+iy)(x-iy)=x^2+y^2=|z|^2$. If $z\ne0$, then $|z|^2>0$, so $z^{-1}=\overline z/|z|^2=(x-iy)/(x^2+y^2)$.

4. The modulus is 2. Since $\cos\\theta=-\sqrt3/2$ and $\sin\\theta=1/2$, the point is in quadrant II with principal argument $5\pi/6$. Hence all arguments are $5\pi/6+2k\pi$, and $-\sqrt3+i=2(\\cos(5\pi/6)+i\\sin(5\pi/6))$.

5. Expand both squared moduli using conjugates:

   $$|1-\overline{z_1}z_2|^2 =(1-\overline{z_1}z_2)(1-z_1\overline{z_2}) =1-\overline{z_1}z_2-z_1\overline{z_2}+|z_1|^2|z_2|^2,$$

   while

   $$|z_1-z_2|^2 =|z_1|^2-z_1\overline{z_2}-\overline{z_1}z_2+|z_2|^2.$$

   Subtracting cancels the mixed terms and gives $(1-|z_1|^2)(1-|z_2|^2)$.

6. Expand $(\cos\\theta+i\sin\\theta)^5=\\cos(5\\theta)+i\\sin(5\\theta)$ by the binomial theorem. The real terms are those with even powers of i, and the imaginary terms are those with odd powers of i:

   $$\cos(5\theta)=\cos^5\theta-10\cos^3\theta\sin^2\theta +5\cos\theta\sin^4\theta,$$ $$\sin(5\theta)=5\cos^4\theta\sin\theta -10\cos^2\theta\sin^3\theta+\sin^5\theta.$$

7. The case $n=1$ is immediate. If the result holds for n, then

   $$\left|\sum_{k=1}^{n+1}z_k\right| = \left|\left(\sum_{k=1}^{n}z_k\right)+z_{n+1}\right| \le \left|\sum_{k=1}^{n}z_k\right|+|z_{n+1}| \le \sum_{k=1}^{n+1}|z_k|.$$

8. Since $e^{i\\theta}\ne1$,

   $$\sum_{k=1}^n e^{ik\theta} =e^{i\theta}\frac{e^{in\theta}-1}{e^{i\theta}-1}.$$

   Rewrite numerator and denominator by factoring half-angles:

   $$e^{in\theta}-1 =e^{in\theta/2}(e^{in\theta/2}-e^{-in\theta/2}),$$

   and similarly for $e^{i\\theta}-1$. This gives

   $$\sum_{k=1}^n e^{ik\theta} = \frac{\sin(n\theta/2)}{\sin(\theta/2)} e^{i(n+1)\theta/2}.$$

   Taking real parts gives the stated cosine-sum formula.

9. Over $\mathbb C$, let $\omega=e^{2\pi i/6}$. Then

   $$z^6-1=\prod_{k=0}^{5}(z-\omega^k).$$

   The roots are 1, $-1$, $e^{\pi i/3}$, $e^{-\pi i/3}$, $e^{2\pi i/3}$, and $e^{-2\pi i/3}$. Pairing conjugates gives

   $$z^6-1=(z-1)(z+1) \left(z^2-2\cos\frac{\pi}{3}z+1\right) \left(z^2-2\cos\frac{2\pi}{3}z+1\right),$$

   so over $\mathbb R$,

   $$z^6-1=(z-1)(z+1)(z^2-z+1)(z^2+z+1).$$

10. Write $16=16e^{2\pi i m}$. The fourth roots have modulus 2 and arguments $(2k\pi)/4=k\pi/2$, so the solutions are

    $$2,\quad 2i,\quad -2,\quad -2i.$$

11. Write $p(z)=a_nz^n+a_{n-1}z^{n-1}+\cdots+a_0$ with all $a_j$ real. If $p(\alpha)=0$, then conjugating both sides gives

    $$a_n\overline\alpha^n+a_{n-1}\overline\alpha^{n-1} +\cdots+a_0=0,$$

    because conjugation respects sums and products and $\overline{a_j}=a_j$. This is exactly $p(\overline\alpha)=0$.

12. If the points are collinear, the vectors $z_1-z_3$ and $z_2-z_3$ have the same or opposite direction, so their argument difference is 0 or $\pi$; the quotient is real. Conversely, if the quotient is real, its argument is 0 or $\pi$, so the two vectors are parallel and the three points lie on a line.

13. Squaring gives

    $$(z-1)(\overline z-1)=9(z+1)(\overline z+1).$$

    Expanding and collecting terms gives

    $$z\overline z+\frac54z+\frac54\overline z+1=0.$$

    This is

    $$\left(z+\frac54\right)\left(\overline z+\frac54\right)=\frac9{16},$$

    hence $|z+5/4|=3/4$.

14. Since $f(z_j)=az_j+b$,

    $$\frac{f(z_1)-f(z_3)}{f(z_2)-f(z_3)} =\frac{a(z_1-z_3)}{a(z_2-z_3)} =\frac{z_1-z_3}{z_2-z_3}.$$

    The translation part cancels, and the non-zero scaling/rotation factor a cancels.