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The Complex Plane

In this lesson you’ll treat complex numbers as points in a plane, learn modulus and argument, convert to polar and exponential form, and use them to take powers and roots almost without calculation.

Write z=x+iyz = x + iy and plot it at (x,y)(x, y). That is the entire idea behind the complex plane, sometimes called the Argand diagram, and once you have it nothing about complex numbers is mysterious.

  • The real axis is the horizontal one, where y=0y = 0.
  • The imaginary axis is the vertical one, where x=0x = 0.
  • Addition is vector addition, exactly as in R2\mathbb{R}^2.
  • Conjugation zzˉz \mapsto \bar z is reflection across the real axis.

Multiplication is where C\mathbb{C} stops behaving like R2\mathbb{R}^2. There is no useful multiplication of plane vectors, but there is for complex numbers, and it has a clean geometric meaning we get to shortly.

The modulus is the distance to the origin:

z=x2+y2=zzˉ|z| = \sqrt{x^2+y^2} = \sqrt{z\bar z}

The argument is the angle from the positive real axis, measured counterclockwise. It is only defined up to multiples of 2π2\pi, since rotating a full turn returns you to the same point. The principal argument Argz\operatorname{Arg} z is the unique choice in (π,π](-\pi, \pi].

Useful properties, all worth knowing cold:

  • z1z2=z1z2|z_1z_2| = |z_1||z_2| and z1z2=z1z2\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}
  • arg(z1z2)=argz1+argz2\arg(z_1z_2) = \arg z_1 + \arg z_2, and arguments subtract on division
  • z2=zzˉ|z|^2 = z\bar z, which is how you divide complex numbers
  • z1+z2=zˉ1+zˉ2\overline{z_1+z_2} = \bar z_1 + \bar z_2 and z1z2=zˉ1zˉ2\overline{z_1z_2} = \bar z_1 \bar z_2
  • z1+z2z1+z2|z_1+z_2| \le |z_1|+|z_2|, the triangle inequality, and it is genuinely the triangle inequality for the picture

Note that z1+z2|z_1 + z_2| has no simple formula. Moduli behave well under multiplication and badly under addition, which is the reverse of what you might expect.

Reading the right triangle off the diagram,

x=rcosθ,y=rsinθ,z=r(cosθ+isinθ)x = r\cos\theta, \qquad y = r\sin\theta, \qquad z = r(\cos\theta + i\sin\theta)

Euler’s formula compresses this to

eiθ=cosθ+isinθ,soz=reiθe^{i\theta} = \cos\theta + i\sin\theta, \qquad\text{so}\qquad z = re^{i\theta}

This is the single most useful identity in the section. Setting θ=π\theta = \pi gives eiπ=1e^{i\pi} = -1, usually written eiπ+1=0e^{i\pi}+1=0 and often called the most beautiful equation in mathematics. It is also just a statement that rotating 1 by half a turn lands on 1-1.

Multiplication becomes rotation and scaling. If z1=r1eiθ1z_1 = r_1e^{i\theta_1} and z2=r2eiθ2z_2 = r_2e^{i\theta_2} then

z1z2=r1r2ei(θ1+θ2)z_1z_2 = r_1r_2\,e^{i(\theta_1+\theta_2)}

Multiply the lengths, add the angles. In particular multiplying by i=eiπ/2i = e^{i\pi/2} is exactly a quarter-turn counterclockwise, which is the honest answer to “what does ii mean.”

De Moivre’s theorem is just the multiplication rule applied nn times:

(reiθ)n=rneinθ\big(re^{i\theta}\big)^n = r^ne^{in\theta}

Roots run the same machinery backwards, and this is where polar form earns its place. The equation zn=wz^n = w with w=ρeiφw = \rho e^{i\varphi} has exactly nn solutions:

zk=ρ1/nei(φ+2πk)/n,k=0,1,,n1z_k = \rho^{1/n}\,e^{i(\varphi+2\pi k)/n}, \qquad k = 0, 1, \ldots, n-1

All have the same modulus, so they sit on a circle, and their arguments differ by 2πn\frac{2\pi}{n}, so they are the vertices of a regular nn-gon. The nnth roots of unity are the case w=1w=1.

Two facts that follow immediately and are used constantly:

  • The nnth roots of unity sum to zero for n2n \ge 2, by symmetry of the polygon.
  • They form a cyclic group of order nn under multiplication, which is the connection back to abstract algebra.

Complex analysis is done on subsets of C\mathbb{C}, so it helps to read set descriptions geometrically:

  • za=r|z - a| = r - the circle of radius rr centred at aa
  • za<r|z - a| < r - the open disc
  • za1=za2|z - a_1| = |z - a_2| - the perpendicular bisector of the segment joining them
  • Rez>0\operatorname{Re} z > 0 - the right half plane
  • α<Argz<β\alpha < \operatorname{Arg} z < \beta - a sector
  • r1<za<r2r_1 < |z-a| < r_2 - an annulus, which is where Laurent series live

Example 1: Put z=1+i3z = -1 + i\sqrt{3} into exponential form.

Solution. Modulus:

z=(1)2+(3)2=4=2|z| = \sqrt{(-1)^2+(\sqrt3)^2} = \sqrt{4} = 2

The point is in the second quadrant, so the reference angle arctan(3/1)=60°\arctan(\sqrt3/1) = 60° gives

Argz=180°60°=120°=2π3\operatorname{Arg} z = 180° - 60° = 120° = \frac{2\pi}{3} z=2e2πi/3z = 2e^{2\pi i/3}

Always check the quadrant. A calculator’s arctan(y/x)\arctan(y/x) returns 60°-60° here, which is the wrong point; the picture is what tells you to add 180°180°.

Example 2: Compute (1+i)10(1+i)^{10}.

Solution. In Cartesian form this is a nightmare. In polar form, 1+i=2|1+i| = \sqrt2 and Arg(1+i)=π/4\operatorname{Arg}(1+i) = \pi/4, so

(1+i)10=(2)10e10iπ/4=32e5iπ/2(1+i)^{10} = \left(\sqrt2\right)^{10} e^{10i\pi/4} = 32\,e^{5i\pi/2}

Reduce the angle mod 2π2\pi: 5π22π=π2\frac{5\pi}{2} - 2\pi = \frac{\pi}{2}, so

(1+i)10=32eiπ/2=32i(1+i)^{10} = 32e^{i\pi/2} = 32i

This is the example that sells polar form. Ten binomial expansions replaced by one multiplication and one angle reduction.

Example 3: Find all solutions of z3=8iz^3 = 8i.

Solution. Write the right side in polar form: 8i=8|8i| = 8 and Arg(8i)=π/2\operatorname{Arg}(8i) = \pi/2, so 8i=8eiπ/28i = 8e^{i\pi/2}.

The three cube roots have modulus 81/3=28^{1/3} = 2 and arguments

π/2+2πk3=π6, 5π6, 3π2\frac{\pi/2 + 2\pi k}{3} = \frac{\pi}{6},\ \frac{5\pi}{6},\ \frac{3\pi}{2}

for k=0,1,2k = 0,1,2. So

z=2eiπ/6,2e5iπ/6,2e3iπ/2z = 2e^{i\pi/6},\quad 2e^{5i\pi/6},\quad 2e^{3i\pi/2}

which in Cartesian form is 3+i\sqrt3 + i, 3+i-\sqrt3+i, and 2i-2i.

Check: the three are 120°120° apart on the circle of radius 2, and (3+i)3=8i(\sqrt3+i)^3 = 8i. ✓

Example 4: Show the nnth roots of unity sum to zero for n2n\ge2.

Solution. They are 1,ω,ω2,,ωn11, \omega, \omega^2,\ldots,\omega^{n-1} where ω=e2πi/n\omega = e^{2\pi i/n}. This is a geometric series:

k=0n1ωk=ωn1ω1=11ω1=0\sum_{k=0}^{n-1}\omega^k = \frac{\omega^n-1}{\omega-1} = \frac{1-1}{\omega-1} = 0

valid because ω1\omega \ne 1 when n2n \ge 2. ∎

Geometrically these are the vertices of a regular polygon centred at the origin, and the centroid of a regular polygon is its centre. Algebra and geometry give the same answer, which is the normal situation in this subject.

Example 5: Divide 3+2i14i\frac{3+2i}{1-4i}.

Solution. Multiply above and below by the conjugate of the denominator:

3+2i14i1+4i1+4i=3+12i+2i+8i21+16=5+14i17\frac{3+2i}{1-4i}\cdot\frac{1+4i}{1+4i} = \frac{3+12i+2i+8i^2}{1+16} = \frac{-5+14i}{17} =517+1417i= -\frac{5}{17}+\frac{14}{17}i

The trick is that zzˉ=z2z\bar z = |z|^2 is real, so conjugating the denominator clears the ii from it. This is the only division algorithm you need.

Example 6: Describe the set z1=2z+1|z-1| = 2|z+1|.

Solution. Square both sides using w2=wwˉ|w|^2 = w\bar w, with z=x+iyz = x+iy:

(x1)2+y2=4[(x+1)2+y2](x-1)^2+y^2 = 4\big[(x+1)^2+y^2\big]

Expanding and collecting,

3x2+10x+3+3y2=0x2+103x+y2=13x^2+10x+3+3y^2 = 0 \quad\Longrightarrow\quad x^2+\tfrac{10}{3}x+y^2 = -1

Completing the square:

(x+53)2+y2=2591=169\left(x+\tfrac53\right)^2+y^2 = \tfrac{25}{9}-1 = \tfrac{16}{9}

A circle centred at 53-\frac53 with radius 43\frac43. ∎

Sets defined by a ratio of distances are circles, called Apollonius circles, and they reappear when we study Möbius transformations.

Example 7: Use polar form to derive a trig identity.

Solution. By De Moivre with n=3n=3 and r=1r=1,

(cosθ+isinθ)3=cos3θ+isin3θ(\cos\theta+i\sin\theta)^3 = \cos3\theta+i\sin3\theta

Expand the left side:

cos3θ+3icos2θsinθ3cosθsin2θisin3θ\cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta

Match real parts:

cos3θ=cos3θ3cosθsin2θ=4cos3θ3cosθ\cos3\theta = \cos^3\theta-3\cos\theta\sin^2\theta = 4\cos^3\theta-3\cos\theta

Multiple-angle formulas are a corollary of complex multiplication. This is the cleanest way to generate them, and it is why trigonometry looks so much easier once you have eiθe^{i\theta}.

Phasors in electrical engineering. A sinusoidal voltage Vcos(ωt+ϕ)V\cos(\omega t+\phi) is stored as the single complex number VeiϕVe^{i\phi}. Adding two sinusoids of the same frequency then becomes adding two complex numbers, which converts a trigonometry problem into arithmetic. This is the standard method in every power systems course.

Rotations in graphics. Multiplying by eiθe^{i\theta} rotates a 2D point about the origin, which is cheaper and more numerically stable than a matrix multiply. The 3D generalization is the quaternion, used in essentially every game engine for smooth camera and character rotation.

Roots of unity and the FFT. The fast Fourier transform is built on the algebraic relations among the nnth roots of unity. It is arguably the most-executed nontrivial algorithm in the world, and it exists because those roots form a group.

Signal representation. Amplitude and phase are precisely modulus and argument. Any tool that displays a spectrum is showing you z|z| and argz\arg z for a family of complex numbers.

Impedance matching. The Smith chart, standard in radio engineering, is a plot in the complex plane using a Möbius transformation, and the circles on it are Apollonius circles like the one in Example 6.

Crystallography and symmetry. Regular polygons of roots of unity describe the rotational symmetries available to a crystal lattice, which is why fivefold symmetry was long thought impossible in crystals.

What is the modulus of z = −3 + 4i?
What is the principal argument of z = −1 + i√3?
Geometrically, what does multiplying by i do to a complex number?
Using polar form, what is (1 + i)^10?
How many solutions does z⁵ = 32 have in ℂ, and where do they lie?
Which identity is used to divide complex numbers?
What does the set |z − 2| = 3 describe?
Why do the nth roots of unity sum to zero for n ≥ 2?