Visualizing Complex Functions

Enjoy life. This is not a dress rehearsal, Friedrich Nietzsche

Visualizing complex functions

Because complex numbers have two components (real and imaginary), visualizing a function $f: \mathbb{C} \to \mathbb{C}$ requires representing both input (domain) and output (range) in a two-dimensional plane. This is often done by using two separate complex planes, a dual-plane visualization:

1. One complex plane for the domain (input z).
2. A second, separate complex plane for the codomain (output w = f(z)).

You then visualize how geometric objects (points, lines, circles, regions) in the z-plane are transformed into objects in the w-plane. Let’s clarify this idea on a few examples.

Examples

• Translation: $f(z) = z + b$, represents a translation (shift) in the complex plane. For $z = x + iy$ and $b = b_1 + ib_2$: $\boxed{f(z) = (x + b_1) + i(y + b_2)}$

This means that every point z in the complex plane is shifted by $b_1$ units in the real direction and $b_2$ units in the imaginary direction.

Property	Value
Fixed points	None (unless $b = 0$)
Preserves	Distances, angles, shapes
Type	Rigid motion (isometry)

As an example, let’s take the unit circle. Draw the output set in another complex plane where w = f(z).

• In the z-plane: circle centered at 0 with radius 1, {z: |z| = 1}.
• In the w-plane under w = z + b: image is $w = f(z) \iff z = w - b, |z| = 1 \iff |w - b| = 1$. So the range is the circle {w : |w - b| = 1}: a circle of radius 1 centered at b.

The transformation f(z) = z + b shifts the unit circle by b (Figure 1).

• Dilation & Rotation: The function f(z) = az where z ∈ ℂ and a is a fixed non-zero constant complex.

Let $a = re^{i\theta}$ where $r = |a|$ and $\theta = \arg(a)$. For $z = \rho(cos(\phi)+ isin(\phi)) = \rho e^{i\phi}$: $f(z) = az = r\rho \cdot e^{i(\theta + \phi)} = r\rho(cos(\theta+\phi)+isin(\theta+\phi))$

When you multiply two complex numbers in polar form, the magnitudes multiply, and the arguments add (Figure 2).

Modulus scaling (dilation) $|f(z)| = r\rho = |a||z|$

The magnitude of the result, rρ, is the product of the magnitudes of a and z. Therefore:

1. If |a| = r > 1, the transformation elongates (stretches) z by a factor of r.
2. If |a| = r < 1, the transformation shrinks (contracts) z by a factor of r.
3. If |a| = r = 1, the transformation preserves the magnitude of z (it’s a pure rotation).

Argument shift $arg(f(z)) = \theta + \phi$

The argument of the result, θ + φ, is the sum of the arguments of a and z. Therefore, the transformation rotates every point counterclockwise by an angle of θ (the argument of a).

Geometric summary

Circles centered at origin |z| = R map to circles {|w| = rR}. Rays (lines from origin at angle $\phi$) map to rays at angle $\theta + \phi$.

Example, $i = e^{i\pi/2}$ is a rotation by $90°$ counterclockwise. $f(z) = iz: \quad 1 \mapsto i, \quad i \mapsto -1, \quad -1 \mapsto -i, \quad -i \mapsto 1$.

• General Linear Transformation: $f(z) = az + b, a \ne 0$. It can be visualized as follows (a three-step process):

1. Scale. z is elongated (stretched, |a | > 1) or contracted (shrunk, |a| < 1) by a factor of |a|.
2. Rotate. Then, the multiplication by a also rotates z counterclockwise around the origin by the angle of arg(a).
3. Finally, the addition of b translates (shifts) the resulting complex number by the vector represented by b. If b = c + id, the translation is c units in the real direction and d units in the imaginary direction.

Property	Value
Fixed points	$z_0 = \frac{b}{1-a}$ (if $a \neq 1$)
Preserves	It is a conformal (angle-preserving) map everywhere
Maps	It maps lines and circles in the z-plane to lines or circles in the w-plane.

Example: $f(z) = (1 + i)z + 2$. Let’s pick z = 1 (the point (1, 0) on the real axis).
(i) $|1 + i| = \sqrt{2}$. Scale by $\sqrt{2}$. The distance from the origin grows from 1 to $\sqrt{2}$.
(ii) $\arg(1 + i) = \frac{\pi}{4}$. Rotate by $45°$. The point moves from the positive real axis to the line y = x in the first quadrant. Current position: $\sqrt{2}e^{\frac{i\pi}{4}} = 1 + i$.
(iii) Translate by $2$ (shift right by 2 units). Final position: (1 + i) + 2 = 3 +i. The point (1, 0) was stretched away from the origin, spun counter-clockwise by $45\degree$, and then slid 2 units to the right.

1. The definition of a fixed point $z_0$ is a point that maps to itself. It does not move. Therefore, $az_0 + b = z_0 \implies b = z_0 - az_0 = z_0(1 -a) \implies[a \ne 1] z_0 = \frac{b}{1-a}$
2. If a = 1, the function is f(z) = z + b (a pure translation). If $b \ne 0$, no point is fixed (everything moves). If b = 0, every point is fixed (f(z) = z).
3. From the fixed point definition $az_0 + b = z_0$. Substitute this expression for b back into the function: $f(z) = az + b = az + (z_0 - az_0) =[\text{Rearrange terms}] a(z-z_0) + z_0$. Then, $f(z) = a(z-z_0) + z_0 \implies f(z) - z_0 = a(z - z_0)$. This shows that the transformation is a rotation and scaling centered at the fixed point $z_0$, not at the origin! The translation effectively shifts the center of rotation from 0 to $z_0$.
   

   a(z-z₀) scales and rotates by a the vector from the fixed point z₀ to z
   

   Local shapes are preserved. A small face drawn on the plane will look like the same face (just rotated and scaled) after the transformation.

• The Cubing Function. f$f(z) = z^3$ ∀z ∈ ℂ, Using the polar form of complex numbers, where z = r(cos(θ) + i sin(θ)) = re^iθ, we have: $f(re^{i\theta}) = r^3 e^{i \cdot 3\theta} = r^3(\cos 3\theta + i\sin 3\theta)$

Key observations:

• Modulus: $|f(z)| = r^3$ (i) If (r > 1): magnitude grows quickly (elongation). (ii) If (0 < r < 1): magnitude shrinks (contraction). (iii) If (r = 1): modulus preserved (point stays on unit circle).
• Argument: $\arg(f(z)) = 3\theta$. So every point is rotated counterclockwise around the origin by tripling its angle (Figure A).

Sector mapping:

Because the argument is multiplied by 3, the map $z \mapsto z^3$ sends each angular sector of width $\frac{2\pi}{3}$ onto the entire angular range $[0, 2\pi)$:

1. Sector $0 \le \theta < \frac{2\pi}{3} \mapsto$ full angle range (from 0 to 2π) after cubing (entire plane).
2. Sector $\frac{2\pi}{3} \le \theta < \frac{4\pi}{3}\mapsto$ full range again (entire plane).
3. Sector $\frac{4\pi}{3} \le \theta < 2\pi \mapsto$ full range again (entire plane).

Roots interpretation: Since $z^3 = w$ has three solutions for each nonzero w (the three cube roots of w), every nonzero complex number w in the codomain has three distinct pre‐images under $f(z) = z^3$.

• Power Functions:, $f(z) = z^n$. Using the polar form z = r(cos(θ) + i sin(θ)) = re^iθ, we have: $f(z) = z^n = r^n e^{in\theta} =[\text{De Moivre’s theorem}] r^n(\cos(n\theta) + i(\sin(n\theta))).$

Effect on modulus: $|f(z)| = r^n$, $r \mapsto r^n$

1. If r > 1: the magnitude grows (stretched or elongated -more dramatic for larger n).
2. If 0 < r < 1: the magnitude shrinks (contracted).
3. If r = 1: the magnitude is preserved or remain unchanged (it is just a pure rotation).

Effect on argument: $\arg(f(z)) = n\theta$. So z is rotated counterclockwise around the origin by an angle $n\theta$.

Sector mapping:

The function $f(z) = z^n$ maps each sector of the complex plane with angle $2\pi/n$ onto the entire complex plane.

The sectors $0 \le \theta < \frac{2\pi}{n}, \frac{2\pi}{n} \le \theta < \frac{4\pi}{n}, \frac{4\pi}{n} \le \theta < \frac{6\pi}{n}, \ldots, \frac{2(n-1)\pi}{n} \le \theta < 2\pi$ are sent (via angle multiplication by n) to the entire angular range $[0, 2\pi)$.

Therefore, every non-zero complex number w in the codomain has n distinct pre-images (its n-th roots) in the domain. This matches the algebraic fact: the equation $z^n = w$ has exactly n distinct solutions in $\mathbb{C}$ for each nonzero w.

• The Exponential: $f(z) = e^z$. For $z = x + iy$: $e^z = e^{x + iy} = e^x\cdot e^y = e^x(\cos(y) + i\sin (y))$. There’s a “division of labor”, the real part (x) controls the magnitude (radius r = $e^x$) and the imaginary part controls the angle (argument $\theta = y$).

Key properties:

Input	Output
Horizontal line $y = c$	Ray from origin at angle $c$
Vertical line $x = c$	Circle of radius $e^c$
Horizontal strip $0 \leq y < 2\pi$	Entire plane minus origin
Left half-plane $x < 0$	Unit disk minus origin
Right half-plane $x > 0$	Exterior of unit disk

1. If y is constant (y = c), then $cos(y) + isin(y)$ is a fixed complex number on the unit circle at angle c (say u). As x goes from from −∞ to +∞, $e^x$ goes from 0 to ∞. We are taking a fixed vector u and stretching it by a scalar factor $e^x$, hence tracing a straight line (a ray) starting at the origin (but never touching it, since $e^x \ne 0$) and pointing in the direction of angle c.
2. If x is contant (x = c), then $e^x$ is a fixed positive number, namely $e^c$. As y varies, cos(y) + isin(y) traces the unit circle. Multiplying by the fixed constant $e^c$ scales the unit circle up to a circle of radius $e^c$.
3. Horizontal strip $0 \leq y < 2\pi$. The bottom edge (y = 0) maps to the positive real axis ($e^x > 0$). The top edge ($y \to 2\pi^-$) also maps to the positive real axis $e^x(cos(2\pi) + isin(2\pi)) = e^x$. Imagine taking the infinite strip of the complex plane and rolling it up like a scroll. The left side ($x \to -\infty$) pinches to a point (the origin, but never reaches it), and the right side ($x \to \infty$) explodes outwards.
4. Left Half-Plane (x < 0): Since x is negative, $e^x$ is a fraction ($0 \lt e^x \lt 1$). This maps to all points whose distance from the origin is less than 1 (the open unit disk), excluding the origin itself.
5. Right Half-Plane (x > 0): Since x is positive, $e^x$ is greater than 1 ($e^x \gt 1$). This maps to all points outside the unit circle.
6. Boundary (x = 0): The imaginary axis maps exactly to the unit circle (∣z∣ = 1).
7. Periodicity: $e^{z + 2\pi i} = e^z$ (period $2\pi i$). $e^{z + 2\pi i}= e^{x} \cdot e^{i(y + 2\pi)} = e^x(\cos(y + 2\pi) + i\sin (y + 2\pi)) = e^x(\cos(y) + i\sin (y)) = e^x$. In the real number, $e^x$ is one-to-one. In complex numbers, $e^z$ is infinite-to-one. There are infinitely many inputs that give the same output.

   The complex plane could be imagine as an infinite parking garage. The exponential function “crushes” the entire garage into a single level (the punctured plane), where every spot on the ground floor corresponds to a spot on every floor of the garage.

• The Logarithm: $f(z) = \log(z)$. Inverse of exponential (multi-valued): $\log(z) = \ln|z| + i(\arg(z) + 2\pi k)$. Principal branch $\text{Log}(z) = \ln|z| + i\text{Arg}(z)$.

1. Magnitude: $u = ln(r) = ln(|z|)$.
2. Angle: $v = \theta \text{ plus multiples of } 2\pi = \text{Arg}(z)$
3. In words, the real part of the log is the log of the distance. The imaginary part of the log is the angle.

Input	Output
Circle $\vert z \vert = r$	Vertical line $x = \ln (r)$
Ray at angle $\theta$	Horizontal line $y = \theta$
Slit plane $\mathbb{C} \setminus (-\infty, 0]$	Strip $-\pi < y \leq \pi$

1. Circle ∣z∣ = r $\to$ Vertical Line x = ln(r). On a circle, the distance from the origin is constant. Since u=ln(r), the real part of the output is fixed at ln(r). The angle θ (which becomes the imaginary part v) sweeps freely from 0 to 2π (or −π to π) as you go around the circle.
2. Ray at angle $\theta \to$ Horizontal Line $y = \theta$. On a ray, the angle θ is constant. Since v = θ, the imaginary part of the output is fixed at θ. The distance r (which becomes the real part u=ln(r)) varies from 0 to ∞ as you move along the ray. It is basically taking a straight ray (spoke of a wheel) and laying it flat. The angle of the spoke determines the height (y-value) of the line (y is constant and x varies from $-\infty$ (as $r \to 0$) to $\infty$ (as $r \to \infty$)).
3. Slit Plane $\mathbb{C} \setminus (-\infty, 0] \to$ Strip $-\pi < y \leq \pi$. In the complex plane, angles wrap around, so we force the imaginary part (the angle) to stay in a specific range, usually(−π, π]. This is the Principal Value Log(z). The positive real axis (θ = 0) maps to the real axis (y = 0). The upper half-plane (0 < θ < π) maps to the top half of the strip (0 < y < π). The lower half-plane (−π < θ < 0) maps to the bottom half of the strip (−π < y < 0).
   The slit (the edge of the cut) maps to the boundaries of the strip (y = π and y = −π). Because the range is (−π, π], the value π is allowed. The “top edge” of the slit maps exactly to the line y = π.