Foundations of Complex Functions and Their Visualization

Beware that, when fighting monsters, you yourself do not become a monster… for when you gaze long into the abyss. The abyss gazes also into you, Friedrich W. Nietzsche

Recall

A complex number is specified by an ordered pair of real numbers (a, b) ∈ ℝ² and expressed or written in the form z = a + bi, where a and b are real numbers, and i is the imaginary unit, defined by the property i² = −1 ⇔ i = $\sqrt{-1}$, e.g., 2 + 5i, $7\pi + i\sqrt{2}.$ ℂ= { a + bi ∣a, b ∈ ℝ}.

Definition. Let D ⊆ ℂ be a set of complex numbers. A complex-valued function f of a complex variable, defined on D, is a rule that assigns to each complex number z belonging to the set D a unique complex number w, f: D ➞ ℂ.

• The set D is called the domain of the complex function f, D = Dom(f).
• The set of all actual outputs {f(z): z ∈ D} is called the range (or image) of the complex function f, also denoted f(D) = {f(z): z ∈ D}.

We often call the elements of D as points. If z = x+ iy ∈ D, then f(z) is called the image of the point z under f. f: D ➞ ℂ means that f is a complex function with domain D. We often write f(z) = u(x ,y) + iv(x, y), where u, v: ℝ² → ℝ are the real and imaginary parts.

In complex analysis, roots, powers, and polynomials behave much like in the real setting —but with extra richness and complexity due to arguments (angles) and multiple values.

Complex square root

For any nonzero complex number z = x + iy = $re^{i\theta}$

$f(z) = \sqrt{z} = \sqrt{|z|e^{iArg(z)}},~k = 0, 1.$ The square roots of $z$ are the two numbers $w$ satisfying $w^2 = z$. In exponential form: w = $\sqrt{|z|}e^{i\frac{θ+2\pi·k}{2}}$, k = 0, 1.

• Principal square root (k = 0): $\sqrt{z} = \sqrt{|z|}e^{i\frac{θ}{2}} = \sqrt{|z|}(cos(\frac{θ}{2})+ isin(\frac{θ}{2})) = \sqrt{x²+y²}cos(\frac{θ}{2}) + i\sqrt{x²+y²}sin(\frac{θ}{2}), u = \sqrt{x²+y²}cos(\frac{θ}{2}), v = \sqrt{x²+y²}sin(\frac{θ}{2})$
• The other root (k = 1) picks up the minus sign: $-\sqrt{z} = \sqrt{|z|}e^{i(\frac{θ}{2}+\pi))} = \sqrt{|z|}(cos(\frac{θ}{2}+\pi)+ isin(\frac{θ}{2}+\pi)) = \sqrt{x²+y²}cos(\frac{θ}{2}+\pi) + i\sqrt{x²+y²}sin(\frac{θ}{2}+\pi), u = \sqrt{x²+y²}cos(\frac{θ}{2}+\pi), v = \sqrt{x²+y²}sin(\frac{θ}{2}+\pi)$

n-th Roots of a Complex Number

More generally, the $n$-th roots of z = x + iy = $re^{i\theta}$, r > 0 are the n distinct solutions of $w^n = z$, namely

$w_k = \sqrt[n]{z} = \sqrt[n]{r}e^{i\frac{\theta + 2\pi·k}{n}}$,k = 0, 1, …, n − 1.

Equivalently, $w_k = \sqrt[n]{r}(cos\frac{\theta + 2\pi·k}{n}+isin\frac{\theta + 2\pi·k}{n})$

• Geometric representation: these $w_k$ lie at the vertices of a regular n-gon inscribed in the circle of radius $r^{1/n}$.

• Real-axis special case: if $z=x$ is positive real ($\theta=0$), then $w_k = x^{1/n} e^{2\pi i k / n}$ are the well‐known “roots of unity” times $x^{1/n}$.

• Branch choice: one often designates $k=0$ as the principal $n$-th root.

Complex Exponentiation

For any complex number z and any complex power p, we can express the exponentiation as: $z^p = e^{log(z^p)} = e^{p∙log(z)}$ where $\textbf{Log(z)}$ is the (multi-valued) complex logarithm.

Recall the Complex Logarithm. The complex logarithm of a complex number z can be written as: $\textbf{Log(z)} = \textbf{ln|z|} + i(Arg(z)+2πk), k \in ℤ$ where ln denotes the natural logarithm of the magnitude (or modulus) of z, denoted as ∣z∣ and Arg represents the principal argument of z (typically in the range (-π, π]) or angle of z in the complex plane, which is the angle formed with the positive real axis. k is an integer that accounts for the multivalued nature of the logarithm due to the periodicity of the argument, as the angle can be expressed in multiple equivalent forms.

Applying to iⁱ, we have (z = i and p = i) $i^i = e^{i∙Log(i)}$

Logarithm of i: For z = i: $Log(i) = ln|i| +i(\pi/2+2\pi k)=i(\pi/2+2\pi k)$ since $|i|=1, ln(1)=0$.

$i^i = e^{i∙Log(i)} = e^{i²(π/2 + 2πk)}$ = {e^-(π/2+2πk) | k ∈ ℤ}

Example: $(1+i)^{2+3i} =$[By definition] $e^{(2+3i)log(1+i)}$ =[Log(z) = ln∣z∣ + i(Arg(z)+2πk)] $e^{(2+3i)[ln|1+i| + i(Arg(1+i)+2πk)]} = e^{(2+3i)[ln(\sqrt{2}) + i(\frac{π}{4}+2πk)]} = e^{2ln(\sqrt{2}) -3(\frac{π}{4}+2πk)+i(3ln(\sqrt{2})+2(\frac{π}{4}+2πk))} = e^{2ln(\sqrt{2}) -3(\frac{π}{4}+2πk)+i(3ln(\sqrt{2})+\frac{π}{2}+4πk)} = e^{ln(2)+\frac{-3π}{4}-6πk+i(3ln(\sqrt{2})+\frac{π}{2}+4πk)} = 2e^{\frac{-3π}{4}-6πk+i(3ln(\sqrt{2})+\frac{π}{2}+4πk)}$, ,k ∈ ℤ.

The Complex Quadratic Formula

Consider the general quadratic in ℂ, $az² + bz + c = 0$ where a, b and c ∈ ℂ, a ≠ 0.

Since a ≠ 0, we can divide by a, $z²+ \frac{b}{a}z + \frac{c}{a} = 0$, and then complete the square, $(z+ \frac{b}{2a})² -(\frac{b}{2a})² + \frac{c}{a} = 0$

Rearrange to isolate the square: $(z+ \frac{b}{2a})² = \frac{b²-4ac}{4a²}$

$z+ \frac{b}{2a} = \plusmn \sqrt{\frac{b²-4ac}{4a²}}$

Recall that $w^{\frac{1}{2}} = $ {$\sqrt{w}, -\sqrt{w}$} where $\sqrt{w} = \sqrt{|w|}e^{i\frac{Arg(w)}{2}}$ [*]

Therefore, z = $-\frac{b}{2a} ± \sqrt{\frac{b²-4ac}{4a²}} = \frac{-b ± \sqrt{b²-4ac}}{2a}$ but take into consideration that $\sqrt{b²-4ac}$ is a complex expression.

Example. Solve $z²+(2−2𝑖)z−(7+26𝑖)=0$

$z = \frac{-b ± \sqrt{b²-4ac}}{2a} = \frac{-(2-2i) ± \sqrt{(2-2i)²+4(7+26i)}}{2} = \frac{-(2-2i) ± \sqrt{−8i+28+104𝑖}}{2} = \frac{-(2-2i) ± \sqrt{28+96i}}{2}$

Let’s compute $\sqrt{28+96i}, |\sqrt{28+96i}| = \sqrt{28²+96²} = 100$.

Since the complex number lies in the first quadrant, its argument is given by 𝜃$ = arctan(\frac{96}{28}) = arctan(\frac{24}{7}) ≈ 1.287$ radians.

The two square roots are given by $±\sqrt{100}e^{\frac{\theta}{2}} = ±10(cos(\frac{\theta}{2})+isin(\frac{\theta}{2})) ≈ ±10(0.8+i0.6) = ±(8+6𝑖)$

$z = \frac{-(2-2i) ± \sqrt{28+96i}}{2} = \frac{-(2-2i) ±(8+6𝑖)}{2} = −1+i±(4+3𝑖)$. Hence, the two roots or solutions of the complex quadratic equation are z = 3 + 4i and z =−5 − 2i.

Visualizing complex functions

Because complex numbers have two components (real and imaginary), visualizing a function f: ℂ → 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)).

Examples

Let’s clarify this idea on a few examples.

• The function f(z) = z + b, where b is a complex constant, represents a translation in the complex plane. If z = x + iy and b = b₁ + ib₂, then: f(z) = (x + b₁) + i(y + b₂)

This means that every point z in the complex plane is shifted by b₁ units in the real direction and b₂ units in the imaginary direction.

As an example, let’s take the unit circle {z: |z| = 1}. Draw the output set in another complex plane where w = f(z). The transformation f(z) = z + b shifts the unit circle by b. So, the range is the circle f({z: |z| = 1}) = {z: |z - b| = 1}, a circle of radius 1 centered at b (Figure 1).

• Dilation & Rotation: The function f(z) = az where z ∈ ℂ and a is a fixed non-zero constant complex, a = r(cos(θ) + isin(θ)) where r = |a| is the magnitude (or modulus) of a, and θ = Arg(a) is the argument (or angle) of a.

z = $\rho(cos(\phi)+ isin(\phi))$ where ρ = |z| is the magnitude of z, and $\phi$ = Arg(z) is the argument of z. When you multiply two complex numbers in polar form, the magnitudes multiply, and the arguments add: $f(z) = r\rho(cos(\theta+\phi)+isin(\theta+\phi))$ (Figure 2).

Modulus scaling

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

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).

• A more generalized example would be the general linear transformation f(z) = az +b, z ∈ ℂ. It can be visualized as follows:

1. z is elongated (stretched, |a | > 1) or contracted (shrunk, |a | < 1) by a factor of |a|.
2. 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.

• Cubing. f(z) = z³ ∀z ∈ ℂ, Using the polar form of complex numbers, where z = r(cos(θ) + i sin(θ)) = re^iθ, we have: $f(r(cos(\theta)+isin(\theta)) = r³(cos(3\theta)+isin(3\theta))$

Geometrically:

The modulus (magnitude) of f(z) is r³. This means:

1. If r > 1, the magnitude is elongated (cubed).
2. If 0 < r < 1, the magnitude is contracted (cubed).
3. If r = 1, the magnitude remains unchanged.

The argument (angle) of f(z) is 3θ. This means that the complex number z is rotated counterclockwise around the origin by an angle of 3θ (Figure A).

The function f(z) = z³ maps each sector of the complex plane with an angle of 2π/3 onto the entire complex plane.

1. The sector 0 ≤ θ < 2π/3 is mapped to the full range of angles from 0 to 2π.
2. The sector 2π/3 ≤ θ < 4π/3 is also mapped to the full range of angles from 0 to 2π.
3. The sector 4π/3 ≤ θ < 2π is also mapped to the full range of angles from 0 to 2π.

This means that every non-zero complex number in the codomain has three distinct pre-images (roots) in the domain.

Roots interpretation: Since z³ = w has three solutions (the three cube roots of w), every nonzero point w in the codomain has three distinct pre‐images in the domain.

• In general, power maps, f(z) = zⁿ.Using the polar form z = r(cos(θ) + i sin(θ)) = re^iθ, we have:

f(z) = zⁿ = (re^iθ)ⁿ = rⁿe^inθ = rⁿ(cos(nθ) + i sin(nθ))

The modulus of f(z) is rⁿ. So:

1. If r > 1, the magnitude grows (stretch); it is raised to the nth power (elongation).
2. If 0 < r < 1, the magnitude shrinks (contract); it is raised to the nth power (contraction).
3. If r = 1, the magnitude is preserved or remains unchanged (it is just a pure rotation).

The argument of f(z) is nθ. So, z is rotated counterclockwise around the origin by an angle of nθ - a winding around the origin.

The function f(z) = zⁿ maps each sector of the complex plane with an angle of 2π/n onto the entire complex plane. The sectors are:

1. 0 ≤ θ < 2π/n
2. 2π/n ≤ θ < 4π/n
3. 4π/n ≤ θ < 6π/n
4. …
5. (2(n-1)π)/n ≤ θ < 2π

Consequently,every non-zero complex number w in the codomain has n distinct pre-images (nth roots) in the domain.