Mathematics is not a careful march down a well-cleared highway, but a journey into a strange wilderness, where the explorers often get lost, – W.S. Anglin.

Complex numbers

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. The principal nth root of w, denoted $\sqrt[n]{w}$, is defined as: $\sqrt[n]{w} = \sqrt[n]{|w|}e^{i(\frac{Arg(w)}{n})}$.

Definition. The set of all nth roots of a complex number w, denoted $w^{\frac{1}{n}}$, consists of n distinct values. It is given by: $w^{\frac{1}{n}}$ = {$\sqrt[n]{w}, w_n\sqrt[n]{w}, w_n²\sqrt[n]{w}, ..., w_n^{n-1}\sqrt[n]{w}$} = $|w| e^{i \text{Arg}(w)}$, where:

|w| is the magnitude (modulus) and $\text{Arg}(w)$ is the argument (angle) of w, $\text{Arg}(w) ∈ (−π,π].$
$w_n=ζ_n=e^{\frac{2πi}{n}}$ is a primitive nth root of unity
The index k ranges over integers from 0 to n − 1. Using k = 0, 1, … , n−1 ensures exactly n distinct roots.

Exercises

Given z = $(1+i\sqrt{3}),\text{ compute } z^{\frac{1}{4}}$

Step 1. Write z in polar form:

|z| = $\sqrt{1² +(\sqrt{3})²} = \sqrt{4} = 2, Arg(z) = arctan(\frac{\sqrt{3}}{1}) = arctan(\sqrt{3}) = \frac{\pi}{3}$. Thus, the polar for of z is $z = 2e^{\frac{\pi}{3}i}$

Step 2. Calculating the Principal Fourth Root:

To find $z^{\frac{1}{n}}$, we use De Moivre’s Theorem, which states that for any complex number z = re ^iθ, its n-th roots are given by: $z^{\frac{1}{n}} = r^{\frac{1}{n}}e^{i(\frac{θ+2\pi k}{n})}$, k = 0, 1, 2, …, n − 1. For the principal (or primary) fourth root (k = 0, n = 4):

$z^{\frac{1}{4}} = (2e^{\frac{\pi}{3}i})^{\frac{1}{4}} = 2^{\frac{1}{4}}e^{\frac{\pi}{3∙4}i}= \sqrt[4]{2}e^{\frac{\pi}{12}i}$

Step 3. Finding All Fourth Roots:

Alternatively, $z^{\frac{1}{n}} = r^{\frac{1}{n}}e^{i(\frac{θ+2\pi k}{n})} = r^{\frac{1}{n}}e^{i(\frac{θ}{n})}e^{i(\frac{2\pi k}{n})}$. $w_n = e^{i(\frac{2\pi}{n})}, \sqrt[n]{w} = \sqrt[n]{|w|}e^{i\frac{Arg(w)}{n}} = \sqrt[4]{2}e^{\frac{\pi}{12}i}$.

To find all four fourth roots of z, $w^{\frac{1}{n}} = ${$r^{\frac{1}{n}}\sqrt[n]{w}, r^{\frac{1}{n}}w_n\sqrt[n]{w}, r^{\frac{1}{n}}w_n²\sqrt[n]{w}, ..., r^{\frac{1}{n}}w_n^{n-1}\sqrt[n]{w}$} =[ n = 4] {$r^{\frac{1}{4}}\sqrt[4]{w}, r^{\frac{1}{4}}w_4\sqrt[4]{w}, r^{\frac{1}{4}}w_n²\sqrt[4]{w}, r^{\frac{1}{4}}w_n^{3}\sqrt[4]{w}$}

n = 4, $w_4 = i = e^{\frac{\pi}{2}i} = e^{\frac{6\pi}{12}i}$

Therefore, the four fourth roots of z are: $z^{\frac{1}{4}} =$ {$\sqrt[4]{2}e^{\frac{\pi}{12}i}, \sqrt[4]{2}e^{\frac{7\pi}{12}i}, \sqrt[4]{2}e^{\frac{13\pi}{12}i}, \sqrt[4]{2}e^{\frac{19\pi}{12}i}$} It’s often helpful to visualize the roots on the complex plane. Each fourth root lies at an angle of π/12, 7π/12, 13π/12, and 19π/12 radians from the positive real axis, each spaced π/2 radians apart (i.e., 90 degrees) scaled by $\sqrt[4]{2}$

What is wrong with $1 = -i² = -i∙i = -\sqrt{-1}∙\sqrt{-1} = \sqrt{(-1)∙(-1)} = \sqrt{1} = 1$?

The rule $\sqrt{a}∗\sqrt{b}=\sqrt{ab}$ is only valid when at least one of a or b is a non-negative real number. The error is in the illicit application of the rule for multiplying square roots to negative numbers.

Complex-Valued Functions: Definitions, Domain, Range

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.

Key examples

Constant functions: $f: \Complex \rarr \Complex$ given by f(z) = c where c ∈ ℂ is a constant. Domain: any chosen D ⊂ ℂ, Range: {c}.
Polynomial functions: $f: \Complex \rarr \Complex$ defined by f(z) = $a_nz^n + a_{n-1}z^{n-1} + ··· + a_0$ where n is a non-negative integer, a_n, a_n-1, ··· , a₀ are complex constants, a_n ≠ 0, e.g., f(z) = 6z² -(2+4i)z + 12 -$\sqrt{3}i$. Domain: ℂ, degree n. The hightest prior of z carrying a non-zero coefficient is 2, so we say that f is a second degree polynomial.
Rational functions. Let p(z) and q(z) be polynomial functions where $q \not\equiv 0$. Then, $\frac{p(z)}{q(z)}$ is called a rational function. The domain of f is the set {z ∈ ℂ : q(z) ≠ 0}, e.g., f(z) = $\frac{az+b}{cz+d}$ where a, b, c, d ∈ ℂ, b and d are not both zeroes is a rational function. f is called a linear fractional transformation. $Domain(f) = \begin{cases} ℂ -{\frac{-d}{c}}, &c ≠ 0 \\\\ ℂ, &c = 0 ~, (d ≠ 0) \end{cases}$

Example: $g(z) = \frac{3z²-(2+i)z}{2z -3i}$ for each z ∈ ℂ - {$\frac{3i}{2}$}

Transcendental Functions

Exponential: $e^z$ is entire. Its domain is all ℂ and it never takes the value zero.
An entire function in complex analysis is a function that is holomorphic (complex differentiable) at every point in the complex plane (ℂ). Essentially, it’s a complex function that is smooth and well-behaved everywhere in the complex plane.
Logarithm: Log{z} has branch cuts; typical domain $\mathbb{C} \setminus (-\infty, 0]$ —excludes that branch to avoid discontinuity. This choice ensures it behaves nicely in the rest of the complex plane.
Trigonometric: sin(z) and cos(z) are entire;
Power and root functions z^α require choices of branch for α ∉ ℤ.

Real and imaginary parts of a complex function

Let f(z) = z³ ∀z ∈ ℂ, f(x + iy) = x³ -3xy² + i(3x²y -y³).

We will call x³ -3xy² to be the real part of f and 3x²y -y³ to be the imaginary part of f.

In general, every complex function f, $f: D \rarr \Complex$ can be written in terms of its real and imaginary parts: f(x, y) = u(x, y) + iv(x, y) ∀x + iy ∈ D, where u and v are real-valued functions on the set of (x, y) corresponding to D, u, v: ℝ² → ℝ; the u(x, y) is called the real part of f and v(x, y) is called the imaginary part.

D ⊆ ℂ. If D is thought of as a subset of ℝ², then $u: D ⊆ ℂ \rarr \Complex \text{ and } v: D ⊆ ℂ \rarr \Complex$ are real valued functions of two real variables.

Domain may exclude points where f isn’t defined/blows up (poles, e.g., f(z) = $\frac{1}{z}$ is undefined at z = 0) or is multivalued (e.g., $\sqrt{z}$).
Range can be all of ℂ (e.g. nonconstant polynomials) or omit values (e.g. exponential never attains 0).

Some examples

Polynomial. f(z) = z². Formula: f(x +iy) = (x +iy)² = x² -y² + (2xy)i. Hence u(x, y) = x² -y², v(x, y) = 2xy. D = ℂ. Concrete evaluation: f(3 + i) = (3 + i)² = 9 - 1 + 6i = 8 + 6i.

We can restrict D to be B(0; 1) ⊆ ℂ, the same formula but with domain the unit disk and write f: B(0, 1) → ℂ given by f(z) = z². This complex function f is a restriction of the original function defined on all of ℂ.

Rational. Formula: $g(z) = \frac{z -1}{iz²}$ $\forall z \in$ ℂ \ {0}. Domain: ℂ \ {0} (and z ≠ 0 already rules out the denominator zero). Singularity: z = 0.
Polynomial degree 3. f(z) = z³, f(x +iy) = (x +iy)³ = $x³+3x²(iy)+3x(iy)²+(iy)³ = x³-3xy²+i(3x²y-y³)$ where $u = x³-3xy², v = 3x²y-y³$. Domain: ℂ.
A function is injective (one‐to‐one) if each w in the range comes from exactly one z.
If more than one value of w corresponds to each value of z, e.g, f(z) = $\sqrt{z}$ we say that is not single‐valued on ℂ∖{0} or is a multiple-valued function.

Complex exponential

If z = x + iy the exponential function e^z, exp: $\mathbb{C} → \mathbb{C}$ may be defined in many equivalent ways:

e^z = e^xe^iy =[Euler’s Formula] e^x(cos(y) + isin(y))
e^z = $\lim_{n \to \infin}(1+\frac{z}{n})^n$
e^z = $\sum_{n=0}^{\infty} \frac{z^n}{n!}$.

e^xcos(y) + e^xsin(y)i where u = e^xcos(y), v = e^xsin(y), and |z| = |e^z| = |e^xe^iy| = |e^x|∙|e^iy| =[e^iy is a point on the unit circle in the complex plane, $|e^{iy}| = |cos(y) + isin(iy)| = \sqrt{cos²(y)+sin²(y)} = 1$, e^x ≥ 0 ∀ x] e^x∙1 = e^x

arg(e^z) = y + 2πk (k = 0, ±1, ±2, …). Re(e^z) = e^xcos(y), Im(e^z) = e^xsin(y)

For any complex number z, the exponential function $e^z$ is never zero.
The range of the exponential function is the set of all non-zero complex numbers, denoted as $\mathbb{C}^*$ or $\mathbb{C} \setminus \${0}. This means that for any non-zero complex number w, there exists a complex number z such that $e^z = w$.
It coincides with the real exponential function when z is real (y = 0).
It has all the properties of power laws, e.g, $e^{-z} = \frac{1}{e^z}, e^{z_1+z_2}=e^{z_1}e^{z_2}$

Complex Logarithm

A multi‐valued inverse of the exponential: log(z) = ln(|z|) + i * arg(z), arg(z) = θ + 2πk, k ∈ ℤ where:

ln(|z|) is the natural logarithm of the magnitude of z
|z| is the modulus (absolute value) of z
arg(z) is the argument (angle) of z

It’s a multi-valued function due to the periodic nature of the argument because arg(z) can take multiple values differing by 2πk, where k is any integer. This means for any complex number z, there are infinitely many complex logarithms that differ by multiples of 2πi, e.g, for z = 1 + i: log(z) = $ln(\sqrt{2}) + i * (π/4 + 2πk)$.

Principal branch: For a complex number z = x + yi, the complex logarithm is defined as: Log(z) = ln(|z|) + i * Arg(z) with Arg(z) ∈ (−π, π] where:

ln(|z|) is the natural logarithm of the magnitude of z
|z| = $\sqrt{(x² + y²)}$ is the modulus (absolute value) of z
Arg(z) is the principal argument (angle) of z, which is the angle between the positive real axis and the line from the origin to the point z in the complex plane. The value of Arg(z) is constrained within the range (−π,π].

Dom(Log(z)) = ℂ*, meaning it encompasses all complex numbers except for zero. This is because the logarithm is undefined at z=0. This definition ensures that exp(Log(z)) = z (consistency). The principal value is chosen to keep the argument within (-π, π], e.g., z = 1 + i, $|z| = \sqrt{1² + 1²} = \sqrt{2}, Arg(z) = arctan(1/1) = π/4, Log(z) = ln(\sqrt{2}) + i * (π/4)$

Sine, Cosine & Hyperbolic Variants

Expanding the complex exponential functions e^iz = cos(z) + isin(z), e^-iz = cos(z) - isin(z). Adding and subtracting these two formulae give respectively the two Euler’s formulae:

$cos(z) = \frac{1}{2}(e^{iz}+e^{-iz}), sin(z) = \frac{1}{2i}(e^{iz}-e^{-iz})$

Hyperbolic: $cosh(z) = \frac{1}{2}(e^{z}+e^{-z}), sinh(z) = \frac{1}{2}(e^{z}-e^{-z})$

$sinh(z) =[\text{By definition}] \frac{1}{2}(e^{z}-e^{-z}) = \frac{1}{2}(e^{x+iy}-e^{-(x+iy)}) = \frac{1}{2}[(e^{x}e^{iy})-(e^{-x}e^{-iy}) ] =[\text{Using Euler's formula}] \frac{1}{2}[e^{x}cos(y)+e^{x}isin(y)-cos(y)e^{-x}+isin(y)e^{-x}]= cos(y)(\frac{1}{2}[e^{x}-e^{-x}]) + sin(y)(\frac{i}{2}[e^{x}+e^{-x}]) = \frac{1}{2}[e^{x}-e^{-x}]cos(y)+i\frac{1}{2}[e^{x}+e^{-x}]sin(y) = sinh(x)cos(y)+icosh(x)sin(y)$, hence u = sinh(x)cos(y) and v = cosh(x)sin(y)

If y = 0, meaning z is a real number, these reduce to the usual real cosh(x), sinh(x), $sinh(z) = sinh(x)cos(y)+icosh(x)sin(y) = sinh(x)∙1+icosh(x)∙ 0 = sinh(x)$. This last mathematical statement clearly demonstrates how the complex hyperbolic sine function reduces to the real hyperbolic sine function for real arguments.

If x = 0, meaning z is a pure imaginary, $sinh(z) = sinh(iy) = sinh(x)cos(y)+icosh(x)sin(y) =[\text{Since sinh(0) = 0 and cosh(0) = 1, this simplifies to}] isin(y)$

$cosh(z) =[\text{By definition}] \frac{1}{2}(e^{z}+e^{-z}) = \frac{1}{2}(e^{x+iy}+e^{-(x+iy)}) = \frac{1}{2}[(e^{x}e^{iy})+(e^{-x}e^{-iy}) ] =[\text{Using Euler's formula}] \frac{1}{2}[e^{x}cos(y)+e^{x}isin(y)+cos(y)e^{-x}-isin(y)e^{-x}]= cos(y)(\frac{1}{2}[e^{x}+e^{-x}]) + sin(y)(\frac{i}{2}[e^{x}-e^{-x}]) = \frac{1}{2}[e^{x}+e^{-x}]cos(y)+i\frac{1}{2}[e^{x}-e^{-x}]sin(y) = cosh(x)cos(y)+isinh(x)sin(y)$, hence u = cosh(x)cos(y) and v = sinh(x)sin(y)

If y = 0, meaning z is a real number, $cosh(z) = cosh(x)cos(y)+isinh(x)sin(y) = cosh(x)∙1+isinh(x)∙0 = cosh(x)$. This last mathematical statement clearly demonstrates how the complex hyperbolic cosine function reduces to the real hyperbolic cosine function for real arguments. If x = 0, meaning z is a pure imaginary, $cosh(z) = cosh(iy) = cosh(x)cos(y)+isinh(x)sin(y) =[\text{Since sinh(0) = 0 and cosh(0) = 1, this simplifies to}] cos(y)$. Therefore, the hyperbolic cosine of a purely imaginary number is equal to the cosine of the real part of the imaginary number.

In summary:

sinh(z) = sinh(x)cos(y)+icosh(x)sin(y), sinh(iy) = isin(y)
cosh(z) = cosh(x)cos(y)+isinh(x)sin(y), cosh(iy) = cos(y)

Another way of calculating the complex cosine function: cos(z) = cos(x + iy) =[Using the formula for the cosine of a sum: cos⁡(a+b) = cos⁡(a)cos⁡(b) − sin⁡(a)sin⁡(b)] cos(x)cos(iy) -sin(x)sin(iy) =[cos(iy) = (e^i(iy) + e^-i(iy)) / 2 = (e^-y + e^y) / 2 = cosh(y); sin(iy) = (e^i(iy) - e^-i(iy)) / 2i = (e^-y - e^y) / 2i = i(e^y - e^-y) / 2 = i sinh(y)] cos(x)cosh(y) -isin(x)sinh(y). Therefore, cos(z) = cos(x + iy) = cos(x)cosh(y) -isin(x)sinh(y)

$tan(z) = \frac{sin(z)}{cos(z)}, sec(z) = \frac{1}{cos(z)}$

The following properties are satisfied:

$cos²(z)+sin²(z) = 1$
$sin(z_1+z_2) = sin(z_1)cos(z_2) + cos(z_1)sin(z_2)$
$cos(z_1+z_2) = cos(z_1)cos(z_2) - sin(z_1)sin(z_2)$
$cos(iz) = \frac{1}{2}(e^{i(iz)}+e^{-i(iz)}) = \frac{1}{2}(e^{-z}+e^{z}) = cosh(z)$
$sin(iz) = \frac{1}{2i}(e^{i(iz)}-e^{-i(iz)})= \frac{1}{2i}(e^{-z}-e^{z}) = [\frac{1}{i} = -i] \frac{1}{2}i(e^{z}-e^{-z}) = isinh(z)$

The addition formulae may be written also as sin(x + iy) =[$sin(z_1+z_2) = sin(z_1)cos(z_2) + cos(z_1)sin(z_2)$] sin(x)cos(iy) + cos(x)sin(iy) = sin(x)cosh(y)+icos(x)sinh(y). |sin(x + iy)| = $\sqrt{sin²(x)cosh²(y)+cos²(x)sinh²(y)}$ = [cos²(x) + sin²(x) = 1] $\sqrt{sin²(x)cosh²(y) + (1-sin²(x))sinh²(y)} = \sqrt{sin²(x)(cosh²(y)-sin²(x)) + sinh²(y)} =$ [cosh²(y) -sin²(y) = 1. This is a fundamental hyperbolic identity] $\sqrt{sin²(x) + sinh²(y)}$

Complex Numbers & Functions: Roots, Exponential, Logarithm, and Trigonometric Identities