Complex functions

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.

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

Step 1. Magnitude and Argument:

|z| = $\sqrt{1² +(\sqrt{3})²} = 2, Arg(z) = arctan(\sqrt{3}) = \frac{∏}{3}$. 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, $z^{\frac{1}{n}} = r^{\frac{1}{n}}e^{i(\frac{θ+2\pi k}{n})} = r^{\frac{1}{n}}e^{i(\frac{θ}{n})}e^{i(\frac{2\pi k}{n})}$. $w_n = e^{i(\frac{2\pi}{n})}, \sqrt[n]{w} = \sqrt[n]{|w|}e^{i\frac{Arg(w)}{n}}$. 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).

• 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 functions

Let D ⊆ ℂ be a set of complex numbers. A function f defined on D is a rule that assigns to each z in D a complex number w. f: D ➞ ℂ. The set D is called the domain of definition of f, D = Dom(f), and the range of f is f(D). f(z) = u(z) + iv(z).

Some examples are:

• f(z) = z², f(x +iy) = (x +iy)² = x² -y² + (2xy)i, hence u(z) = u(x, y) = x² -y² and v(z) = v(x, y) = (2xy)i

• f(z) = z³, f(x +iy) = (x +iy)³ = $x³+3x²(iy)+3x(iy)²+(iy)³ = x³-3xy²+i(3x²y-y³)$ where $u = x³-3xy², v = 3x²y-y³$

If more than one value of w corresponds to each value of z, e.g, f(z) = $z^{\frac{1}{2}}$ we say that is not a function or is a multiple-valued function.

Is f invertible? If not, how can we restrict its domain to make it injective (one-to-one) f|_D

Complex exponential

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

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

e^xcos(x) + e^xsin(y)i where u = e^xcos(x), 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 + 2nπ (n = 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

For a complex number z = x + yi, the complex logarithm is defined as: Log(z) = ln(|z|) + i * Arg(z) 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. The principal value is chosen to keep the argument within (-π, π], e.g. For z = 1 + i, |z| = √(1² + 1²) = √2, Arg(z) = arctan(1/1) = π/4, Log(z) = ln(√2) + i * (π/4)

log(z) = ln(|z|) + i * arg(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(√2) + i * (π/4 + 2πk).

Complex sine and cosine functions

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

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

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

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

$cosh(z) =[\text{By definition}] \frac{1}{2}(e^{z}+e^{-z}) = \frac{1}{2}(e^{x+iy}+e^{-(x+iy)}) = \frac{1}{2}[(e^{x}e^{iy})+(e^{-x}e^{-iy}) ] =[\text{Using Euler’s formula}] \frac{1}{2}[e^{x}cos(y)+e^{x}isin(y)+cos(y)e^{-x}-isin(y)e^{-x}]= cos(y)(\frac{1}{2}[e^{x}+e^{-x}]) + sin(y)(\frac{i}{2i}[e^{x}-e^{-x}]) = \frac{1}{2}[e^{x}+e^{-x}]cos(y)+i\frac{1}{2}[e^{x}-e^{-x}]sin(y) = cosh(x)cos(y)+isinh(x)sin(y)$, 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 -sin² = 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] $\sqrt{sin²(x) + sinh²(y)}$