Complex functions II

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

Complex square root

$f(z) = \sqrt{z} = \sqrt{|z|e^{Arg(z)}} = \sqrt{|z|}e^{\frac{Arg(z)}{2}} = \sqrt{|z|}(cos(\frac{θ}{2})+ isin(\frac{θ}{2})) = \sqrt{x²+y²}cos(\frac{θ}{2}) + i\sqrt{x²+y²}sin(\frac{θ}{2}), u = \sqrt{x²+y²}cos(\frac{θ}{2}), v = \sqrt{x²+y²}sin(\frac{θ}{2})$

$f(z) = \sqrt[n]{z}$ where Re(z) = 0,

$f(z) = \sqrt[n]{z} = \sqrt[n]{|z|e^{Arg(z)}} = \sqrt[n]{|z|}e^{\frac{Arg(z)}{n}} = \sqrt[n]{|z|}(cos(\frac{θ}{n})+ isin(\frac{θ}{n})) = \sqrt[n]{x²+y²}cos(\frac{θ}{n}) + i\sqrt[n]{x²+y²}sin(\frac{θ}{n}) = \sqrt[n]{x²+y²}cos(\frac{1}{n}tan^{-1}(\frac{y}{x})) + i\sqrt[n]{x²+y²}sin(\frac{1}{n}tan^{-1}(\frac{y}{x})), u = \sqrt[n]{x²+y²}cos(\frac{1}{n}tan^{-1}(\frac{y}{x})), v = \sqrt[n]{x²+y²}sin(\frac{1}{n}tan^{-1}(\frac{y}{x}))$

Complex power

For any complex number z and any complex power p, we can express the exponentiation as: $z^p = e^{log(z^p)} = e^{p∙log(z)}$. Here, “log” represents the complex logarithm, which is multi-valued.

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

Recall the Complex Logarithm. The complex logarithm of a complex number z can be written as: Log(z) = ln|z| + iArg(z) where ln denotes the natural logarithm of the magnitude of z and arg denotes the argument (angle) of z in the complex plane.

Because the argument is periodic with a period of 2π, we can write the general form of the complex logarithm as: log(z) = ln∣z∣ + i(Arg(z)+2πk) where Arg(z) is the principal argument of z (typically in the range (-π, π]), and k is any integer (k ∈ ℤ).

Logarithm of i: For z = i: Log(i) = ln|i| + iArg(i) =[ln|i| = ln(1) = 0] 0 + iπ/2.

Therefore, the complex logarithm of i is: $log(i) = 0 + i(π/2 + 2πk) = i(π/2 + 2πk)$

$i^i = e^{i∙log(i)} = e^{i²(π/2 + 2πk)}$ =[Since k can be any integer, we can rewrite this as:] {e^(-π/2-2πk) | k ∈ ℤ}

$(1+i)^{2+3i} =$[By definition] $e^{(2+3i)log(1+i)}$ =[log(z) = ln∣z∣ + i(Arg(z)+2πk)] $e^{(2+3i)[ln|1+i| + i(Arg(1+i)+2πk)]} = e^{(2+3i)[ln(\sqrt{2}) + i(\frac{π}{4}+2πk)]} = e^{2ln(\sqrt{2}) -3(\frac{π}{4}+2πk)+i(3ln(\sqrt{2})+2(\frac{π}{4}+2πk))} = e^{2ln(\sqrt{2}) -3(\frac{π}{4}+2πk)+i(3ln(\sqrt{2})+\frac{π}{2}+4πk)} = e^{ln(2)+\frac{-3π}{4}-6πk+i(3ln(\sqrt{2})+\frac{π}{2}+4πk)} = 2e^{\frac{-3π}{4}-6πk+i(3ln(\sqrt{2})+\frac{π}{2}+4πk)}$ where k is any integer.

Quadratic formula

Let $az² + bz + c = 0$ 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: $(z+ \frac{b}{2a})² = \frac{b²-4ac}{4a²}$

$z+ \frac{b}{2a} ∈ (\frac{b²-4ac}{4a²})^{\frac{1}{2}}$

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 that should be calculated according to [*]

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 root of the complex quadratic equation is 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 the input (domain) and the output (range) in a two-dimensional plane. This is often done by using two separate complex planes.

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 c units in the real direction and d 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).

• The function f(z) = az where z ∈ ℂ where 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).

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).
4. The argument of the result, θ + φ, is the sum of the arguments of a and z. Therefore, the transformation rotates z 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.

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

Scaling:

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.

2.0 Flash Experimental. Might not work as expected.

• In general, 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 is raised to the nth power (elongation).
2. If 0 < r < 1, the magnitude is raised to the nth power (contraction).
3. If r = 1, the magnitude remains unchanged (pure rotation).

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

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.