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.