Complex Numbers and Loci in Advanced Mathematics II

Find the general complex logarithm, $\ln(z)$, and principal complex logarithm, $\operatorname{Ln}(z)$, for the following $z$:

Modulus: $|z| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \frac{1}{\sqrt{2}}$
Argument: $\arg(z) = \frac{\pi}{4}$ (principal argument)
$\ln(z) = \ln|z| + i(\arg(z) + 2k\pi) = \ln\left(\frac{1}{\sqrt{2}}\right) + i\left(\frac{\pi}{4} + 2k\pi\right)$, $k \in \mathbb{Z}$
$\operatorname{Ln}(z) = \ln|z| + i\operatorname{Arg}(z) = \ln\left(\frac{1}{\sqrt{2}}\right) + i\frac{\pi}{4}$

Modulus: $|z| = \sqrt{4 + 3} = \sqrt{7}$
Argument (principal): $\tan \theta = \frac{-\sqrt{3}}{2} \implies \operatorname{Arg}(z) = -\arctan\left(\frac{\sqrt{3}}{2}\right)$
$\ln(z) = \ln\sqrt{7} + i\left(-\arctan\left(\frac{\sqrt{3}}{2}\right) + 2k\pi\right)$, $k \in \mathbb{Z}$
$\operatorname{Ln}(z) = \ln\sqrt{7} - i\arctan\left(\frac{\sqrt{3}}{2}\right)$

Modulus: $|z| = 2\sqrt{3}$, Argument: $\arg(z) = \frac{\pi}{3} + 2k\pi$
$\ln(z) = \ln(2\sqrt{3}) + i\left(\frac{\pi}{3} + 2k\pi\right)$, $k \in \mathbb{Z}$
$\operatorname{Ln}(z) = \ln(2\sqrt{3}) + i\frac{\pi}{3}$

(a) Find the Cartesian form of $\sin i$ and $\cos i$.

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

(b) Compute $\cos\left(\frac{\pi}{6} - i\frac{\pi}{4}\right)$ and $\sin\left(\frac{\pi}{6} - i\frac{\pi}{4}\right)$ by:

Method 1: Using definition
- Let $z = \frac{\pi}{6} - i\frac{\pi}{4}$
- $\cos z = \frac{e^{i(\pi/6 - i\pi/4)} + e^{-i(\pi/6 - i\pi/4)}}{2} = \frac{e^{i\pi/6}e^{\pi/4} + e^{-i\pi/6}e^{-\pi/4}}{2}$
- $\sin z = \frac{e^{i\pi/6}e^{\pi/4} - e^{-i\pi/6}e^{-\pi/4}}{2i}$
Method 2: Using compound angles
- $\cos(A - B) = \cos A \cos B + \sin A \sin B$
  - $\cos\left(\frac{\pi}{6} - i\frac{\pi}{4}\right) = \cos\frac{\pi}{6}\cos\left(i\frac{\pi}{4}\right) + \sin\frac{\pi}{6}\sin\left(i\frac{\pi}{4}\right)$
  - Using $\cos(iy) = \cosh y$ and $\sin(iy) = i\sinh y$:
  - $= \frac{\sqrt{3}}{2}\cosh\left(\frac{\pi}{4}\right) + \frac{1}{2}\cdot i\sinh\left(\frac{\pi}{4}\right)$
- $\sin(A - B) = \sin A \cos B - \cos A \sin B$
  - $\sin\left(\frac{\pi}{6} - i\frac{\pi}{4}\right) = \sin\frac{\pi}{6}\cos\left(i\frac{\pi}{4}\right) - \cos\frac{\pi}{6}\sin\left(i\frac{\pi}{4}\right)$
  - $= \frac{1}{2}\cosh\left(\frac{\pi}{4}\right) - \frac{\sqrt{3}}{2}\cdot i\sinh\left(\frac{\pi}{4}\right)$
Both methods give the same result.

Find $\operatorname{Re}(z)$, $\operatorname{Im}(z)$, and $\arg(z)$ for $z = 2^{\sin\theta + i\cos\theta}$.

$z = 2^{\sin\theta} \cdot 2^{i\cos\theta} = 2^{\sin\theta} e^{i\cos\theta \ln 2}$
$\operatorname{Re}(z) = 2^{\sin\theta} \cos(\cos\theta \ln 2)$
$\operatorname{Im}(z) = 2^{\sin\theta} \sin(\cos\theta \ln 2)$
$\arg(z) = \cos\theta \ln 2 + 2k\pi$, $k \in \mathbb{Z}$

Standard equation: $|z - (a + bi)| = r$, centre at $(a, b)$, radius $r$.

(a) $|z - (4 + i)| = 3$

(b) $|z - 1 - i| = 5$

(c) $|2z + 6 - 4i| = 6$

Rewrite: $|2(z + 3 - 2i)| = 6 \implies 2|z + 3 - 2i| = 6 \implies |z - (-3 + 2i)| = 3$
Centre: $(-3, 2)$, Radius: $3$
Cartesian: $(x + 3)^2 + (y - 2)^2 = 9$

Standard equation: $|z - (a_1 + b_1i)| = |z - (a_2 + b_2i)|$

(a) $|z - (2 + i)| = |z - (1 + 3i)|$

Midpoint: $\left(\frac{2+1}{2}, \frac{1+3}{2}\right) = (1.5, 2)$
Slope of segment: $\frac{3-1}{1-2} = -2$
Perpendicular slope: $\frac{1}{2}$
Cartesian: $y - 2 = \frac{1}{2}(x - 1.5) \implies y = \frac{1}{2}x + \frac{5}{4}$

(b) $|z + 2 - i| = |z - 1 + 3i|$

Rewrite: $|z - (-2 + i)| = |z - (1 - 3i)|$
Midpoint: $\left(\frac{-2+1}{2}, \frac{1 + (-3)}{2}\right) = (-0.5, -1)$
Slope of segment: $\frac{-3-1}{1-(-2)} = -\frac{4}{3}$
Perpendicular slope: $\frac{3}{4}$
Cartesian: $y + 1 = \frac{3}{4}(x + 0.5) \implies y = \frac{3}{4}x - \frac{5}{8}$

(c) $\left| \frac{z+1+i}{z-2} \right| =

Source: FAC1004 Tutorial 3 25-26.pdf