Complex Logarithms and Loci

Complex Logarithm

For $z = x + iy = |z|e^{i\theta}$, the logarithm is:

$\ln z = \ln|z| + i\theta$

Since $\theta$ is defined up to multiples of $2\pi$:

$\ln z = \ln|z| + i(\theta + 2n\pi)$

for integer $n$.

Principal value: Choose $-\pi < \theta \leq \pi$, giving:

$\text{Ln } z = \ln|z| + i\text{Arg}(z)$

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Example: Evaluating Powers

Evaluate $2^i$ and $i^i$ in form $x + iy$.

For $z^w = e^{w\ln z}$, using principal values:

$2^i$:

• $\ln 2 = \ln 2 + i \cdot 0$ (real, so $\theta = 0$)
• $2^i = e^{i\ln 2} = \cos(\ln 2) + i\sin(\ln 2)$
• $= 0.769 + 0.639i$ (approximately)

$i^i$:

• $i = e^{i\pi/2}$, so $\ln i = i\pi/2$
• $i^i = e^{i \cdot i\pi/2} = e^{-\pi/2}$
• $= 0.208$ (real!)

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Multi-Valued Nature

Unlike real logarithms, $\ln z$ is multi-valued:

• Different branches correspond to different choices of $n$
• The principal branch has $-\pi < \theta \leq \pi$
• Moving around the origin by $2\pi$ returns to the same point but adds $2\pi i$ to the logarithm

Branch cut: To make $\ln z$ single-valued, we cut the complex plane (usually along negative real axis).

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Loci in the Argand Diagram

A locus is a set of points satisfying a geometric condition.

Circle

$|z - a| = r$ describes a circle:

• Centre: $a$ (complex number)
• Radius: $r$

Example: $|z - 1 - 2i| = 3$ is circle centred at $(1, 2)$ with radius 3.

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Perpendicular Bisector

$|z - z_1| = |z - z_2|$ describes the perpendicular bisector of the line segment joining $z_1$ and $z_2$.

Example: $|z| = |z - i|$ describes the line $y = \frac{1}{2}$.

Derivation: Let $z = x + iy$: $|z|^2 = |z - i|^2$ $x^2 + y^2 = x^2 + (y-1)^2$ $y^2 = y^2 - 2y + 1$ $y = \frac{1}{2}$

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Half-Line (Ray)

$\arg(z - a) = \theta$ describes a half-line from point $a$ making angle $\theta$ with positive real axis.

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Region Inside a Circle

$|z - a| < r$ describes the interior of the circle.

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Example: Combined Loci

Question: What shape is described by $|z| = 3|z - i|$?

Let $z = x + iy$:

$x^2 + y^2 = 9(x^2 + (y-1)^2)$

$x^2 + y^2 = 9x^2 + 9y^2 - 18y + 9$

$8x^2 + 8y^2 - 18y + 9 = 0$

Dividing by 8 and completing the square:

$x^2 + y^2 - \frac{9y}{4} + \frac{9}{8} = 0$

$x^2 + \left(y - \frac{9}{8}\right)^2 = \frac{81}{64} - \frac{9}{8} = -\frac{9}{64}$

This would be impossible (negative radius squared), so let me recalculate:

$x^2 + y^2 = 9(x^2 + y^2 - 2y + 1)$

$x^2 + y^2 = 9x^2 + 9y^2 - 18y + 9$

$0 = 8x^2 + 8y^2 - 18y + 9$

$x^2 + y^2 - \frac{9y}{4} + \frac{9}{8} = 0$

$x^2 + \left(y - \frac{9}{8}\right)^2 = \frac{81}{64} - \frac{72}{64} = \frac{9}{64}$

Circle centred at $(0, \frac{9}{8})$ with radius $\frac{3}{8}$.

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General Powers

For complex $z$ and $w$:

$z^w = e^{w\ln z}$

Since $\ln z$ is multi-valued, so is $z^w$ (unless $w$ is an integer).