The Argand Diagram

(i) $z = 10 + 5\mathrm{i}$ → $(10, 5)$

(ii) $w = 1 + 2\mathrm{i}$ → $(1, 2)$

(iii) $z + w = (10+1) + (5+2)\mathrm{i} = 11 + 7\mathrm{i}$ → $(11, 7)$

(iv) $z - w = (10-1) + (5-2)\mathrm{i} = 9 + 3\mathrm{i}$ → $(9, 3)$

(v) $w - z = (1-10) + (2-5)\mathrm{i} = -9 - 3\mathrm{i}$ → $(-9, -3)$

Plot all five points on a single Argand diagram:

$\bullet$ $z$ at $(10, 5)$ and $w$ at $(1, 2)$ are both in quadrant I.

$\bullet$ $z + w$ at $(11, 7)$ is the diagonal of the parallelogram formed by $O$, $z$, $w$.

$\bullet$ $z - w$ at $(9, 3)$ is the vector from $w$ to $z$ (transferred to start at $O$).

$\bullet$ $w - z$ at $(-9, -3)$ is the point opposite $z - w$ through the origin, i.e., $w - z = -(z - w)$.

The four points $O(0,0)$, $z(10,5)$, $z+w(11,7)$, $w(1,2)$ form a parallelogram: the vector from $O$ to $z$ equals the vector from $w$ to $z+w$ (both are $(10,5)$), and the vector from $O$ to $w$ equals the vector from $z$ to $z+w$ (both are $(1,2)$).

$\boxed{z+w = 11+7\mathrm{i}, \quad z-w = 9+3\mathrm{i}, \quad w-z = -9-3\mathrm{i}}$

(i) $|z| = |3 + 4\mathrm{i}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = \boxed{5}$

(ii) $|w| = |5 - 12\mathrm{i}| = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = \boxed{13}$

Method 1 (using the property): $|zw| = |z| \cdot |w| = 5 \times 13 = \boxed{65}$.

Method 2 (direct computation): $zw = (3+4\mathrm{i})(5-12\mathrm{i}) = 15 - 36\mathrm{i} + 20\mathrm{i} - 48\mathrm{i}^2 = 15 - 16\mathrm{i} + 48 = 63 - 16\mathrm{i}$.

Then $|zw| = \sqrt{63^2 + (-16)^2} = \sqrt{3969 + 256} = \sqrt{4225} = 65$. ✓ (Both methods agree.)

(iv) $\left|\dfrac{z}{w}\right| = \dfrac{|z|}{|w|} = \dfrac{5}{13} = \boxed{\dfrac{5}{13}}$

(v) $\left|\dfrac{w}{z}\right| = \dfrac{|w|}{|z|} = \dfrac{13}{5} = \boxed{\dfrac{13}{5}}$

The key property of moduli under multiplication and division:

$$|zw| = |z| \cdot |w|, \qquad \left|\dfrac{z}{w}\right| = \dfrac{|z|}{|w|}$$

Multiplying two complex numbers scales the modulus (the distances multiply). Dividing scales the modulus by the reciprocal. This is the geometric meaning of multiplication: it scales the distance from the origin (and also rotates, which we will study next lesson via the argument).

First, $|z| = |1+\mathrm{i}| = \sqrt{2}$, so $|z^n| = |z|^n = (\sqrt{2})^n = 2^{n/2}$.

$n = -1$: $z^{-1} = \dfrac{1}{1+\mathrm{i}} = \dfrac{1-\mathrm{i}}{(1+\mathrm{i})(1-\mathrm{i})} = \dfrac{1-\mathrm{i}}{2}$, so $z^{-1} = \dfrac{1}{2} - \dfrac{1}{2}\mathrm{i}$, and $|z^{-1}| = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$.

$n = 0$: $z^0 = 1$, $|z^0| = 1$.

$n = 1$: $z^1 = 1 + \mathrm{i}$, $|z^1| = \sqrt{2}$.

$n = 2$: $z^2 = (1+\mathrm{i})^2 = 1 + 2\mathrm{i} + \mathrm{i}^2 = 2\mathrm{i}$, $|z^2| = 2$.

$n = 3$: $z^3 = z^2 \cdot z = 2\mathrm{i}(1+\mathrm{i}) = 2\mathrm{i} + 2\mathrm{i}^2 = -2 + 2\mathrm{i}$, $|z^3| = \sqrt{4+4} = 2\sqrt{2}$.

$n = 4$: $z^4 = (z^2)^2 = (2\mathrm{i})^2 = 4\mathrm{i}^2 = -4$, $|z^4| = 4$.

$n = 5$: $z^5 = z^4 \cdot z = -4(1+\mathrm{i}) = -4 - 4\mathrm{i}$, $|z^5| = \sqrt{16+16} = 4\sqrt{2}$.

The seven points on the Argand diagram are:

$z^{-1} = (0.5, -0.5)$, $z^0 = (1, 0)$, $z^1 = (1, 1)$, $z^2 = (0, 2)$, $z^3 = (-2, 2)$, $z^4 = (-4, 0)$, $z^5 = (-4, -4)$.

Joining each point to its predecessor and to the origin creates a spiral pattern.

Three key observations:

1. The moduli grow geometrically: $|z^n| = (\sqrt{2})^n$, so each power multiplies the distance from the origin by $\sqrt{2}$. The points spiral outward.

2. Each power rotates by $45^\circ$ anticlockwise: since $\arg(1+\mathrm{i}) = 45^\circ$, multiplying by $z$ rotates by $45^\circ$ each time. So $z^0, z^1, z^2, \ldots$ each turn $45^\circ$ anticlockwise from the previous.

3. The points lie on a spiral: combining the scaling ($\times\sqrt{2}$) and rotation ($+45^\circ$), the points $z^{-1}, z^0, z^1, \ldots, z^5$ trace out a spiral emanating from the origin. After 8 multiplications ($z^8$), the point has rotated by $8 \times 45^\circ = 360^\circ$ — back to the positive real axis, but scaled by $(\sqrt{2})^8 = 16$.

Let $z = x + \mathrm{i}y$ be represented by the point $P(x, y)$ on the Argand diagram, with origin $O$.

We use two geometric facts:

$\bullet$ Negation ($z \mapsto -z$): rotates the point $180^\circ$ about $O$ (i.e., reflects through the origin). The point $(x, y)$ maps to $(-x, -y)$.

$\bullet$ Conjugation ($z \mapsto z^*$): reflects the point across the real axis. The point $(x, y)$ maps to $(x, -y)$.

Start with $z$ at $P(x, y)$.

Step 1 — negate: $-z = -x - \mathrm{i}y$ is at $P'(-x, -y)$, obtained by rotating $P$ by $180^\circ$ about $O$.

Step 2 — conjugate: $(-z)^* = -x + \mathrm{i}y$ is at $Q(-x, y)$, obtained by reflecting $P'(-x, -y)$ across the real axis (the $y$-coordinate changes sign: $-y \to y$).

So $(-z)^*$ is at the point $(-x, y)$.

Start with $z$ at $P(x, y)$.

Step 1 — conjugate: $z^* = x - \mathrm{i}y$ is at $P''(x, -y)$, obtained by reflecting $P(x, y)$ across the real axis.

Step 2 — negate: $-(z^*) = -x + \mathrm{i}y$ is at $Q'(-x, y)$, obtained by rotating $P''(x, -y)$ by $180^\circ$ about $O$ (both coordinates change sign: $x \to -x$, $-y \to y$).

So $-(z^*)$ is at the point $(-x, y)$.

Both $(-z)^*$ and $-(z^*)$ map the original point $P(x, y)$ to the same point $(-x, y)$.

Geometrically, $(-x, y)$ is the reflection of $z$ across the imaginary axis (the $x$-coordinate changes sign, the $y$-coordinate stays the same).

Since both sequences of transformations produce the same final point, we conclude:

$$\boxed{(-z)^* = -(z^*)}$$

Geometric insight: rotating $180^\circ$ about $O$ and reflecting across the real axis commute — the order does not matter. Both combinations are equivalent to a single reflection across the imaginary axis.