Quick Recall
Quick check — pick the correct option in each question. These cover the Week 10 toolkit.
For addition/subtraction we handle the two components independently: $(a + b\mathrm{i}) \pm (c + d\mathrm{i}) = (a \pm c) + (b \pm d)\mathrm{i}$.
Expanding with $\mathrm{i}^2 = -1$: $(a + b\mathrm{i})(c + d\mathrm{i}) = ac + ad\mathrm{i} + bc\mathrm{i} + bd\mathrm{i}^2 = (ac - bd) + (ad + bc)\mathrm{i}$.
Multiplying by $\dfrac{c - d\mathrm{i}}{c - d\mathrm{i}}$ makes the denominator a real number $c^2 + d^2$, so the result is again in Cartesian form.
For a polynomial with real coefficients, non-real roots come in conjugate pairs: if $a + b\mathrm{i}$ is a root, so is $a - b\mathrm{i}$.
The Wall
Simplify the following expression and give your answer in the form $x + \mathrm{i}y$:
$$\dfrac{(1 + \mathrm{i})^4}{(1 - \mathrm{i})^2}$$
Try it by direct expansion first: compute $(1+\mathrm{i})^2 = 2\mathrm{i}$, then $(2\mathrm{i})^2 = -4$, then $(1-\mathrm{i})^2 = -2\mathrm{i}$, then divide. It works, but it is easy to drop a sign. Is there a pattern you can exploit to do it in one line?
The Need
Brute-force expansion works for small problems, but it is slow, error-prone, and hides the geometry. We need three big strategies to bring order to mixed operations:
Strategy 1 — Power patterns: recognise $(a + b\mathrm{i})^2$ and $(a + b\mathrm{i})^3$ so you never expand from scratch twice.
Strategy 2 — Conjugate & modulus: use $z \cdot z^* = |z|^2$ to clear denominators and the modulus–argument form as a uniform simplifier.
Strategy 3 — Equate real and imaginary parts: to solve $z^2 = a + b\mathrm{i}$, write $z = p + q\mathrm{i}$ and match components.
💡 The Bridge (click to reveal)
For The Wall problem: $(1+\mathrm{i})^2 = 1 + 2\mathrm{i} + \mathrm{i}^2 = 2\mathrm{i}$, so $(1+\mathrm{i})^4 = (2\mathrm{i})^2 = -4$. Also $(1-\mathrm{i})^2 = -2\mathrm{i}$. Hence $\dfrac{(1+\mathrm{i})^4}{(1-\mathrm{i})^2} = \dfrac{-4}{-2\mathrm{i}} = \dfrac{2}{\mathrm{i}} = \dfrac{2\mathrm{i}}{\mathrm{i}^2} = -2\mathrm{i}$.
The pattern $z \cdot z^* = |z|^2$ is the master key for division, and writing $z = r(\cos\theta + \mathrm{i}\sin\theta)$ turns multiplication into “scale by $r$, rotate by $\theta$”.
Complex arithmetic gets powerful when you combine operations and look for shortcuts. The three strategies below — power patterns, conjugate/modulus tricks, and equating real/imaginary parts — turn long, error-prone computations into quick, geometrically meaningful steps.
For $z = a + b\mathrm{i}$:
$z^2 = (a^2 - b^2) + 2ab\,\mathrm{i}$
$z^3 = (a^3 - 3ab^2) + (3a^2b - b^3)\,\mathrm{i}$
The real part uses even powers of $\mathrm{i}$ ($\mathrm{i}^2 = -1$); the imaginary part uses odd powers ($\mathrm{i}^3 = -\mathrm{i}$). Memorising these two expansions saves time on every mixed problem.
Conjugate trick: $z \cdot z^* = a^2 + b^2 = |z|^2$ (always real and positive). Use this to clear any complex denominator.
Modulus–argument form: write $z = r(\cos\theta + \mathrm{i}\sin\theta)$ where
$r = |z| = \sqrt{a^2 + b^2}$, $\theta = \arg(z) = \arctan\!\left(\dfrac{b}{a}\right)$ (adjusted for quadrant).
This form makes multiplication a “scale by $r$, rotate by $\theta$” operation and is the bridge to powers in higher lessons.
To solve $z^2 = w$ where $w = a + b\mathrm{i}$, write $z = p + q\mathrm{i}$ and equate real and imaginary parts:
$z^2 = (p^2 - q^2) + 2pq\,\mathrm{i} = a + b\,\mathrm{i}$
$\Rightarrow p^2 - q^2 = a$, $2pq = b$.
Combined with $p^2 + q^2 = |w| = \sqrt{a^2 + b^2}$, this gives
$p = \sqrt{\dfrac{|w| + a}{2}}$, $q = \operatorname{sgn}(b)\sqrt{\dfrac{|w| - a}{2}}$.
The two square roots are $z = \pm(p + q\mathrm{i})$. This is the prototype for solving any polynomial equation with complex coefficients.
The teacher poses a mixed-operation problem and the class works through it together, surfacing as many different strategies as possible. Students vote on which strategy is “most elegant” for each step.
Brute force: expand $(2+\mathrm{i})^2 = 3+4\mathrm{i}$, then $(2+\mathrm{i})^3 = (3+4\mathrm{i})(2+\mathrm{i}) = 2+11\mathrm{i}$, then divide by $2-\mathrm{i}$.
Conjugate trick: spot that $\dfrac{1}{2-\mathrm{i}} = \dfrac{2+\mathrm{i}}{5}$, so $\dfrac{(2+\mathrm{i})^3}{2-\mathrm{i}} = \dfrac{(2+\mathrm{i})^4}{5}$ — one fewer division step.
Modulus form: $|2+\mathrm{i}| = \sqrt{5}$, so $|(2+\mathrm{i})^3| = 5\sqrt{5}$ and $|2-\mathrm{i}| = \sqrt{5}$, giving $|$answer$| = 5$.
Quick-response matching game: match each complex expression (left) to its simplified form (right). Click an expression, then click the matching simplified form. The right column is shuffled each time you load the page.
👁 Show Answer Key
$(1 + \mathrm{i})^2 = 2\mathrm{i}$: expand $1 + 2\mathrm{i} + \mathrm{i}^2 = 1 + 2\mathrm{i} - 1 = 2\mathrm{i}$.
$(1 + \mathrm{i})(1 - \mathrm{i}) = 2$: the conjugate trick gives $|1+\mathrm{i}|^2 = 1 + 1 = 2$.
$\dfrac{1}{1 + \mathrm{i}} = \tfrac{1}{2} - \tfrac{1}{2}\mathrm{i}$: multiply by $\dfrac{1-\mathrm{i}}{1-\mathrm{i}}$ to get $\dfrac{1-\mathrm{i}}{2}$.
$(2 + \mathrm{i})^2 = 3 + 4\mathrm{i}$: expand $4 + 4\mathrm{i} + \mathrm{i}^2 = 3 + 4\mathrm{i}$.
For each complex number, find the modulus and principal argument, and hence write the complex number in polar form.
Give the argument in radians, either as a simple rational multiple of $\pi$ or correct to 3 decimal places.
(i) $1$ (ii) $-2$ (iii) $3\mathrm{i}$ (iv) $-4\mathrm{i}$ (v) $1+\mathrm{i}$ (vi) $-5-5\mathrm{i}$
(vii) $1-\sqrt{3}\mathrm{i}$ (viii) $6\sqrt{3}+6\mathrm{i}$ (ix) $3-4\mathrm{i}$ (x) $-12+5\mathrm{i}$ (xi) $4+7\mathrm{i}$ (xii) $-58-93\mathrm{i}$
👁 Reveal Solution
(i) $z = 1$: $r = 1$, $\theta = 0$. Polar form: $\cos 0 + \mathrm{i}\sin 0$.
(ii) $z = -2$: $r = 2$, $\theta = \pi$. Polar form: $2(\cos\pi + \mathrm{i}\sin\pi)$.
(iii) $z = 3\mathrm{i}$: $r = 3$, $\theta = \pi/2$. Polar form: $3(\cos\tfrac{\pi}{2} + \mathrm{i}\sin\tfrac{\pi}{2})$.
(iv) $z = -4\mathrm{i}$: $r = 4$, $\theta = -\pi/2$. Polar form: $4(\cos(-\tfrac{\pi}{2}) + \mathrm{i}\sin(-\tfrac{\pi}{2}))$.
(v) $z = 1 + \mathrm{i}$: $r = \sqrt{1+1} = \sqrt{2}$, $\theta = \arctan(1) = \pi/4$. Polar form: $\sqrt{2}(\cos\tfrac{\pi}{4} + \mathrm{i}\sin\tfrac{\pi}{4})$.
(viii) $z = 6\sqrt{3} + 6\mathrm{i}$: $r = \sqrt{108 + 36} = \sqrt{144} = 12$, $\theta = \arctan\!\left(\tfrac{6}{6\sqrt{3}}\right) = \arctan\!\left(\tfrac{1}{\sqrt{3}}\right) = \pi/6$. Polar form: $12(\cos\tfrac{\pi}{6} + \mathrm{i}\sin\tfrac{\pi}{6})$.
(xi) $z = 4 + 7\mathrm{i}$: $r = \sqrt{16+49} = \sqrt{65}$, $\theta = \arctan(7/4) \approx 1.052$ rad. Polar form: $\sqrt{65}(\cos 1.052 + \mathrm{i}\sin 1.052)$.
(vi) $z = -5 - 5\mathrm{i}$: $r = \sqrt{25+25} = 5\sqrt{2}$, Q3 so $\theta = -\pi + \arctan(1) = -\tfrac{3\pi}{4}$. Polar form: $5\sqrt{2}(\cos(-\tfrac{3\pi}{4}) + \mathrm{i}\sin(-\tfrac{3\pi}{4}))$.
(vii) $z = 1 - \sqrt{3}\mathrm{i}$: $r = \sqrt{1+3} = 2$, Q4 so $\theta = -\arctan(\sqrt{3}) = -\pi/3$. Polar form: $2(\cos(-\tfrac{\pi}{3}) + \mathrm{i}\sin(-\tfrac{\pi}{3}))$.
(ix) $z = 3 - 4\mathrm{i}$: $r = \sqrt{9+16} = 5$, Q4 so $\theta = -\arctan(4/3) \approx -0.927$ rad. Polar form: $5(\cos(-0.927) + \mathrm{i}\sin(-0.927))$.
(x) $z = -12 + 5\mathrm{i}$: $r = \sqrt{144+25} = 13$, Q2 so $\theta = \pi - \arctan(5/12) \approx 2.747$ rad. Polar form: $13(\cos 2.747 + \mathrm{i}\sin 2.747)$.
(xii) $z = -58 - 93\mathrm{i}$: $r = \sqrt{58^2 + 93^2} = \sqrt{3364 + 8649} = \sqrt{12013} \approx 109.604$.
Q3 so $\theta = -\pi + \arctan(93/58) \approx -\pi + 1.015 \approx -2.127$ rad.
Polar form: $109.604(\cos(-2.127) + \mathrm{i}\sin(-2.127))$.
$$\text{All 12 parts have been converted to polar form } r(\cos\theta + \mathrm{i}\sin\theta).$$
The pattern: compute $r$ by Pythagoras, compute the reference angle $\arctan(|b/a|)$, then attach the sign of $\theta$ based on the quadrant. Special-axis points ($a = 0$ or $b = 0$) skip the arctan step entirely.
Write each complex number with the given modulus and argument in the form $x + \mathrm{i}y$, giving surds in your answer where appropriate.
(i) $|z|=2, \arg z = \dfrac{\pi}{2}$ (ii) $|z|=3, \arg z = \dfrac{\pi}{3}$ (iii) $|z|=7, \arg z = \dfrac{5\pi}{6}$
(iv) $|z|=1, \arg z = -\dfrac{\pi}{4}$ (v) $|z|=5, \arg z = -\dfrac{2\pi}{3}$ (vi) $|z|=6, \arg z = -2$
👁 Reveal Solution
(i) $|z|=2$, $\arg z = \pi/2$: $z = 2(\cos\tfrac{\pi}{2} + \mathrm{i}\sin\tfrac{\pi}{2}) = 2(0 + \mathrm{i}\cdot 1) = \boxed{2\mathrm{i}}$.
(ii) $|z|=3$, $\arg z = \pi/3$: $z = 3(\cos\tfrac{\pi}{3} + \mathrm{i}\sin\tfrac{\pi}{3}) = 3\!\left(\tfrac{1}{2} + \mathrm{i}\tfrac{\sqrt{3}}{2}\right) = \boxed{\tfrac{3}{2} + \tfrac{3\sqrt{3}}{2}\mathrm{i}}$.
(iii) $|z|=7$, $\arg z = 5\pi/6$: $z = 7(\cos\tfrac{5\pi}{6} + \mathrm{i}\sin\tfrac{5\pi}{6}) = 7\!\left(-\tfrac{\sqrt{3}}{2} + \mathrm{i}\tfrac{1}{2}\right) = \boxed{-\tfrac{7\sqrt{3}}{2} + \tfrac{7}{2}\mathrm{i}}$.
(iv) $|z|=1$, $\arg z = -\pi/4$: $z = \cos(-\tfrac{\pi}{4}) + \mathrm{i}\sin(-\tfrac{\pi}{4}) = \tfrac{\sqrt{2}}{2} - \mathrm{i}\tfrac{\sqrt{2}}{2} = \boxed{\tfrac{\sqrt{2}}{2} - \tfrac{\sqrt{2}}{2}\mathrm{i}}$.
(v) $|z|=5$, $\arg z = -2\pi/3$: $z = 5(\cos(-\tfrac{2\pi}{3}) + \mathrm{i}\sin(-\tfrac{2\pi}{3})) = 5\!\left(-\tfrac{1}{2} - \mathrm{i}\tfrac{\sqrt{3}}{2}\right) = \boxed{-\tfrac{5}{2} - \tfrac{5\sqrt{3}}{2}\mathrm{i}}$.
(vi) $|z|=6$, $\arg z = -2$: $z = 6(\cos(-2) + \mathrm{i}\sin(-2))$.
Using $\cos(-2) \approx -0.4161$ and $\sin(-2) \approx -0.9093$:
$z \approx 6(-0.4161 - 0.9093\mathrm{i}) \approx \boxed{-2.497 - 5.456\mathrm{i}}$ (to 3 d.p.).
$$\text{(i) } 2\mathrm{i} \quad \text{(ii) } \tfrac{3}{2} + \tfrac{3\sqrt{3}}{2}\mathrm{i} \quad \text{(iii) } -\tfrac{7\sqrt{3}}{2} + \tfrac{7}{2}\mathrm{i}$$
$$\text{(iv) } \tfrac{\sqrt{2}}{2} - \tfrac{\sqrt{2}}{2}\mathrm{i} \quad \text{(v) } -\tfrac{5}{2} - \tfrac{5\sqrt{3}}{2}\mathrm{i} \quad \text{(vi) } -2.497 - 5.456\mathrm{i}$$
The pattern: $x = r\cos\theta$ and $y = r\sin\theta$. For non-special angles, give numerical answers to 3 decimal places.
For each of parts (i) to (viii), draw an Argand diagram showing the set of points $z$ for which the given condition is true.
(i) $|z|=2$ (ii) $|z - 4| \leqslant 3$ (iii) $|z - 5\mathrm{i}| = 6$ (iv) $|z + 3 - 4\mathrm{i}| < 5$
(v) $|6 - \mathrm{i} - z| \geqslant 2$ (vi) $|z + 2 + 4\mathrm{i}| = 0$ (vii) $2 \leqslant |z - 1 + \mathrm{i}| \leqslant 3$ (viii) $\mathrm{Re}(z) = -2$
💡 Hint
Recall: $|z - z_0|$ is the distance from $z$ to $z_0$ on the Argand diagram. So:
$|z - z_0| = r$ → circle of radius $r$ centred at $z_0$.
$|z - z_0| \leqslant r$ → closed disk of radius $r$ centred at $z_0$.
$|z - z_0| < r$ → open disk (boundary not included, use a dashed circle).
$|z - z_0| \geqslant r$ → exterior of the disk (boundary included).
$r_1 \leqslant |z - z_0| \leqslant r_2$ → annulus (ring) between two circles.
Rewrite each condition in the form $|z - z_0| \circ r$ to identify the centre $z_0$ and radius $r$.
👁 Reveal Solution
(i) $|z| = 2$: centre $0$, radius $2$ → circle.
(ii) $|z - 4| \leqslant 3$: centre $4$, radius $3$ → closed disk (solid boundary).
(iii) $|z - 5\mathrm{i}| = 6$: centre $5\mathrm{i}$, radius $6$ → circle.
(iv) $|z + 3 - 4\mathrm{i}| < 5$: rewrite as $|z - (-3 + 4\mathrm{i})| < 5$ → open disk centre $-3 + 4\mathrm{i}$, radius $5$ (dashed boundary).
(v) $|6 - \mathrm{i} - z| \geqslant 2$: rewrite as $|z - (6 - \mathrm{i})| \geqslant 2$ → exterior of disk centre $6 - \mathrm{i}$, radius $2$ (solid boundary, shaded outside).
(vi) $|z + 2 + 4\mathrm{i}| = 0$: rewrite as $|z - (-2 - 4\mathrm{i})| = 0$ → single point $-2 - 4\mathrm{i}$.
(vii) $2 \leqslant |z - 1 + \mathrm{i}| \leqslant 3$: rewrite as $2 \leqslant |z - (1 - \mathrm{i})| \leqslant 3$ → annulus centre $1 - \mathrm{i}$, inner radius $2$, outer radius $3$.
(viii) $\mathrm{Re}(z) = -2$: vertical line $x = -2$.
Draw a separate Argand diagram for each part. Mark the centre $z_0$ as a dot, draw the circle(s) of the required radius, and shade the region indicated by the inequality. For (viii), draw the vertical line $x = -2$.
Key conventions: solid circle = boundary included ($\leqslant$, $\geqslant$, $=$); dashed circle = boundary excluded ($<$, $>$); single point for $= 0$.
$$\text{Eight Argand diagrams as described above.}$$
Verification: each condition has the form $|z - z_0| \circ r$ (or $\mathrm{Re}(z) = c$), and the geometry follows directly from the definition of modulus as distance.
Draw an Argand diagram showing the set of points $z$ for which $|z - 12 + 5\mathrm{i}| \leqslant 7$. Use the diagram to prove that, for these $z$, $6 \leqslant |z| \leqslant 20$.
💡 Hint
Rewrite the condition as $|z - (12 - 5\mathrm{i})| \leqslant 7$: a closed disk centre $12 - 5\mathrm{i}$, radius $7$.
$|z|$ is the distance from $z$ to the origin. The greatest and least values of $|z|$ on the disk occur on the line through the origin and the centre.
Compute the distance from the origin to the centre: $|12 - 5\mathrm{i}| = \sqrt{144 + 25} = \sqrt{169} = 13$.
The least $|z|$ is $13 - 7$ (centre minus radius); the greatest is $13 + 7$ (centre plus radius).
👁 Reveal Solution
Rewrite $|z - 12 + 5\mathrm{i}| \leqslant 7$ as $|z - (12 - 5\mathrm{i})| \leqslant 7$.
This is a closed disk with centre $z_0 = 12 - 5\mathrm{i}$ (the point $(12, -5)$) and radius $7$.
$|z_0| = |12 - 5\mathrm{i}| = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13$.
So the centre of the disk is at distance $13$ from the origin.
The least $|z|$ occurs at the point on the disk closest to the origin, which is on the line from the origin through the centre, at distance $|z_0| - 7 = 13 - 7 = 6$.
The greatest $|z|$ occurs at the point on the disk farthest from the origin, at distance $|z_0| + 7 = 13 + 7 = 20$.
$$\boxed{6 \leqslant |z| \leqslant 20}$$
The disk of radius $7$ centred at $12 - 5\mathrm{i}$ lies entirely at distances between $6$ and $20$ from the origin.
(i) On an Argand diagram, show the region R for which $|z - 5 + 4\mathrm{i}| \leqslant 3$.
(ii) Find the greatest and least values of $|z + 3 - 2\mathrm{i}|$ in the region R.
💡 Hint
Part (i): rewrite $|z - 5 + 4\mathrm{i}| \leqslant 3$ as $|z - (5 - 4\mathrm{i})| \leqslant 3$. This is a closed disk centre $5 - 4\mathrm{i}$ (the point $(5, -4)$), radius $3$.
Part (ii): $|z + 3 - 2\mathrm{i}| = |z - (-3 + 2\mathrm{i})|$ is the distance from $z$ to the fixed point $-3 + 2\mathrm{i}$ (the point $(-3, 2)$).
The greatest and least distances from a fixed point to a disk occur on the line through the fixed point and the centre of the disk.
Compute the distance from $(-3, 2)$ to the centre $(5, -4)$ of the disk.
👁 Reveal Solution
Rewrite $|z - 5 + 4\mathrm{i}| \leqslant 3$ as $|z - (5 - 4\mathrm{i})| \leqslant 3$.
Region R is a closed disk with centre $z_0 = 5 - 4\mathrm{i}$ (point $(5, -4)$) and radius $3$.
$|z + 3 - 2\mathrm{i}| = |z - (-3 + 2\mathrm{i})|$ is the distance from $z$ to the fixed point $w = -3 + 2\mathrm{i}$ (point $(-3, 2)$).
$|w - z_0| = |(-3 + 2\mathrm{i}) - (5 - 4\mathrm{i})| = |-8 + 6\mathrm{i}| = \sqrt{64 + 36} = \sqrt{100} = 10$.
So the fixed point $w$ is at distance $10$ from the centre of the disk R.
Least value: $|w - z_0| - r = 10 - 3 = 7$ (closest point on the disk).
Greatest value: $|w - z_0| + r = 10 + 3 = 13$ (farthest point on the disk).
$$\boxed{\text{Least value of } |z + 3 - 2\mathrm{i}| = 7, \quad \text{Greatest value} = 13}$$
Verification: $7 + 13 = 20 = 2 \times 10 = 2|w - z_0|$ ✓ and $13 - 7 = 6 = 2 \times 3 = 2r$ ✓.
By using an Argand diagram see if it is possible to find values of $z$ for which $|z - 2 + \mathrm{i}| \geqslant 10$ and $|z + 4 + 2\mathrm{i}| \leqslant 2$ simultaneously.
💡 Hint 1
Rewrite each condition in the form $|z - z_k| \circ r$:
$|z - 2 + \mathrm{i}| \geqslant 10 \Leftrightarrow |z - (2 - \mathrm{i})| \geqslant 10$: exterior of disk centre $z_1 = 2 - \mathrm{i}$, radius $10$.
$|z + 4 + 2\mathrm{i}| \leqslant 2 \Leftrightarrow |z - (-4 - 2\mathrm{i})| \leqslant 2$: closed disk centre $z_2 = -4 - 2\mathrm{i}$, radius $2$.
Draw both regions on the same Argand diagram and check whether they overlap.
💡 Hint 2
Compute the distance between the two centres:
$|z_1 - z_2| = |(2 - \mathrm{i}) - (-4 - 2\mathrm{i})| = |6 + \mathrm{i}| = \sqrt{36 + 1} = \sqrt{37} \approx 6.08$.
The closest point of the second disk to $z_1$ is at distance $|z_1 - z_2| - r_2 = \sqrt{37} - 2 \approx 4.08$, which is less than $10$.
So the entire second disk lies inside the inner boundary of the first region, meaning the two regions do not overlap.
👁 Reveal Solution
$|z - 2 + \mathrm{i}| \geqslant 10 \Leftrightarrow |z - (2 - \mathrm{i})| \geqslant 10$.
This is the exterior (including boundary) of a disk centre $z_1 = 2 - \mathrm{i}$, radius $r_1 = 10$.
$|z + 4 + 2\mathrm{i}| \leqslant 2 \Leftrightarrow |z - (-4 - 2\mathrm{i})| \leqslant 2$.
This is a closed disk centre $z_2 = -4 - 2\mathrm{i}$, radius $r_2 = 2$.
$|z_1 - z_2| = |(2 - \mathrm{i}) - (-4 - 2\mathrm{i})| = |6 + \mathrm{i}| = \sqrt{6^2 + 1^2} = \sqrt{37} \approx 6.08$.
The greatest distance from $z_1$ to any point of the second disk is $|z_1 - z_2| + r_2 = \sqrt{37} + 2 \approx 8.08$.
Since $8.08 < 10 = r_1$, every point of the second disk is strictly inside the inner boundary of the first region.
Therefore the second disk lies entirely in the region $|z - z_1| < 10$, which is the complement of the first region. The two regions do not overlap.
$$\boxed{\text{There are no values of } z \text{ satisfying both conditions simultaneously.}}$$
The disk $|z + 4 + 2\mathrm{i}| \leqslant 2$ is entirely contained inside the open disk $|z - 2 + \mathrm{i}| < 10$, so no point satisfies $|z - 2 + \mathrm{i}| \geqslant 10$ and $|z + 4 + 2\mathrm{i}| \leqslant 2$ at the same time.
Exit Ticket
Before you leave, tick the objectives you are confident you can achieve. Be honest — this guides next lesson's review.