60-second diagnostic
Test your recall across all the complex number topics we have covered. Pick the correct option in each question without looking at notes.
A student is asked to find the principal argument of $z = -3$. They write:
$\arg(-3) = \tan^{-1}\left(\dfrac{0}{-3}\right) = \tan^{-1}(0) = 0$
Why is this wrong? The point $(-3, 0)$ lies on the negative real axis, not the positive real axis. The principal argument of a negative real number is $\pi$ (or $180^\circ$), not $0$. The formula $\arg(x + \mathrm{i}y) = \tan^{-1}(y/x)$ only gives the correct quadrant when the point is in the first or fourth quadrant. For points in the second or third quadrant, you must add or subtract $\pi$.
Correct: $\arg(-3) = \pi$ (or $\arg(-3) = -\pi$, depending on the principal range used). The point $(-3,0)$ is on the negative real axis, so its argument is $\pi$ radians.
This mistake is very common under exam pressure. Always check which quadrant your point lies in before giving the argument!
A complex number $z = x + \mathrm{i}y$ has real part $x$ and imaginary part $y$. Addition and subtraction are component-wise. Multiplication follows normal algebra with $\mathrm{i}^2 = -1$.
The conjugate of $z = x + \mathrm{i}y$ is $z^* = x - \mathrm{i}y$. Division uses the conjugate to clear the denominator: $\dfrac{z_1}{z_2} = \dfrac{z_1 z_2^*}{|z_2|^2}$.
Modulus $|z| = \sqrt{x^2 + y^2}$ is the distance from the origin. Argument $\arg z$ is the angle measured anticlockwise from the positive real axis. The principal argument lies in $(-\pi, \pi]$.
Any complex number can be written as $z = r(\cos\theta + \mathrm{i}\sin\theta)$ where $r = |z|$ and $\theta = \arg z$. This is also written as $z = re^{\mathrm{i}\theta}$ using Euler's formula.
To find $\sqrt{a + b\mathrm{i}}$, set $(p + \mathrm{i}q)^2 = a + b\mathrm{i}$ and equate real/imaginary parts. For polynomial equations with real coefficients, complex roots occur in conjugate pairs (Conjugate Root Theorem).
Loci describe sets of points satisfying geometric conditions. Key types:
- $|z - a| = k$ — circle centre $a$, radius $k$
- $|z - a| = |z - b|$ — perpendicular bisector of $a$ and $b$
- $\arg(z - a) = \alpha$ — half-line from $a$ at angle $\alpha$
- $\alpha \leqslant \arg(z - a) \leqslant \beta$ — sector (wedge) from $a$
- Combined: intersection of modulus and argument conditions
In groups of 4, each member starts at a different station. Work through the problem at your station, then rotate clockwise to the next station. Each station builds on a different topic. After all rotations, compare answers as a group.
Time: 4 minutes per station, then rotate. Total: 16 minutes + 5 minutes group discussion.
Given $z = 2 - 3\mathrm{i}$ and $w = 4 + \mathrm{i}$, compute:
(a) $z + w$ (b) $z - w$ (c) $zw$ (d) $z^2$
Express all answers in the form $x + \mathrm{i}y$.
(a) $z + w = (2 + 4) + (-3 + 1)\mathrm{i} = 6 - 2\mathrm{i}$
(b) $z - w = (2 - 4) + (-3 - 1)\mathrm{i} = -2 - 4\mathrm{i}$
(c) $zw = (2 - 3\mathrm{i})(4 + \mathrm{i}) = 8 + 2\mathrm{i} - 12\mathrm{i} - 3\mathrm{i}^2 = 8 - 10\mathrm{i} + 3 = 11 - 10\mathrm{i}$
(d) $z^2 = (2 - 3\mathrm{i})^2 = 4 - 12\mathrm{i} + 9\mathrm{i}^2 = 4 - 12\mathrm{i} - 9 = -5 - 12\mathrm{i}$
Given $z = 2 - 3\mathrm{i}$, find:
(a) $z^*$ (b) $|z|$ (c) $\dfrac{1}{z}$ in $x + \mathrm{i}y$ form
(a) $z^* = 2 + 3\mathrm{i}$
(b) $|z| = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}$
(c) $\dfrac{1}{z} = \dfrac{1}{2 - 3\mathrm{i}} \times \dfrac{2 + 3\mathrm{i}}{2 + 3\mathrm{i}} = \dfrac{2 + 3\mathrm{i}}{4 + 9} = \dfrac{2 + 3\mathrm{i}}{13} = \dfrac{2}{13} + \dfrac{3}{13}\mathrm{i}$
(a) Express $z = -5 + 12\mathrm{i}$ in polar form (find $|z|$ and $\arg z$).
(b) Find the two square roots of $z$ in Cartesian form.
(a) $|z| = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$.
$\arg z = \pi - \tan^{-1}(12/5) = \pi - 1.176 \approx 1.966$ rad (or $112.6^\circ$).
(b) Let $(p + \mathrm{i}q)^2 = -5 + 12\mathrm{i}$. Then $p^2 - q^2 = -5$ and $2pq = 12$, so $q = 6/p$.
Substituting: $p^2 - 36/p^2 = -5 \implies p^4 + 5p^2 - 36 = 0 \implies (p^2 + 9)(p^2 - 4) = 0$.
$p^2 = 4$ (since $p^2 \geq 0$), so $p = \pm 2$, $q = 6/p = \pm 3$.
The square roots are $2 + 3\mathrm{i}$ and $-2 - 3\mathrm{i}$.
(a) Describe the set of points satisfying $|z - 2 + 3\mathrm{i}| = 5$.
(b) What does $\arg(z + 1 - 2\mathrm{i}) = \dfrac{\pi}{3}$ represent?
(c) Sketch the region given by $|z - \mathrm{i}| \leqslant 3$ and $0 \leqslant \arg(z - \mathrm{i}) \leqslant \dfrac{\pi}{2}$.
(a) A circle with centre $(2, -3)$ and radius $5$.
(b) A half-line from $(-1, 2)$ at $60^\circ$ to the positive real axis.
(c) A quarter-circular disc: the intersection of a disc centre $(0,1)$ radius $3$ with the first-quadrant sector (angles $0^\circ$ to $90^\circ$) from $(0,1)$.
After all rotations: Compare your answers as a group. Discuss any station where group members disagree. Use the Check buttons to verify.
The complex number $z$ is given by $z = \dfrac{4 - 3\mathrm{i}}{1 - 2\mathrm{i}}$.
(i) (a) Express $z$ in the form $x + \mathrm{i}y$, where $x$ and $y$ are real.
(b) Find the modulus and argument of $z$.
(ii) Find the two square roots of the complex number $5 - 12\mathrm{i}$, giving your answers in the form $x + \mathrm{i}y$, where $x$ and $y$ are real.
$$z = \frac{4 - 3\mathrm{i}}{1 - 2\mathrm{i}} \times \frac{1 + 2\mathrm{i}}{1 + 2\mathrm{i}} = \frac{(4 - 3\mathrm{i})(1 + 2\mathrm{i})}{1^2 + 2^2} = \frac{4 + 8\mathrm{i} - 3\mathrm{i} - 6\mathrm{i}^2}{5}$$
$$= \frac{4 + 5\mathrm{i} + 6}{5} = \frac{10 + 5\mathrm{i}}{5} = 2 + \mathrm{i}$$
$|z| = |2 + \mathrm{i}| = \sqrt{2^2 + 1^2} = \sqrt{5}$
$\arg z = \tan^{-1}\left(\frac{1}{2}\right)$ (since $z$ is in the first quadrant)
$\arg z \approx 0.4636$ rad (or $26.565^\circ$)
Let $(p + \mathrm{i}q)^2 = 5 - 12\mathrm{i}$.
$(p^2 - q^2) + 2pq\mathrm{i} = 5 - 12\mathrm{i}$.
Equating: $p^2 - q^2 = 5$ (1) and $2pq = -12$ (2).
From (2): $q = -\dfrac{6}{p}$. Substitute into (1):
$p^2 - \dfrac{36}{p^2} = 5 \implies p^4 - 5p^2 - 36 = 0$.
$(p^2 - 9)(p^2 + 4) = 0$, so $p^2 = 9$ (since $p^2 \geq 0$).
$p = \pm 3$, and $q = -\dfrac{6}{p} = \mp 2$.
The square roots of $5 - 12\mathrm{i}$ are $3 - 2\mathrm{i}$ and $-3 + 2\mathrm{i}$.
Verification: $(3 - 2\mathrm{i})^2 = 9 - 12\mathrm{i} + 4\mathrm{i}^2 = 9 - 12\mathrm{i} - 4 = 5 - 12\mathrm{i}$.
Solve these equations.
(i) $(1 + \mathrm{i})z = 3 + \mathrm{i}$
(ii) $(3 - 4\mathrm{i})(z - 1) = 10 - 5\mathrm{i}$
$z = \dfrac{3 + \mathrm{i}}{1 + \mathrm{i}} = \dfrac{(3 + \mathrm{i})(1 - \mathrm{i})}{(1 + \mathrm{i})(1 - \mathrm{i})} = \dfrac{3 - 3\mathrm{i} + \mathrm{i} - \mathrm{i}^2}{1 + 1} = \dfrac{3 - 2\mathrm{i} + 1}{2} = \dfrac{4 - 2\mathrm{i}}{2} = 2 - \mathrm{i}$.
$z - 1 = \dfrac{10 - 5\mathrm{i}}{3 - 4\mathrm{i}} = \dfrac{(10 - 5\mathrm{i})(3 + 4\mathrm{i})}{3^2 + 4^2} = \dfrac{30 + 40\mathrm{i} - 15\mathrm{i} - 20\mathrm{i}^2}{25} = \dfrac{30 + 25\mathrm{i} + 20}{25} = \dfrac{50 + 25\mathrm{i}}{25} = 2 + \mathrm{i}$.
Therefore $z = (2 + \mathrm{i}) + 1 = 3 + \mathrm{i}$.
(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.
Rewrite $|z - 5 + 4\mathrm{i}|$ as $|z - (5 - 4\mathrm{i})|$. This is a disc centre $(5, -4)$ radius $3$. The expression $|z + 3 - 2\mathrm{i}| = |z - (-3 + 2\mathrm{i})|$ is the distance from $z$ to $(-3, 2)$. The greatest/least distance from a point to a disc is the distance to the centre $\pm$ the radius.
$|z - 5 + 4\mathrm{i}| = |z - (5 - 4\mathrm{i})| \leqslant 3$ is a closed disc (including boundary) with centre $C = (5, -4)$ and radius $3$.
$|z + 3 - 2\mathrm{i}| = |z - (-3 + 2\mathrm{i})|$ is the distance from $z$ to point $P = (-3, 2)$.
Distance $PC = \sqrt{(-3 - 5)^2 + (2 - (-4))^2} = \sqrt{(-8)^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$.
The greatest distance from $P$ to a point in the disc is $PC + r = 10 + 3 = 13$.
The least distance from $P$ to a point in the disc is $PC - r = 10 - 3 = 7$.
Greatest value: $13$, Least value: $7$.
Check that $2 + \mathrm{i}$ is a root of $z^3 - z^2 - 7z + 15 = 0$, and find the other roots.
Substitute $z = 2 + \mathrm{i}$:
$(2 + \mathrm{i})^3 - (2 + \mathrm{i})^2 - 7(2 + \mathrm{i}) + 15$
$(2 + \mathrm{i})^2 = 4 + 4\mathrm{i} + \mathrm{i}^2 = 3 + 4\mathrm{i}$
$(2 + \mathrm{i})^3 = (2 + \mathrm{i})(3 + 4\mathrm{i}) = 6 + 8\mathrm{i} + 3\mathrm{i} + 4\mathrm{i}^2 = 2 + 11\mathrm{i}$
Expression: $(2 + 11\mathrm{i}) - (3 + 4\mathrm{i}) - 7(2 + \mathrm{i}) + 15$
$= 2 + 11\mathrm{i} - 3 - 4\mathrm{i} - 14 - 7\mathrm{i} + 15$
$= (2 - 3 - 14 + 15) + (11 - 4 - 7)\mathrm{i} = 0 + 0\mathrm{i} = 0$
Therefore $2 + \mathrm{i}$ is a root. $\checkmark$
Since coefficients are real, $2 - \mathrm{i}$ is also a root.
The quadratic factor is $(z - (2 + \mathrm{i}))(z - (2 - \mathrm{i})) = z^2 - 4z + 5$.
Divide $z^3 - z^2 - 7z + 15$ by $z^2 - 4z + 5$:
$z^3 - z^2 - 7z + 15 = (z^2 - 4z + 5)(z + 3)$
Verification: $(z^2 - 4z + 5)(z + 3) = z^3 - 4z^2 + 5z + 3z^2 - 12z + 15 = z^3 - z^2 - 7z + 15$.
The three roots are: $2 + \mathrm{i}$, $2 - \mathrm{i}$, and $-3$.
Given that $1 - \mathrm{i}$ is a root of $z^3 + pz^2 + qz + 12 = 0$, find the real numbers $p$ and $q$, and the other roots.
Since coefficients are real, $1 + \mathrm{i}$ is also a root. Form the quadratic factor from these two roots, then divide the cubic by this quadratic to find the third root and the values of $p$ and $q$.
Since coefficients are real, $1 + \mathrm{i}$ is also a root.
The quadratic factor from these roots is:
$(z - (1 - \mathrm{i}))(z - (1 + \mathrm{i})) = z^2 - 2z + 2$.
Let the third root be $r$. Then $z^3 + pz^2 + qz + 12 = (z^2 - 2z + 2)(z - r)$.
Expanding: $(z^2 - 2z + 2)(z - r) = z^3 - (r + 2)z^2 + (2r + 2)z - 2r$.
Comparing with $z^3 + pz^2 + qz + 12$:
Constant term: $-2r = 12 \implies r = -6$.
$z^2$ coefficient: $-(r + 2) = p \implies -(-6 + 2) = p \implies p = 4$.
$z^1$ coefficient: $2r + 2 = q \implies 2(-6) + 2 = q \implies q = -10$.
$p = 4$, $q = -10$. The three roots are: $1 - \mathrm{i}$, $1 + \mathrm{i}$, and $-6$.
(i) It is given that $(1 + 3\mathrm{i})w = 2 + 4\mathrm{i}$. Showing all necessary working, prove that the exact value of $|w^2|$ is $2$ and find $\arg(w^2)$ correct to 3 significant figures.
(ii) On a single Argand diagram sketch the loci $|z| = 5$ and $|z - 5| = |z|$. Hence determine the complex numbers represented by points common to both loci, giving each answer in the form $re^{\mathrm{i}\theta}$.
For (i), find $w$ first, then $|w|$, then $|w^2| = |w|^2$. For $\arg(w^2)$, compute $\arg(w)$ first and double it. For (ii), $|z| = 5$ is a circle centred at origin. $|z - 5| = |z|$ is the perpendicular bisector of $0$ and $5$.
$w = \dfrac{2 + 4\mathrm{i}}{1 + 3\mathrm{i}} = \dfrac{(2 + 4\mathrm{i})(1 - 3\mathrm{i})}{1^2 + 3^2} = \dfrac{(2 + 4\mathrm{i})(1 - 3\mathrm{i})}{10}$
$= \dfrac{2 - 6\mathrm{i} + 4\mathrm{i} - 12\mathrm{i}^2}{10} = \dfrac{2 - 2\mathrm{i} + 12}{10} = \dfrac{14 - 2\mathrm{i}}{10} = \dfrac{7 - \mathrm{i}}{5}$
$|w| = \left|\dfrac{7 - \mathrm{i}}{5}\right| = \dfrac{|7 - \mathrm{i}|}{5} = \dfrac{\sqrt{7^2 + (-1)^2}}{5} = \dfrac{\sqrt{50}}{5} = \dfrac{5\sqrt{2}}{5} = \sqrt{2}$
$|w^2| = |w|^2 = (\sqrt{2})^2 = 2$
$\arg(w) = \arg(7 - \mathrm{i}) = -\tan^{-1}(1/7)$ (fourth quadrant)
$\arg(w) \approx -0.1419$ rad
$\arg(w^2) = 2\arg(w) \approx -0.284$ rad (3 s.f.)
$|z| = 5$: circle centre $0$, radius $5$.
$|z - 5| = |z|$: perpendicular bisector of $0$ and $5$, i.e. the vertical line $x = 2.5$.
Points of intersection: substitute $x = 2.5$ into $|z| = 5$:
$\sqrt{(2.5)^2 + y^2} = 5 \implies 6.25 + y^2 = 25 \implies y^2 = 18.75 = \frac{75}{4} \implies y = \pm \frac{5\sqrt{3}}{2}$.
Points: $z = \frac{5}{2} \pm \frac{5\sqrt{3}}{2}\mathrm{i}$
In polar form: $z = 5e^{\pm \mathrm{i}\pi/3}$
Verification: $|z| = 5$, $\arg z = \pm \pi/3$, and $|z - 5| = \sqrt{(\frac{5}{2} - 5)^2 + (\frac{5\sqrt{3}}{2})^2} = \sqrt{(-\frac{5}{2})^2 + \frac{75}{4}} = \sqrt{\frac{25}{4} + \frac{75}{4}} = \sqrt{\frac{100}{4}} = \sqrt{25} = 5 = |z|$.
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