Chapter 4 · Complex Numbers and Quadratic Equations

4.3 The Argand Plane

Every complex number $a+ib$ is a point $(a,b)$ of a plane. There the modulus becomes a distance, the conjugate a reflection, addition a parallelogram — and multiplication by $i$ a quarter turn.

Real numbers live on a line. Complex numbers need two real numbers each — a real part and an imaginary part — and a pair of real numbers is exactly what labels a point of a plane. Around 1800 two amateurs, the Norwegian surveyor Caspar Wessel and the Paris bookshop manager Jean-Robert Argand, independently had the idea of drawing $a+ib$ as the point $(a,b)$. It was the moment complex numbers stopped looking "imaginary": suddenly they had a place to live, and every algebraic rule from Sections 4.1 and 4.2 turned out to have a geometric meaning.

In this section you will plot complex numbers, and see that the modulus $|z|$ is a distance, that $\bar z$ and $-z$ are reflections, that $z_1+z_2$ completes a parallelogram, that $|z_1-z_2|$ is the distance between two points, and that multiplying by $i$ rotates the whole plane through a right angle about the origin. These facts are why engineers use complex numbers for anything that turns — alternating currents, robot arms, the rotation of an image on your phone screen.

Plotting Complex Numbers

📘 Definition 4.9 — The Argand plane

In the Argand plane (or complex plane) the complex number $z=a+ib$ is represented by the point $P(a,b)$. The $x$-axis, whose points represent the real numbers $a+0i$, is called the real axis; the $y$-axis, whose points represent the numbers $0+ib$, is called the imaginary axis. A drawing of complex numbers in this plane is called an Argand diagram.

This is the ordered pair of Section 2.1 in disguise: $a+ib\leftrightarrow(a,b)$. Two complex numbers are equal exactly when they give the same point, which is Definition 4.3 again. And by the distance formula, the distance from the origin $O$ to $P(a,b)$ is $$OP=\sqrt{a^2+b^2}=|z|.$$ The modulus of $z$ is the distance of its point from the origin. So the complex numbers with $|z|=5$ are exactly the points of the circle of radius $5$ centred at $O$.

✏️ Example 1 — Plotting points

Plot $z_1=3+2i$, $z_2=-2+i$, $z_3=-1-3i$, $z_4=4i$ and $z_5=-3$ in the Argand plane. For each, say where it lies and find its distance from the origin.

💡 Show solution▶

Points $z_1\to(3,2)$, $z_2\to(-2,1)$, $z_3\to(-1,-3)$, $z_4\to(0,4)$, $z_5\to(-3,0)$.

Where $z_1$ in the first quadrant, $z_2$ in the second, $z_3$ in the third; $z_4$ on the positive imaginary axis; $z_5$ on the negative real axis.

Distances $|z_1|=\sqrt{13}$, $|z_2|=\sqrt5$, $|z_3|=\sqrt{10}$, $|z_4|=4$, $|z_5|=3$.

📐 Theorem 4.8 — Conjugate and negative as reflections

Let $z=a+ib$ be represented by $P(a,b)$. Then:

$\bar z=a-ib$ is represented by $(a,-b)$, the reflection of $P$ in the real axis;
$-z=-a-ib$ is represented by $(-a,-b)$, the reflection of $P$ in the origin (a half turn about $O$);
$-\bar z=-a+ib$ is represented by $(-a,b)$, the reflection of $P$ in the imaginary axis.

Proof. Reflecting $(a,b)$ in the $x$-axis changes only the sign of the $y$-coordinate; reflecting in the origin changes both signs; reflecting in the $y$-axis changes only the sign of the $x$-coordinate. Since reflections do not change distances from $O$, this also shows $|\bar z|=|-z|=|z|$ (Theorem 4.6). $\blacksquare$

✏️ Example 2 — Four reflections make a rectangle

For $z=3+2i$, plot $z$, $\bar z$, $-z$ and $-\bar z$. Show that they are the vertices of a rectangle centred at $O$, and find its area.

💡 Show solution▶

Points $z\to(3,2)$, $\bar z\to(3,-2)$, $-z\to(-3,-2)$, $-\bar z\to(-3,2)$.

Shape Two sides are vertical ($x=\pm3$) and two horizontal ($y=\pm2$), so the four points form a rectangle with sides $6$ and $4$. Its diagonals meet at the midpoint of $z$ and $-z$, which is $O$.

Answer Area $=6\times4=24$ square units.

Addition, Subtraction and Distance

📐 Theorem 4.9 — The parallelogram law of addition

Let $z_1$ and $z_2$ be represented by the points $P_1$ and $P_2$, where $O$, $P_1$, $P_2$ do not lie on one line. Then $z_1+z_2$ is represented by the fourth vertex $P$ of the parallelogram $OP_1PP_2$.

Proof. Let $z_1=a+ib$ and $z_2=c+id$, so $z_1+z_2$ is the point $P(a+c,\,b+d)$. The midpoint of the diagonal $OP$ is $\big(\tfrac{a+c}{2},\tfrac{b+d}{2}\big)$, and the midpoint of the other diagonal $P_1P_2$ is also $\big(\tfrac{a+c}{2},\tfrac{b+d}{2}\big)$. A quadrilateral whose diagonals bisect each other is a parallelogram. $\blacksquare$

Another way to see this: to get from $O$ to $z_1+z_2$, walk along the arrow from $O$ to $z_1$, then along an arrow of the same length and direction as the one from $O$ to $z_2$. Adding complex numbers is adding arrows "nose to tail" — the same rule that adds displacements and forces in physics.

📐 Theorem 4.10 — Distance between two points

If $z_1$ and $z_2$ are represented by the points $P_1$ and $P_2$, then $$P_1P_2=|z_1-z_2|.$$

Proof. With $z_1=a+ib$ and $z_2=c+id$: $z_1-z_2=(a-c)+i(b-d)$, so $|z_1-z_2|=\sqrt{(a-c)^2+(b-d)^2}$, which is the distance formula for $P_1(a,b)$ and $P_2(c,d)$. $\blacksquare$

✏️ Example 3 — A parallelogram and a distance

Let $z_1=1+2i$ and $z_2=3-i$. (a) Find $z_1+z_2$ and check that it is the fourth vertex of the parallelogram with vertices $O$, $z_1$, $z_2$. (b) Find the distance between the points $z_1$ and $z_2$.

💡 Show solution▶

(a) $z_1+z_2=4+i$, the point $(4,1)$.

(a) Check Midpoint of $O$ and $(4,1)$: $(2,\tfrac12)$. Midpoint of $(1,2)$ and $(3,-1)$: $(2,\tfrac12)$ ✓ — the diagonals bisect each other.

(b) $z_1-z_2=-2+3i$, so the distance is $|{-2+3i}|=\sqrt{4+9}=\sqrt{13}\approx3.61$.

Drag the point $z$ (and $w$, when it is shown) with your mouse or finger; points snap to halves. Choose what to see: the reflections $\bar z$ and $-z$; the parallelogram that adds $z+w$; the distance $|z-w|$; or the quarter turn $z\mapsto iz$. Press Play to send $z$ on a tour while the readouts update.

Show:

z—

w—

result—

modulus—

✏️ Example 4 — What kind of triangle?

Show that the points $A=2+i$, $B=5+5i$ and $C=-2+4i$ are the vertices of a right-angled isosceles triangle, and find its area.

💡 Show solution▶

Sides $AB=|(5+5i)-(2+i)|=|3+4i|=5$; $\ AC=|(-2+4i)-(2+i)|=|{-4+3i}|=5$; $\ BC=|(-2+4i)-(5+5i)|=|{-7-i}|=\sqrt{50}$.

Type $AB=AC$, so the triangle is isosceles; and $AB^2+AC^2=25+25=50=BC^2$, so by the converse of Pythagoras' theorem the angle at $A$ is a right angle.

Area $\tfrac12\cdot AB\cdot AC=\tfrac12\cdot5\cdot5=12.5$ square units.

✏️ Example 5 — Describing sets of points

Describe the set of points $z$ in the Argand plane for which (a) $|z|=5$ (b) $|z-(1+i)|=2$ (c) (challenge) $|z-2|=|z+2i|$.

💡 Show solution▶

(a) Distance $5$ from $O$: the circle with centre $O$ and radius $5$.

(b) By Theorem 4.10, distance $2$ from the point $1+i$: the circle with centre $(1,1)$ and radius $2$.

(c) Meaning $z$ is equally far from the points $2$ and $-2i$, i.e. from $(2,0)$ and $(0,-2)$: the perpendicular bisector of the segment joining them.

(c) Equation With $z=x+iy$: $(x-2)^2+y^2=x^2+(y+2)^2$, so $-4x+4=4y+4$, i.e. the line $y=-x$. Check: $z=1-i$ is at distance $\sqrt2$ from both $(2,0)$ and $(0,-2)$ ✓.

Multiplication by i: a Quarter Turn

Multiplying by a positive real number $k$ simply stretches: $kz=ka+ikb$ is the point $(ka,kb)$, on the same ray from $O$ as $z$ but $k$ times as far out. Multiplying by $i$ does something far more interesting. For $z=a+ib$, $$iz=ia+i^2b=-b+ia,$$ so the point $(a,b)$ moves to $(-b,a)$. Try $z=1$: it goes to $i$ — from the positive real axis to the positive imaginary axis, a quarter turn anticlockwise.

📐 Theorem 4.11 — Multiplication by $i$ is a rotation through $90^\circ$

Let $z\ne0$ be represented by $P$ and $iz$ by $P'$. Then $OP'=OP$ and $\angle POP'=90^\circ$: the point $P'$ is obtained by rotating $P$ through $90^\circ$ anticlockwise about $O$.

Proof. Let $z=a+ib$, so $P=(a,b)$ and $P'=(-b,a)$. First, $OP'=\sqrt{b^2+a^2}=OP$. Next, $$\begin{aligned}PP'^2&=(a+b)^2+(b-a)^2\\&=2a^2+2b^2=OP^2+OP'^2,\end{aligned}$$ so by the converse of Pythagoras' theorem the angle at $O$ is a right angle. Finally, the turn is anticlockwise: $1\mapsto i\mapsto-1\mapsto-i\mapsto1$ goes round the axes from east to north to west to south. $\blacksquare$

So the four-step cycle of powers of $i$ in Section 4.1 is really a rotation: multiplying by $i^2=-1$ is a half turn (Theorem 4.8 again), by $i^3=-i$ a quarter turn clockwise, and by $i^4=1$ a full turn. Multiplying by $ki$ ($k\gt0$) first turns through $90^\circ$ and then stretches by $k$.

✏️ Example 6 — A square from quarter turns

Let $z=3+i$. (a) Find $iz$, $i^2z$ and $i^3z$, and plot all four points. (b) Show that they are the vertices of a square centred at $O$, and find its area. (c) Find the point obtained by rotating $z$ through $90^\circ$ anticlockwise about $O$ and then doubling its distance from $O$.

💡 Show solution▶

(a) $iz=3i+i^2=-1+3i$; $\ i^2z=-z=-3-i$; $\ i^3z=-iz=1-3i$. Points: $(3,1)$, $(-1,3)$, $(-3,-1)$, $(1,-3)$.

(b) Each point is the previous one turned through $90^\circ$ about $O$ (Theorem 4.11), so all four are at distance $|z|=\sqrt{10}$ from $O$ and consecutive ones subtend right angles there: a square centred at $O$. Its side is $|z-iz|=|4-2i|=\sqrt{20}$, so its area is $20$ square units.

(b) Check The diagonal is $|z-(-z)|=2\sqrt{10}$, and a square with diagonal $d$ has area $\tfrac12d^2=\tfrac12\cdot40=20$ ✓.

Choose a multiplier $w$ (or drag $w$ and $z$). The blue triangle has corners $0$, $1$, $w$; the purple triangle has corners $0$, $z$, $zw$. They always have the same shape: multiplying by $w$ turns $z$ through the same angle that takes $1$ to $w$, and stretches it by the factor $|w|$. For $w=i$ this is the quarter turn of Theorem 4.11. Press Play to swing $w$ round a circle.

Multiplier:

$z$ and $w$—

$zw$—

$|z|\times|w|$—

$|zw|$—

⚠️ Common mistakes in the Argand plane

1. The imaginary axis is marked with $b$, not $ib$. $3-4i$ is the point $(3,-4)$, not $(3,-4i)$.

2. Reflection in the real axis gives $\bar z$, not $-z$. $-z$ is the reflection in the origin.

3. Distance needs a difference. The distance between $z_1$ and $z_2$ is $|z_1-z_2|$, not $|z_1|-|z_2|$: for $z_1=3$ and $z_2=3i$ the distance is $3\sqrt2$, while $|z_1|-|z_2|=0$.

4. Multiplying by $i$ turns anticlockwise. $i\cdot1=i$ (east to north). A quarter turn clockwise is multiplication by $-i$.

5. Use the same scale on both axes. Otherwise circles $|z|=r$ look like ellipses and quarter turns stop looking like right angles.

📘 In the CBSE syllabus — school assessment only (not in the year-end board exam)

Taught and tested in school, but not in the year-end board exam; JEE Main and JEE Advanced both list it and use the polar (modulus–argument) form of a complex number all the time.

Polar form. If $z\ne0$ is represented by $P$, let $r=|z|=OP$ and let $\theta$ be the angle from the positive real axis to $OP$. Then $a=r\cos\theta$ and $b=r\sin\theta$ (sine and cosine of any angle, Chapter 3), so $$z=r(\cos\theta+i\sin\theta).$$ The angle $\theta$ is an argument of $z$, written $\arg z$; the principal argument is the one with $-\pi\lt\theta\le\pi$. Using the sum formulae for $\cos(\alpha+\beta)$ and $\sin(\alpha+\beta)$ one finds that $|z_1z_2|=|z_1||z_2|$ and $\arg(z_1z_2)=\arg z_1+\arg z_2$ (up to a multiple of $2\pi$) — the "turn and stretch" of Interactive 2.

Worked example. Write $z=-1+i\sqrt3$ in polar form.
Solution. $r=\sqrt{1+3}=2$. We need $\cos\theta=-\tfrac12$ and $\sin\theta=\tfrac{\sqrt3}{2}$: the point $(-1,\sqrt3)$ is in the second quadrant, so $\theta=\tfrac{2\pi}{3}$. Hence $z=2\Big(\cos\tfrac{2\pi}{3}+i\sin\tfrac{2\pi}{3}\Big)$. Check: $2\cos\tfrac{2\pi}{3}=-1$ and $2\sin\tfrac{2\pi}{3}=\sqrt3$ ✓.

Exercises

Plotting. In parts 1–4 plot the number, say where it lies, and find its distance from the origin.

1 $4-3i$
2 $-5+12i$
3 $-6$
4 $-2i$
5 Which of $3i$, $3$, $0$, $-i\sqrt2$ are represented by points of the imaginary axis?
6 Describe the set of points $z$ with (a) $\operatorname{Re}z=2$ (b) $\operatorname{Im}z=-1$.
7 For $z=-4+3i$, give $\bar z$, $-z$ and $-\bar z$, and the quadrant in which each lies.
8 Find every complex number $z$ with $|z|=5$ whose real and imaginary parts are integers and whose point lies inside the first quadrant.

💡 Show solutions▶

1 $(4,-3)$, fourth quadrant; distance $5$.

2 $(-5,12)$, second quadrant; distance $13$.

3 $(-6,0)$, on the negative real axis; distance $6$.

4 $(0,-2)$, on the negative imaginary axis; distance $2$.

5 $3i$, $0$ and $-i\sqrt2$ (the origin lies on both axes); $3$ is on the real axis.

6 (a) The vertical line $x=2$. (b) The horizontal line $y=-1$.

7 $\bar z=-4-3i$ (third quadrant); $-z=4-3i$ (fourth); $-\bar z=4+3i$ (first).

8 We need $a^2+b^2=25$ with $a,b$ positive integers: $3+4i$ and $4+3i$. ($5$ and $5i$ lie on the axes, not inside the quadrant.)

Sums, differences and distances.

1 For $z_1=2+3i$ and $z_2=4-i$, find the fourth vertex of the parallelogram with vertices $O$, $z_1$, $z_2$ (with $O$ opposite the new vertex).
2 Find the distance between the points $z_1$ and $z_2$ of part 1.
3 Find the distance between $-3+2i$ and $1-i$.
4 $ABCD$ is a parallelogram with $A=1+i$, $B=4+2i$ and $C=5+5i$. Find $D$.
5 Show that $A=1+2i$, $B=4+6i$, $C=7+2i$ form an isosceles triangle.
6 Use distances to show that $1+i$, $3+3i$ and $5+5i$ lie on one straight line.
7 Find the centre and radius of the circle $|z-3+4i|=5$. Does it pass through $O$?
8 Find the points of the real axis at distance $5$ from $1+3i$.
9 For $z_1=3+4i$ and $z_2=5-12i$, compare $|z_1+z_2|$ with $|z_1|+|z_2|$, and explain the result using the parallelogram.

💡 Show solutions▶

1 $z_1+z_2=6+2i$, the point $(6,2)$.

2 $|z_1-z_2|=|{-2+4i}|=\sqrt{20}=2\sqrt5$.

3 $|(-3+2i)-(1-i)|=|{-4+3i}|=5$.

4 The diagonals $AC$ and $BD$ bisect each other, so $A+C=B+D$: $D=A+C-B=(1+i)+(5+5i)-(4+2i)=2+4i$.

5 $AB=|3+4i|=5$, $BC=|3-4i|=5$, $AC=|6|=6$: two equal sides.

6 The distances are $|2+2i|=2\sqrt2$, $|2+2i|=2\sqrt2$ and $|4+4i|=4\sqrt2$, and $2\sqrt2+2\sqrt2=4\sqrt2$: the middle point lies on the segment joining the other two.

7 $|z-(3-4i)|=5$: centre $(3,-4)$, radius $5$. $|0-(3-4i)|=5$, so yes, it passes through $O$.

8 $z=x$: $(x-1)^2+9=25$, $x-1=\pm4$, so $x=5$ or $x=-3$: the points $5$ and $-3$.

9 $|8-8i|=8\sqrt2\approx11.31$ and $5+13=18$. In the parallelogram, $|z_1+z_2|$ is a diagonal and $|z_1|$, $|z_2|$ are two sides of a triangle with that diagonal; one side of a triangle is shorter than the sum of the other two.

III

Turning with $i$.

1 For $z=2+5i$, find $iz$, $-z$ and $-iz$.
2 The point $P(4,1)$ is rotated about $O$ through $90^\circ$ (a) anticlockwise (b) clockwise. Find its new position.
3 Where does $P(4,1)$ go under a rotation of $180^\circ$ about $O$?
4 A square has centre $O$ and one vertex at $3+2i$. Find the other three vertices and the area.
5 Prove that for every $z\ne0$ the points $0$, $z$, $iz$ form a right-angled isosceles triangle.
6 Find $z$ if $iz=3-2i$.
7 Drones. A drone hovers at $30+40i$ metres (east and north of its base). It flies round the base, keeping its distance, until its position has turned through $90^\circ$ anticlockwise. Find its new position and its distance from the base.

💡 Show solutions▶

1 $iz=2i+5i^2=-5+2i$; $\ -z=-2-5i$; $\ -iz=5-2i$.

2 (a) $i(4+i)=-1+4i$: $(-1,4)$. (b) $-i(4+i)=1-4i$: $(1,-4)$.

3 Multiply by $-1$: $-4-i$, the point $(-4,-1)$.

4 Multiply by $i$ repeatedly: $-2+3i$, $-3-2i$, $2-3i$. Side $|z-iz|=|5-i|=\sqrt{26}$, area $26$ square units.

5 $|iz|=|i||z|=|z|$, and $|z-iz|=|z||1-i|=\sqrt2\,|z|$. Then $|z|^2+|iz|^2=2|z|^2=|z-iz|^2$: two equal sides with a right angle between them (converse of Pythagoras).

6 $z=\dfrac{3-2i}{i}=-i(3-2i)=-3i+2i^2=-2-3i$. Check: $i(-2-3i)=-2i+3=3-2i$ ✓.

7 $i(30+40i)=-40+30i$: $40$ m west and $30$ m north of the base, still $|{-40+30i}|=50$ m away.

Modelling.

1 Navigation. A boat leaves port $O$ and sails $12+5i$ km (east, north), then $4-8i$ km. Where does it end up, how far is it from port, and how far did it sail?
2 Screen graphics. A triangle on a phone screen has vertices $1$, $3$ and $2+2i$ (centre of the screen at $O$). Find its vertices after the screen image is rotated $90^\circ$ anticlockwise about $O$.

💡 Show solutions▶

1 End point $(12+5i)+(4-8i)=16-3i$; distance from port $|16-3i|=\sqrt{265}\approx16.28$ km; distance sailed $|12+5i|+|4-8i|=13+\sqrt{80}\approx21.94$ km.

2 Multiply each vertex by $i$: $i$, $3i$ and $i(2+2i)=-2+2i$.

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case studyJEE MainJEE Advanced

How this is examined: Board papers ask you to locate a number or its conjugate in the Argand plane (1 mark) and to use $|z_1-z_2|$ as a distance (2–3 marks). JEE turns conditions on $|z-z_0|$ into circles and lines, and uses multiplication by $i$ for right angles and squares.

E1. Board · 1 mark The point representing the conjugate of $-3+2i$ lies in the (A) first quadrant (B) second quadrant (C) third quadrant (D) fourth quadrant

💡 Show solution▶

Idea $\overline{-3+2i}=-3-2i$, the point $(-3,-2)$.

Trap (B) is where $-3+2i$ itself lies; (D) comes from changing the sign of the real part instead.

Answer (C).

E2. Board · 2 marks Find the distance between the points representing $z_1=2-3i$ and $z_2=-1+i$, and the complex number representing the midpoint of the segment joining them.

💡 Show solution▶

Distance $|z_1-z_2|=|3-4i|=5$. (1 mark)

Answer Midpoint $\Big(\tfrac{2-1}{2},\tfrac{-3+1}{2}\Big)=\big(\tfrac12,-1\big)$, i.e. $\dfrac{z_1+z_2}{2}=\dfrac12-i$. (1 mark)

E3. Board · 3 marks Show that the points $A=3+4i$, $B=-4+3i$, $C=-3-4i$ and $D=4-3i$ are the vertices of a square, and find its area.

💡 Show solution▶

Sides $AB=|7+i|=\sqrt{50}$, $BC=|{-1+7i}|=\sqrt{50}$, $CD=|{-7-i}|=\sqrt{50}$, $DA=|1-7i|=\sqrt{50}$. (1 mark)

Diagonals $AC=|6+8i|=10$ and $BD=|{-8+6i}|=10$. Four equal sides and equal diagonals: a square. (1 mark)

Answer Area $=(\sqrt{50})^2=50$ square units. (Shortcut: $B=iA$, $C=i^2A$, $D=i^3A$ — successive quarter turns, Theorem 4.11.) (1 mark)

E4. Board · Case study A delivery drone is based at the origin $O$ of an Argand plane in which $1$ unit is $100$ m (east along the real axis, north along the imaginary axis). Its first customer is at $P=6+8i$. (i) How far is $P$ from the base, in metres? (1 mark) (ii) A second customer $Q$ is at the position obtained by rotating $P$ through $90^\circ$ anticlockwise about $O$. Find $Q$. (1 mark) (iii) Find the distance between the two customers, in metres. (2 marks)

💡 Show solution▶

(i) $|6+8i|=10$ units $=1000$ m. (1 mark)

(ii) $Q=iP=6i+8i^2=-8+6i$. (1 mark)

Answer (iii) $|P-Q|=|14+2i|=\sqrt{200}=10\sqrt2\approx14.14$ units $\approx1414$ m. (Check: $OPQ$ is right-angled isosceles with legs $10$, so $PQ=10\sqrt2$.) (2 marks)

E5. JEE Main The set of points $z$ with $|z-1|=|z-i|$ is (A) the line $y=x$ (B) the line $y=-x$ (C) a circle through $1$ and $i$ (D) the line $x+y=1$

💡 Show solution▶

Idea $z$ is equally far from $(1,0)$ and $(0,1)$: the perpendicular bisector of that segment. With $z=x+iy$: $(x-1)^2+y^2=x^2+(y-1)^2$, so $-2x=-2y$.

Trap (D) is the line through $1$ and $i$, not the line equidistant from them.

Answer (A) $y=x$.

⏱ Fast route The perpendicular bisector passes through the midpoint $\big(\tfrac12,\tfrac12\big)$ and through $O$ (which is at distance $1$ from both points) — only (A) contains both.

E6. JEE Advanced Numerical answer. Let $|z|=4$. Find the area of the triangle whose vertices are the points $z$, $iz$ and $z+iz$.

💡 Show solution▶

Shape $O$, $z$, $z+iz$, $iz$ form a parallelogram (Theorem 4.9) whose sides $Oz$ and $O(iz)$ are equal ($|iz|=|z|=4$) and perpendicular (Theorem 4.11): a square of side $4$.

Triangle The triangle with vertices $z$, $iz$, $z+iz$ is the half of the square cut off by the diagonal from $z$ to $iz$.

Answer $\tfrac12\times16=8$.

🔬 Research–Teaching NexusMini project · about 45 minutes

Research question. How can a phone rotate or zoom a picture using nothing but complex multiplication — and which multipliers rotate a shape without distorting it?

Why it matters. Screen graphics, computer games, robot arms and drone flight controllers must turn shapes thousands of times a second. Treating every point $(x,y)$ as $x+iy$ turns "rotate and resize" into a single multiplication.

What to do.

On squared paper (or in a spreadsheet with an XY-scatter chart) draw a capital letter F using about eight grid points, e.g. $0$, $2i$, $4i$, $4i+2$, $2i+1$, …, and write each point as a complex number.
Multiply every point by $i$, by $-1$ and by $-i$, and plot each new letter. Describe each transformation.
Now multiply every point by $w=0.8+0.6i$ (note $|w|=1$) and then by $w=1+i$. Measure the new distances from $O$ and the side lengths of the letter. Which multiplier only turns, and which also enlarges? By what factor?
Multiply the original letter by $w=0.8+0.6i$ again and again. Does it ever come back exactly to where it started?

Tools: Squared paper, or a spreadsheet / free graphing app

Hand in: Your five plots, a table of moduli before and after, and a rule that tells you from $w$ alone how much a shape is enlarged.

Going further. Find a multiplier $w$ that rotates through $90^\circ$ and halves every distance.

Instructor note: by Theorem 4.6, $|wz|=|w||z|$, so every distance is multiplied by $|w|$ — by $1$ for $0.8+0.6i$ and by $\sqrt2$ for $1+i$ — while the shape is not distorted. The turn for $0.8+0.6i$ is about $36.87^\circ$, not a divisor of $360^\circ$, so repeated multiplication never returns exactly to the start. The "going further" answer is $w=\tfrac12i$.

🧪 Self-check quiz

0/5 correct

Q1. The point $(-2,5)$ of the Argand plane represents

Q2. The reflection of the point $z$ in the real axis represents

Q3. The distance between the points $1+i$ and $4+5i$ is

Q4. Multiplying every complex number by $i$

Q5. $O$, $2+i$ and $1+3i$ are three vertices of a parallelogram in which $O$ is opposite the fourth vertex. The fourth vertex is

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