CP 2.4 - Argand diagrams

CP 2.4 - Argand diagrams

An Argand diagram gives each complex number a position in a plane. You will plot numbers, interpret conjugation and show sums and differences as vectors (CP 2.4), so that the geometry explains the arithmetic.

Complex numbers as points and vectors

If z=3+2iz=3+2\mathrm{i} and w=−1+3iw=-1+3\mathrm{i}, where should z+wz+w appear on a diagram, and which arrow represents z−wz-w? You can answer both questions without measuring an angle. The trick is to give the real and imaginary parts different axes.

The number 3+2i3+2\mathrm{i} sits three units right and two units up from the origin. The number −1+3i-1+3\mathrm{i} sits one unit left and three units up. A complex number records two real coordinates in one expression.

You already know

From A-level Mathematics, a two-dimensional vector (xy)\binom{x}{y} records a horizontal displacement xx and vertical displacement yy. From complex arithmetic, x+iyx+\mathrm{i}y has real part xx, imaginary part yy, and conjugate x−iyx-\mathrm{i}y.

Use the same coordinate idea, but give the axes their complex-number names.

Argand diagram

An Argand diagram is a plane whose horizontal axis is the real axis and whose vertical axis is the imaginary axis. The complex number z=x+iyz=x+\mathrm{i}y is represented by the point Z(x,y)Z(x,y), or by the position vector OZ→=(xy)\overrightarrow{OZ}=\binom{x}{y}. For example, −2+5i-2+5\mathrm{i} is plotted at (−2,5)(-2,5). It is not plotted at (5,−2)(5,-2): the real part goes first.

The point and the arrow represent the same complex number in different ways. The point shows its position; the arrow from the origin shows the displacement needed to reach it. An arrow copied elsewhere keeps its displacement, although it is no longer a position vector from the origin.

[DIAGRAM: asset_name: Coordinates and conjugation on an Argand diagram; asset_slug: 9fm0-argand-coordinates; caption: Real parts run horizontally, imaginary parts vertically, and conjugation reflects in the real axis; description: Equal-scale real and imaginary axes show z equals 3 plus 2i at (3,2), its conjugate at (3,minus 2), minus z at (minus 3,minus 2) and minus z star at (minus 3,2). The four points form a pale rectangle. Dashed projections from z label its real part 3 and imaginary part 2, and an arrow goes from the origin to z.]
Real parts run horizontally, imaginary parts vertically, and conjugation reflects in the real axis

Worked example 1: plotting a number and its conjugate

Question. Describe the positions of z=3+2iz=3+2\mathrm{i} and z∗z^* on an Argand diagram, and state the geometric relationship between them.

Plan. Read the two coordinates, then change the sign of only the imaginary coordinate for the conjugate.

z=3+2i⟷Z(3,2)real part firstz∗=3−2i⟷Z∗(3,−2)reverse vertical coordinate\begin{aligned} z=3+2\mathrm{i}&\quad\longleftrightarrow\quad Z(3,2) && \text{real part first} \\ z^*=3-2\mathrm{i}&\quad\longleftrightarrow\quad Z^*(3,-2) && \text{reverse vertical coordinate} \end{aligned}

Why this step? Conjugation changes the imaginary part alone, so the point crosses the real axis without moving sideways.

Answer. Plot ZZ at (3,2)(3,2) and Z∗Z^* at (3,−2)(3,-2). They are reflections in the real axis.

Check. Their horizontal coordinates agree, while their signed vertical coordinates are opposites.

Now change the real part to zero. A point on an axis is still a perfectly ordinary complex number.

Worked example 2: a purely imaginary number

Question. Show u=−4iu=-4\mathrm{i} and its conjugate on an Argand diagram. State their position vectors.

Plan. Insert the missing real part as zero, then plot each vertical coordinate.

u=0−4i⟷U(0,−4)zero real partu∗=0+4i⟷U∗(0,4)conjugateOU→=(0−4),OU∗→=(04)position vectors\begin{aligned} u=0-4\mathrm{i}&\quad\longleftrightarrow\quad U(0,-4) && \text{zero real part} \\ u^*=0+4\mathrm{i}&\quad\longleftrightarrow\quad U^*(0,4) && \text{conjugate} \\ \overrightarrow{OU}&=\binom0{-4},\qquad \overrightarrow{OU^*}=\binom04 && \text{position vectors} \end{aligned}

Why this step? Writing the zero prevents −4i-4\mathrm{i} from being mistaken for a point on the real axis.

Answer. Both points lie on the imaginary axis: UU four units below the origin and U∗U^* four units above it.

Check. Reflection in the real axis takes (0,−4)(0,-4) to (0,4)(0,4), just as conjugation requires.

There is a neat shape hiding in four sign choices. If z=x+iyz=x+\mathrm{i}y with x,y≠0x,y\ne0, the points for z,z∗,−z,−z∗z,z^*,-z,-z^* are the four corners (±x,±y)(\pm x,\pm y) of a rectangle centred at the origin. If either coordinate is zero, two pairs coincide and the rectangle collapses to a segment. The algebra tells you both the shape and its exceptional case.

Watch out

Wrong: labelling the imaginary coordinate of 3+2i3+2\mathrm{i} as 2i2\mathrm{i} in the ordered pair. The coordinate is the real number 22, so the point is (3,2)(3,2). The axis name supplies its imaginary meaning.

Use the reflection rule to describe another pair, without relying on the picture above.

Adding with a triangle or parallelogram

Adding complex numbers adds their horizontal and vertical displacements, so the sum is found by placing the second vector at the end of the first.

For 3+2i3+2\mathrm{i} followed by −1+3i-1+3\mathrm{i}, move three right and two up, then one left and three up. You finish two right and five up. This is the same component addition you use for vectors.

The vector for a sum

For z=x+iyz=x+\mathrm{i}y and w=u+ivw=u+\mathrm{i}v,

z+w=(x+u)+i(y+v)z+w=(x+u)+\mathrm{i}(y+v)

Copy the vector for ww to the endpoint of the vector for zz. The vector from the origin to the final endpoint represents z+wz+w. Equivalently, z+wz+w is the opposite vertex in the parallelogram with sides zz and ww, when those sides are not parallel.

Not in the formula booklet: learn it.

The second arrow must keep both its length and its direction when copied. Sliding it is allowed; rotating it would change the number.

[DIAGRAM: asset_name: The triangle and parallelogram rules for a complex sum; asset_slug: 9fm0-argand-sum; caption: Copy the w arrow to the end of z to reach z plus w; description: Equal-scale Argand axes show O, Z at (3,2), W at (minus 1,3) and S at (2,5). Blue arrows go from O to Z and from W to S, sage arrows go from O to W and from Z to S, and a clay arrow from O to S is labelled z plus w equals 2 plus 5i. Labels sit clear of the parallelogram sides.]
Copy the w arrow to the end of z to reach z plus w

Worked example 3: a sum above both starting points

Question. Given z=3+2iz=3+2\mathrm{i} and w=−1+3iw=-1+3\mathrm{i}, find z+wz+w and explain its triangle construction.

Plan. Add corresponding components, then copy the displacement for ww from the endpoint of zz.

z+w=(3−1)+i(2+3)add components=2+5icollect parts(3,2)+(−1,3)=(2,5)endpoint of copied arrow\begin{aligned} z+w&=(3-1)+\mathrm{i}(2+3) && \text{add components} \\ &=2+5\mathrm{i} && \text{collect parts} \\ (3,2)+(-1,3)&=(2,5) && \text{endpoint of copied arrow} \end{aligned}

Why this step? The copied ww arrow starts at ZZ, so its endpoint is reached by adding its displacement to ZZ's coordinates.

Answer. z+w=2+5iz+w=2+5\mathrm{i}. Draw the ww arrow one unit left and three units up from ZZ; the arrow from OO to its endpoint is the sum.

Check. Starting with ww and then copying zz reaches the same endpoint: (−1,3)+(3,2)=(2,5)(-1,3)+(3,2)=(2,5).

The second example changes only the sign of the imaginary part of ww. The copied arrow now goes down, and the sum can lie below the real axis.

Worked example 4: opposing vertical displacements

Question. Given z=3+2iz=3+2\mathrm{i} and w=−1−3iw=-1-3\mathrm{i}, find z+wz+w and describe its position.

Plan. Add components, keeping the downward displacement negative.

z+w=(3−1)+i(2−3)signed displacements=2−icollect partsZ+(−1−3)=(2,−1)copy w from Z\begin{aligned} z+w&=(3-1)+\mathrm{i}(2-3) && \text{signed displacements} \\ &=2-\mathrm{i} && \text{collect parts} \\ Z+\binom{-1}{-3}&=(2,-1) && \text{copy }w\text{ from }Z \end{aligned}

Why this step? A downward movement subtracts from the vertical coordinate, even when the arrow starts above the axis.

Answer. The sum is 2−i2-\mathrm{i}, represented by (2,−1)(2,-1) in the fourth quadrant.

Check. From the final point (2,−1)(2,-1), subtracting the displacement (−1,−3)(-1,-3) returns (3,2)(3,2).

The parallelogram rule explains why the order of addition makes no difference: the two routes use the same two displacements.

Watch out

Wrong: joining the endpoints ZZ and WW and calling that arrow z+wz+w. Addition uses the arrow from the origin to the opposite parallelogram vertex. The arrow joining ZZ and WW represents a difference.

A sum does not have to lie between its two starting points.

The direction of a difference

The difference z−wz-w is the displacement from the point for ww to the point for zz: destination minus starting point.

To see why, let ZZ and WW have position vectors z\mathbf{z} and w\mathbf{w}. The route OO to WW followed by WW to ZZ must end at ZZ, so w+WZ→=z\mathbf{w}+\overrightarrow{WZ}=\mathbf{z}. Rearranging gives the direction as well as the components.

The vector for a difference

If ZZ represents zz and WW represents ww,

WZ→ represents z−w,ZW→ represents w−z\overrightarrow{WZ}\text{ represents }z-w, \qquad \overrightarrow{ZW}\text{ represents }w-z

As a position vector from the origin, z−wz-w is obtained by adding zz to −w-w. Multiplication by −1-1 reverses a vector.

Not in the formula booklet: learn it.

These are two drawings of the same displacement: the connecting arrow from WW to ZZ, and a parallel copy starting at the origin. Their endpoints need not coincide.

[DIAGRAM: asset_name: A difference as a connecting arrow and a position vector; asset_slug: 9fm0-argand-difference; caption: z minus w points from W to Z; its copy from O has endpoint (4,minus 1); description: Equal-scale axes show Z at (3,2) and W at (minus 1,3). A clay arrow points from W to Z and is labelled z minus w. A second parallel clay arrow goes from the origin to D at (4,minus 1), labelled 4 minus i. Faint position arrows show O to Z and O to W.]
z minus w points from W to Z

Worked example 5: destination minus start

Question. Given z=3+2iz=3+2\mathrm{i} and w=−1+3iw=-1+3\mathrm{i}, find z−wz-w and identify its connecting arrow.

Plan. Subtract the coordinates of WW from those of ZZ, then state the arrow's start and finish.

z−w=(3−(−1))+i(2−3)destination minus start=4−isimplify signs(−1,3)+(4,−1)=(3,2)follow the displacement\begin{aligned} z-w&=(3-(-1))+\mathrm{i}(2-3) && \text{destination minus start} \\ &=4-\mathrm{i} && \text{simplify signs} \\ (-1,3)+(4,-1)&=(3,2) && \text{follow the displacement} \end{aligned}

Why this step? The check starts at WW. Adding the difference must take you to ZZ, which fixes the direction unambiguously.

Answer. z−w=4−iz-w=4-\mathrm{i}, represented by the arrow WW to ZZ.

Check. w+(z−w)=(−1+3i)+(4−i)=3+2i=zw+(z-w)=(-1+3\mathrm{i})+(4-\mathrm{i})=3+2\mathrm{i}=z.

Now reverse the subtraction and change nothing else. The connecting arrow must reverse too.

Worked example 6: reversing a difference

Question. For the same numbers, find w−zw-z and explain its relationship to z−wz-w.

Plan. Reverse the start and destination, or negate the previous difference.

w−z=(−1−3)+i(3−2)new destination minus start=−4+icollect partsw−z=−(4−i)=−(z−w)opposite displacement\begin{aligned} w-z&=(-1-3)+\mathrm{i}(3-2) && \text{new destination minus start} \\ &=-4+\mathrm{i} && \text{collect parts} \\ w-z&=-(4-\mathrm{i})=-(z-w) && \text{opposite displacement} \end{aligned}

Why this step? Negating both components reverses an arrow while leaving its length unchanged.

Answer. w−z=−4+iw-z=-4+\mathrm{i}, the arrow ZZ to WW. It is the reverse of the arrow for z−wz-w.

Check. Starting at Z=(3,2)Z=(3,2), the displacement (−4,1)(-4,1) reaches (−1,3)=W(-1,3)=W.

The arrow's direction is part of the answer, as well as its components.

Watch out

Wrong: the arrow from ZZ to WW represents z−wz-w. Its destination is WW, so it represents w−zw-z. The correct arrow for z−wz-w runs from WW to ZZ: destination minus start.

You can now answer both opening questions: the sum reaches (2,5)(2,5), and the difference points from (−1,3)(-1,3) to (3,2)(3,2). The same two starting points produce two quite different constructions.

Exam tip

Label the real and imaginary axes, put points in sensible proportions, and give every connecting arrow a direction. An unlabelled segment alone cannot distinguish z−wz-w from w−zw-z.

Keep “destination minus start” in mind for this final check.

Quick Recap

  • Plot x+iyx+\mathrm{i}y at (x,y)(x,y), with real parts across and imaginary parts up.
  • Conjugation reflects a point in the real axis; negation reverses its position vector.
  • Add vectors head to tail, or complete a parallelogram, to represent a sum.
  • The arrow from WW to ZZ represents z−wz-w: destination minus start.

For the opening pair, z+w=2+5iz+w=2+5\mathrm{i}, while z−w=4−iz-w=4-\mathrm{i} is the arrow from WW to ZZ.