Complex Numbers · HL
Argand Diagrams
Plotting complex numbers, real multiples, modulus and argument, the conjugate, and adding with the parallelogram.
Section 1
Re(z) and Im(z) on the diagram
A complex number $z = a + bi$ is plotted as the point $(a,\,b)$ on the Argand diagram — the real part across, the imaginary part up.
$z = -3 + 2i$ (our number)
$\text{Re}(z) = -3$
$\text{Im}(z) = 2$
Watch this
$\text{Re}(z)$ and $\text{Im}(z)$ are real numbers. So both of them sit on the real axis — not the imaginary axis.
$\text{Im}(z) = 2$ is just the number $2$. It lives at $2$ on the real axis, the same place the number $2$ always lives.
Section 2
Real multiples: 2z, ½z and −z
Must learn
Multiplying $z$ by a real number keeps it on the same line through the origin. It only stretches, shrinks or flips it.
$2z$ — twice as far out from $O$, same direction.
$2z = 2(-3 + 2i) = -6 + 4i$
$\tfrac{1}{2}z$ — half as far out, same direction.
$\tfrac{1}{2}z = \tfrac{1}{2}(-3 + 2i) = -\tfrac{3}{2} + i$
$-z$ — same distance, opposite direction (through $O$).
$-z = 3 - 2i$
Here is the whole picture. Notice $2z$, $z$, $\tfrac{1}{2}z$, $O$ and $-z$ all lie on one straight line through the origin:
The slope of that line is $\dfrac{2}{-3} = -\dfrac{2}{3}$. Every real multiple of $z$ keeps the same slope, which is why they all line up.
You try
For $z = -3 + 2i$, where does $3z$ plot?
Three times as far out from $O$ as $z$, same direction. Multiply both parts by 3.
$3z = 3(-3 + 2i)$
$3z = -9 + 6i \;\to\; (-9,\,6)$
$(-9,\,6)$, further along the same line through $O$.
You try
Where does $-\tfrac{1}{2}z$ plot, for $z = -3 + 2i$?
Half the distance ($\tfrac{1}{2}$) and flipped to the opposite side (minus). Same line.
$-\tfrac{1}{2}z = -\tfrac{1}{2}(-3 + 2i)$
$= \tfrac{3}{2} - i \;\to\; (1.5,\,-1)$
$\left(\tfrac{3}{2},\,-1\right)$ — opposite side of $O$, half as far.
You try
$w = 4 - i$. Plot $w$, $\text{Re}(w)$ and $\text{Im}(w)$ — which axis does each of $\text{Re}(w)$ and $\text{Im}(w)$ sit on?
$\text{Re}(w)$ and $\text{Im}(w)$ are both real numbers. Where do real numbers live?
$w = 4 - i \;\to\; (4,\,-1)$
$\text{Re}(w) = 4$, $\text{Im}(w) = -1$
Both are real, so both sit on the real axis: at $4$ and at $-1$.
Both $\text{Re}(w)$ and $\text{Im}(w)$ are on the real axis.
Section 3
Modulus and argument
Any complex number can be written in polar form:
$z = r(\cos\theta + i\sin\theta)$
Two words to know
Modmodulus $= r$ — the distance from $O$.
Argargument $= \theta$ — the angle from the positive real axis.
A turning example
Take a number on the unit circle, so the modulus is $1$:
$r = 1 \quad\Rightarrow\quad z = \cos\theta + i\sin\theta$
Cubing it adds the angle three times:
$z^{3} = \cos 3\theta + i\sin 3\theta$
If $z^{3}$ lands back on the positive real axis after one full turn, then $3\theta$ is a full revolution:
$3\theta = 360^{\circ}$
$\theta = 120^{\circ}$
So $z$, $z^{2}$ and $z^{3}$ sit on the unit circle, exactly $120^{\circ}$ apart:
The amber dashed loop is the “turn budget”: each step from $z$ to $z^{2}$ to $z^{3}$ turns another $120^{\circ}$, and three of them make the full $360^{\circ}$.
You try
If $z^{4}$ comes back to the positive real axis after one full turn (with $r = 1$), what is $\theta$?
Same idea: cubing tripled the angle, so the power tells you how many $\theta$'s make $360^{\circ}$.
$z^{4}$ means $4\theta = 360^{\circ}$
$\theta = 90^{\circ}$
$\theta = 90^{\circ}$
You try
Write down the modulus and argument of $z = 2\big(\cos 30^{\circ} + i\sin 30^{\circ}\big)$.
Read it straight off the polar form $r(\cos\theta + i\sin\theta)$. What is $r$? What is $\theta$?
Compare with $r(\cos\theta + i\sin\theta)$
Mod $= r = 2$, Arg $= \theta = 30^{\circ}$
$\text{Mod}(z) = 2$, $\text{Arg}(z) = 30^{\circ}$
Section 4
Conjugate and addition
Must learn
The conjugate $\bar{z}$ is the reflection of $z$ in the real axis. The real part stays; the imaginary part flips sign.
Must learn
To add $z + w$, complete the parallelogram on $z$ and $w$ from the origin. The diagonal from $O$ lands on $z + w$.
Here both ideas at once — the conjugate $\bar{z}$ mirrored below the real axis, and $z + w$ as the far corner of the parallelogram:
The conjugate $\bar{z}$ is directly below $z$ — same real part, imaginary part flipped. And $z + w$ is the corner you reach by sliding $w$ along to the tip of $z$.
You try
$z = 5 - 3i$. Write down $\bar{z}$ and say how its point relates to $z$.
Keep the real part. Flip the sign of the imaginary part.
$z = 5 - 3i$
$\bar{z} = 5 + 3i$ — the reflection of $z$ in the real axis.
$\bar{z} = 5 + 3i$, mirrored in the real axis.
You try
$z = 1 + 3i$ and $w = 4 + i$. Find $z + w$, and what shape do $O$, $z$, $w$ and $z+w$ make?
Add real to real, imaginary to imaginary. Then think about the four points.
$z + w = (1+4) + (3+1)i$
$z + w = 5 + 4i$ — and $O$, $z$, $w$, $z+w$ form a parallelogram.
$z + w = 5 + 4i$; the four points make a parallelogram.
You try
$z = -2 + 4i$. Find $z + \bar{z}$. Where does the answer sit on the diagram?
Find $\bar{z}$ first, then add. The imaginary parts are opposite — watch what happens.
$\bar{z} = -2 - 4i$
$z + \bar{z} = (-2 + 4i) + (-2 - 4i)$
$= -4$ — a real number, sitting on the real axis.
$z + \bar{z} = -4$, on the real axis. (A number plus its conjugate is always real.)
Section 5
Exam question: four complex numbers
Four complex numbers $z_1$, $z_2$, $z_3$, $z_4$ are shown on an Argand diagram (same scale on both axes) and satisfy:
$z_2 = i\,z_1$
$z_3 = k\,z_1, \quad k \in \mathbb{R}$
$z_4 = z_2 + z_3$
(i) Identify which point is which. (ii) Write down the approximate value of $k$.
Reading each clue
$z_2 = i\,z_1$: multiplying by $i$ turns a number $90^{\circ}$ anticlockwise. So $z_2$ is $z_1$ rotated a quarter turn — at right angles to $z_1$.
$z_3 = k\,z_1$ with $k$ real: a real multiple stays on the same line through $O$ as $z_1$.
$z_4 = z_2 + z_3$: the parallelogram corner. Because $z_2 \perp z_3$, that parallelogram is a rectangle.
(i) $z_1$ is the lone point off on its own; $z_2$ is the one at a right angle to it; $z_3$ is on the same line as $z_1$ but closer in; $z_4$ is the far corner of the rectangle.
(ii) $z_3$ is about half as far out as $z_1$ on the same ray, so $k = \tfrac{1}{2}$.
You try
$z_1 = 3 - i$. Work out $z_2 = i\,z_1$ in the form $a + bi$.
Multiply out $i(3 - i)$ and remember $i^{2} = -1$.
$z_2 = i(3 - i) = 3i - i^{2}$
$= 3i - (-1)$
$z_2 = 1 + 3i$
$z_2 = 1 + 3i$ (which is $z_1$ turned $90^{\circ}$).
You try
Using $z_1 = 3 - i$ and the same $k = \tfrac{1}{2}$, find $z_3 = k\,z_1$, then $z_4 = z_2 + z_3$.
$z_3 = \tfrac{1}{2}(3 - i)$. You already have $z_2 = 1 + 3i$ from the last one. Add them.
$z_3 = \tfrac{1}{2}(3 - i) = \tfrac{3}{2} - \tfrac{1}{2}i$
$z_4 = z_2 + z_3 = (1 + 3i) + \left(\tfrac{3}{2} - \tfrac{1}{2}i\right)$
$z_4 = \tfrac{5}{2} + \tfrac{5}{2}i$
$z_3 = \tfrac{3}{2} - \tfrac{1}{2}i$, $z_4 = \tfrac{5}{2} + \tfrac{5}{2}i$
That's the diagrams
Plotting, real multiples, modulus and argument, the conjugate, and adding by parallelogram — all on the Argand diagram.