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AI HL · AHL 6

Complex Numbers

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AHL 6.12

Geometry on the Argand diagram

AHL 6.12

Geometry on the Argand diagram

Target: I can plot \(z + w\) on an Argand diagram using the parallelogram rule. (+5 more)

By the end of this section

Key points

Adding and subtracting

  • On an Argand diagram, the real and imaginary parts of \(z\) are the components of a position vector from the origin to the point.
  • \(z + w\) is the fourth corner of the parallelogram with corners at \(0\), \(z\) and \(w\).
  • \(-w\) is \(w\) rotated through \(\pi\) about the origin. \(z - w = z + (-w)\), so complete the parallelogram with \(0\), \(z\) and \(-w\).
  • \(z - \alpha\) is the vector from the point \(\alpha\) to the point \(z\). \(|z - \alpha|\) is the distance between the two points.

Multiplying turns and scales

  • Multiplying by \(r\,\mathrm{cis}\,\theta\) rotates a point through \(\theta\) anticlockwise about the origin and multiplies its distance from the origin by \(r\): add the angles, scale by \(r\).
  • Multiply every vertex of a shape by the same number to rotate and enlarge the whole shape about the origin. Angles stay the same.
  • A rotation with no enlargement has modulus 1. Multiplying by \(\mathrm{cis}\left(\frac{\pi}{3}\right)\) rotates through \(\frac{\pi}{3}\); multiplying by \(i\) rotates through \(\frac{\pi}{2}\).
  • To find the third vertex of an equilateral triangle with a vertex at the origin, rotate the second vertex through \(\frac{\pi}{3}\) one way or the other: multiply by \(\mathrm{cis}\left(\pm\frac{\pi}{3}\right)\). There are two possible points, one on each side of the line.

Dividing, and turning about another point

  • \(\left|\dfrac{z}{w}\right| = \dfrac{|z|}{|w|}\) and \(\arg\dfrac{z}{w} = \arg z - \arg w\). Dividing by \(w\) rotates a point through \(\arg w\) clockwise and divides its distance from the origin by \(|w|\): the reverse of multiplying.
  • To turn a point \(z\) through \(\theta\) anticlockwise about a point \(\alpha\): find \(z - \alpha\) (the vector from \(\alpha\) to \(z\)), multiply it by \(\mathrm{cis}\,\theta\), then add \(\alpha\) back. \(z' = \alpha + (z - \alpha)\,\mathrm{cis}\,\theta\). For a clockwise turn use \(\mathrm{cis}(-\theta)\).
  • A turn does not change the distance from the centre: \(|z' - \alpha| = |z - \alpha|\).

The mapping w = az + b, and areas

  • In \(w = az + b\), the first step is to multiply by \(a\): a rotation through \(\arg a\) and an enlargement by \(|a|\), both about the origin. The second step is to add \(b\): a translation by the vector \(b\). The order matters.
  • To find \(z\) from \(w\), undo the steps in reverse order: \(z = \dfrac{w - b}{a}\).
  • The triangle with vertices \(0\), \(z\) and \(w\) has area \(\tfrac{1}{2}|z||w|\sin\varphi\), where \(\varphi\) is the angle between the two sides: \(\varphi = |\arg w - \arg z|\). Use radian mode.

Explore: adding complex numbers as vectors

Drag z (black) and w (teal). The orange arrow is z + w, the fourth corner of the parallelogram. Predict where it lands before you read the working.

|z|
|w|
|z + w|

The points snap to whole numbers. Select one (Tab) and use the arrow keys to move it.

  1. Drag w so that z + w = 0. What is w in terms of z? Where does w sit compared with z?
  2. Is w + z the same point as z + w? Where do you see this on the parallelogram?
  3. Show z − w. Which two sides of the parallelogram make its other diagonal? Why does z − w point from w to z?
  4. Compare |z + w| with |z| + |w|. When are they equal? Can |z + w| ever be larger?

Explore: multiplying turns and scales

Drag z (black) and w (teal). The orange arrow is zw. Predict its distance from O and its angle before you read the working. Or press a button to see what multiplying does to a whole shape.

The points snap to whole and half numbers. Select one (Tab) and use the arrow keys to move it. Angles are in radians, to 3 significant figures; the sums are worked from the unrounded values. The dashed circle has radius 1.

  1. Press "Multiply by i". Where do A, B and C go? What happens to the size of the triangle, and to its angles?
  2. Drag w onto the dashed circle. What does multiplying by w do to the distance of every point from O? What does it still do to the angle?
  3. Press "Multiply by i" twice in your head: i × i. Which button gives the same result? Check with the arguments: π/2 + π/2.
  4. Show z ÷ w. Predict its distance and angle first. Why is dividing by w the same as turning back and shrinking?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Adding with the parallelogram rule

The complex numbers \(z = 5 + i\) and \(w = -2 + 3i\) are plotted on an Argand diagram. Mark the complex number \(z + w\).

−3−2−1123456−2−1123456ReImzwz + w
Working
  1. Position vectors

    \(z = 5 + i \text{ is the point } (5, 1) \text{ and } w = -2 + 3i \text{ is the point } (-2, 3)\)

    Think of each number as a position vector: the real part is the horizontal component, the imaginary part the vertical component.

  2. Add

    \(z + w = (5 - 2) + (1 + 3)i = 3 + 4i\)

    Vectors add component by component.

  3. −3−2−1123456−2−1123456ReImzwz + w
    The parallelogram

    Corners at \(0\), \(z\), \(w\) and \(z + w\): \((0, 0),\ (5, 1),\ (-2, 3),\ (3, 4)\)

    Complete the parallelogram with sides \(0z\) and \(0w\). The fourth corner is \(z + w\).

Video walkthrough coming
Subtracting by adding the negative

The complex numbers \(z = 4 + 2i\) and \(w = -3 - 2i\) are plotted on an Argand diagram. Mark the complex number \(z - w\).

−4−3−2−112345678−4−3−2−1123456ReImzw-wz - w
Working
  1. Find −w

    \(-w = 3 + 2i\)

    \(-w\) is \(w\) rotated through \(\pi\) about the origin: change the sign of both parts. This is the point \((3, 2)\).

  2. Add −w to z

    \(z - w = z + (-w) = (4 + 3) + (2 + 2)i = 7 + 4i\)

    Subtracting is adding the negative.

  3. −4−3−2−112345678−4−3−2−1123456ReImzw-wz - w
    The parallelogram

    Corners at \(0\), \(z\), \(-w\) and \(z - w\): \((0, 0),\ (4, 2),\ (3, 2),\ (7, 4)\)

    Complete the parallelogram with sides \(0z\) and \(0(-w)\). The fourth corner is \(z - w\).

Video walkthrough coming
z − α as a vector

The diagram shows the points that represent \(z\) and \(\alpha\). Write down \(z\) and \(\alpha\), find \(z - \alpha\) and explain what it represents on the diagram.

−2−112345678−1123456ReImzα
Working
  1. −2−112345678−1123456ReImzα
    Read the points

    \(z = 1 + 5i, \quad \alpha = 6 + 2i\)

    The real part is across and the imaginary part is up.

  2. Subtract

    \(z - \alpha = (1 - 6) + (5 - 2)i = -5 + 3i\)

    Subtract real parts, then imaginary parts.

  3. What it means

    \(z - \alpha\) is the vector from \(\alpha\) to \(z\): \(5 \text{ left and } 3 \text{ up}\). Its length is \(|z - \alpha| = \sqrt{34} = 5.83\), the distance between the points.

    To go from \(\alpha\) to \(z\), add \(z - \alpha\) to \(\alpha\).

Video walkthrough coming
Multiplying every vertex

The triangle \(ABC\) has vertices \(A(1, 0)\), \(B(3, 0)\) and \(C(3, 1)\). Every vertex is multiplied by \(z = -1 + \sqrt{3}\,i\) to give the triangle \(A'B'C'\). Find the vertices of \(A'B'C'\) and describe the transformation from \(ABC\) to \(A'B'C'\).

−11234−2−1123ReImABC
Working
  1. The multiplier

    \(|z| = \sqrt{(-1)^2 + (\sqrt{3})^2} = 2,\quad \arg z = \tfrac{2\pi}{3}\)

    The point \((-1, \sqrt{3})\) is in the second quadrant, and the angle with the negative real axis is \(\frac{\pi}{3}\).

  2. −7−6−5−4−3−2−112345−11234567ReImABCA'B'C'
    The new vertices

    \(A' = -1 + \sqrt{3} i\)<br>\(B' = -3 + 3 \sqrt{3} i\)<br>\(C' = (3 + i)(-1 + \sqrt{3}i) = -3 - \sqrt{3} - \left(1 - 3 \sqrt{3}\right) i\)

    Multiply each vertex by \(z\), expanding the brackets and using \(i^2 = -1\).

  3. Describe it

    A rotation of \(\tfrac{2\pi}{3}\) anticlockwise about the origin, then an enlargement of scale factor 2 from the origin.

    Multiplying by \(r\,\mathrm{cis}\,\theta\) rotates by \(\theta\) and scales by \(r\): add the angles, scale by \(r\).

Video walkthrough coming
The third vertex of an equilateral triangle

An equilateral triangle \(OAB\) has a vertex \(O\) at the origin and a vertex \(A\) at \((4, 2)\). The vertex \(B\) is above the line \(OA\). Find the coordinates of \(B\).

−112345−2−11234ReImOA
Working
  1. −112345−2−11234ReImOA
    Choose the rotation

    \(A = 4 + 2i\). Rotating \(A\) through \(\frac{\pi}{3}\) anticlockwise about \(O\) gives \(B\). That is multiplication by \(\mathrm{cis}\left(\tfrac{\pi}{3}\right) = \tfrac{1}{2} + \tfrac{\sqrt{3}}{2}i\).

    The modulus of \(\mathrm{cis}\left(\frac{\pi}{3}\right)\) is 1, so there is no enlargement.

  2. Multiply

    \(B = (4 + 2i)\left(\tfrac{1}{2} + \tfrac{\sqrt{3}}{2}i\right) = 2 + 2\sqrt{3}\,i + i + \sqrt{3}\,i^2\)

    Expand, then use \(i^2 = -1\).

  3. −3−2−11234567−1123456ReImOAB
    Collect

    \(B = 2 - \sqrt{3} + \left(1 + 2 \sqrt{3}\right) i\)

    Real part and imaginary part.

  4. Coordinates

    \(B\left(2 - \sqrt{3},\ 1 + 2 \sqrt{3}\right) = (0.268,\ 4.46)\)

    The point is above the line \(OA\), as required.

Video walkthrough coming
Dividing on the Argand diagram

\(z = 3\sqrt{3} + 3i\) and \(w = 1 + \sqrt{3}\,i\). (a) Find \(|z|\), \(\arg z\), \(|w|\) and \(\arg w\). (b) Find the modulus and argument of \(\dfrac{z}{w}\). (c) Describe the transformation that takes the point \(z\) to the point \(\dfrac{z}{w}\), and write \(\dfrac{z}{w}\) in the form \(x + iy\).

−1123456−2−11234ReImzwz/w
Working
  1. Modulus and argument of each

    \(|z| = \sqrt{27 + 9} = 6,\quad \arg z = \tan^{-1}\dfrac{3}{3\sqrt{3}} = \dfrac{\pi}{6}\)<br>\(|w| = \sqrt{1 + 3} = 2,\quad \arg w = \tan^{-1}\sqrt{3} = \dfrac{\pi}{3}\)

    Both points are in the first quadrant, so the inverse tangent gives the argument.

  2. The quotient

    \(\left|\dfrac{z}{w}\right| = \dfrac{6}{2} = 3,\quad \arg\dfrac{z}{w} = \dfrac{\pi}{6} - \dfrac{\pi}{3} = -\dfrac{\pi}{6}\)

    Divide the moduli and subtract the arguments.

  3. The transformation

    Rotation through \(\dfrac{\pi}{3}\) clockwise about the origin, then an enlargement of scale factor \(\dfrac{1}{2}\) from the origin.

    Dividing by \(w\) undoes multiplying by \(w\): turn back by \(\arg w\) and divide the distance by \(|w|\).

  4. −1123456−2−11234ReImzwz/w
    Cartesian form

    \(\dfrac{z}{w} = 3e^{-i\frac{\pi}{6}} = 3\cos\dfrac{\pi}{6} - 3\sin\dfrac{\pi}{6}\,i = \dfrac{3\sqrt{3}}{2} - \dfrac{3}{2}i\)

    Check with the calculator in complex mode.

Video walkthrough coming
Turning about a point

The point \(z = 5 + 3i\) is turned through \(\frac{\pi}{2}\) anticlockwise about the point \(\alpha = 2 + i\). Find the image \(z'\).

−3−2−11234567−3−2−1123456ReImαzz'
Working
  1. Vector from the centre

    \(z - \alpha = (5 + 3i) - (2 + i) = 3 + 2i\)

    Move the centre to the origin: this is the vector from \(\alpha\) to \(z\).

  2. Turn it

    \((3 + 2i) \times i = 3i + 2i^2 = -2 + 3i\)

    Multiplying by \(\mathrm{cis}\frac{\pi}{2} = i\) turns a vector through \(\frac{\pi}{2}\) anticlockwise.

  3. Add the centre back

    \(z' = \alpha + (-2 + 3i) = (2 - 2) + (1 + 3)i = 4i\)

    The image is the point \((0, 4)\).

  4. −3−2−11234567−3−2−1123456ReImαzz'
    Check

    \(|z' - \alpha| = |-2 + 3i| = \sqrt{13} = |z - \alpha|\)

    A turn keeps the distance from the centre.

Video walkthrough coming
Mapping w = az + b

The point \(z\) is mapped to the point \(w\) by \(w = 2iz + 3\). (a) Find \(w\) when \(z = 2 + i\). (b) Describe the two transformations that map the point \(z\) onto the point \(w\), in the order they are applied. (c) Find \(z\) when \(w = 7 + 2i\).

Working
  1. Find w

    \(w = 2i(2 + i) + 3 = 4i + 2i^2 + 3 = 4i - 2 + 3 = 1 + 4i\)

    Multiply first, then add 3.

  2. Describe the steps

    Step 1: multiply by \(2i\), which has modulus 2 and argument \(\frac{\pi}{2}\): a rotation of \(\frac{\pi}{2}\) anticlockwise and an enlargement of scale factor 2, both about the origin.<br>Step 2: add 3: a translation of 3 units to the right.

    Multiplying by \(a\) turns and scales. Adding \(b\) slides the whole picture.

  3. Undo the steps

    \(w - 3 = 4 + 2i\), then \(z = \dfrac{4 + 2i}{2i} = \dfrac{(4 + 2i)(-i)}{2} = \dfrac{-4i + 2}{2} = 1 - 2i\)

    Work backwards: subtract 3 first, then divide by \(2i\).

  4. Check

    \(2i(1 - 2i) + 3 = 2i + 4 + 3 = 7 + 2i\)

    Put \(z\) back into the rule.

Video walkthrough coming
Area of a triangle

The complex numbers are \(z = 3 + 4i\) and \(w = 4e^{2.1i}\). Find the area of the triangle on an Argand diagram with vertices at \(0\), \(z\) and \(w\).

−4−3−2−112345−1123456ReIm0zw
Working
  1. The two sides

    \(|z| = \sqrt{3^2 + 4^2} = 5\)<br>\(|w| = 4\)

    The sides from the origin have lengths \(|z|\) and \(|w|\).

  2. The angle between them

    \(\arg z = \tan^{-1}\dfrac{4}{3} = 0.9273,\quad \arg w = 2.1\)<br>\(\varphi = 2.1 - 0.9273 = 1.173\)

    The angle between the sides is the difference of the arguments. Radian mode.

  3. −4−3−2−112345−1123456ReIm0zw
    The area

    \(\text{Area} = \tfrac{1}{2}(5)(4)\sin 1.173 = 9.22\)

    The area is \(\frac{1}{2}ab\sin C\) with the two moduli as \(a\) and \(b\).

Video walkthrough coming