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IB Mathematics: Applications and Interpretation HL · AHL 6: Complex Numbers · AHL 6.12

Complex numbers on the Argand diagram Problem set B One question at a time

Try each question on paper first. Open the working a line at a time, then say whether you had it.

Skills

  1. The Argand diagram shows two complex numbers, \(z\) and \(w\), as points.

    −4−3−2−112345−3−2−1123456ReImzw

    1. (a) Write down \(z\) and \(w\) as complex numbers.

      • \(z\) is the point \((-3, 2)\) and \(w\) is the point \((4, 3)\).

    2. (b) Find \(z + w\) and mark it on a copy of the diagram, using the parallelogram rule.

      • \(z + w = (-3 + 2 i) + (4 + 3 i) = 1 + 5 i\)
      • Complete the parallelogram with corners \(0\), \(z\), \(w\). The fourth corner is the point \((1, 5)\).

  2. Two complex numbers, \(z\) and \(w\), are plotted on the Argand diagram.

    −6−5−4−3−2−11234567−2−1123456ReImzw

    1. (a) Write down \(-w\) and mark it on a copy of the diagram.

      • \(w = 5 - i \Rightarrow -w = -5 + i\)
      • \(-w\) is \(w\) reflected through the origin (rotated through \(\pi\) about the origin).

    2. (b) Find \(z - w\) and mark it on the diagram.

      • \(z - w = z + (-w) = (2 + 4 i) + (-5 + i) = -3 + 5 i\)
      • Complete the parallelogram with corners \(0\), \(z\), \(-w\). The fourth corner is \((-3, 5)\).

  3. Two points on the Argand diagram represent \(z\) and \(\alpha\). A dashed line joins them.

    −7−6−5−4−3−2−1123−3−2−11234ReImzα

    1. (a) Write down \(z\) and \(\alpha\) as complex numbers.

      • \(z\) is the point \((-4, 3)\). \(\alpha\) is the point \((-1, -2)\).

    2. (b) Find \(z - \alpha\).

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

    3. (c) The vector from the point \(\alpha\) to the point \(z\) is \(z - \alpha\). Find the distance between the two points, to 3 significant figures.

      • \(|z - \alpha| = \sqrt{(-3)^2 + (5)^2} = \sqrt{34} = 5.83\)

  4. The triangle PQR has vertices P(0, 1), Q(0, 3) and R(2, 3).

    −3−2−112345−11234ReImPQR

    1. (a) Find the modulus and argument of \(\sqrt{3} + i\).

      • \(|\sqrt{3} + i| = \sqrt{(\sqrt{3})^2 + (1)^2} = 2\)
      • \(\arg = \frac{\pi}{6}\) (use the quadrant and the reference angle).

    2. (b) Every vertex is multiplied by \(\sqrt{3} + i\) to give the triangle P′Q′R′. Find the new vertices.

      • \(P' = (i)(\sqrt{3} + i) = -1 + \sqrt{3} i\)
      • \(Q' = (3 i)(\sqrt{3} + i) = -3 + 3 \sqrt{3} i\)
      • \(R' = (2 + 3 i)(\sqrt{3} + i) = -3 + 2 \sqrt{3} + \left(2 + 3 \sqrt{3}\right) i\)

    3. (c) Describe the transformation that takes the first triangle to the second.

      • Multiplying by a complex number rotates a point by its argument and scales its distance from the origin by its modulus.
      • So the triangle is rotated through \(\frac{\pi}{6}\) anticlockwise about the origin and enlarged by scale factor \(2\).

  5. \(OAB\) is an equilateral triangle. \(O\) is the origin and \(A\) is the point \((5, 1)\).

    −1123456−3−2−11234ReImOA

    1. (a) Write the rotation of \(A\) through \(\frac{\pi}{3}\) anticlockwise about \(O\) as a multiplication, and find the third vertex \(B\) above the line \(OA\).

      • \(B = (5 + i)\,\mathrm{cis}\left(\tfrac{\pi}{3}\right) = (5 + i)\left(\tfrac{1}{2} + \tfrac{\sqrt{3}}{2}i\right)\)
      • \(= \frac{5}{2} - \frac{\sqrt{3}}{2} + \left(\frac{1}{2} + \frac{5 \sqrt{3}}{2}\right) i\)
      • \(B\) is the point \((\frac{5}{2} - \frac{\sqrt{3}}{2}, \frac{1}{2} + \frac{5 \sqrt{3}}{2})\).

    2. (b) Find the other point \(C\) that makes \(OAC\) equilateral, below the line \(OA\).

      • \(C = (5 + i)\,\mathrm{cis}\left(-\tfrac{\pi}{3}\right) = (5 + i)\left(\tfrac{1}{2} - \tfrac{\sqrt{3}}{2}i\right) = \frac{\sqrt{3}}{2} + \frac{5}{2} + \left(\frac{1}{2} - \frac{5 \sqrt{3}}{2}\right) i\)
      • \(C\) is the point \((\frac{\sqrt{3}}{2} + \frac{5}{2}, \frac{1}{2} - \frac{5 \sqrt{3}}{2})\).

  6. The points \(z\) and \(w\) are shown on the diagram. Give each argument in the range \(-\pi < \theta \le \pi\).

    −11234567−7−6−5−4−3−2−112345ReImzw

    1. (a) Write down \(z\) and \(w\), and find their moduli and arguments.

      • \(z = 6 - 6i\) is in the fourth quadrant: \(|z| = \sqrt{36 + 36} = 6\sqrt{2}\), \(\arg z = -\frac{\pi}{4}\).
      • \(w = 3i\) is on the positive imaginary axis: \(|w| = 3\), \(\arg w = \frac{\pi}{2}\).

    2. (b) Find the modulus and argument of \(\dfrac{z}{w}\).

      • \(\left|\frac{z}{w}\right| = \frac{6\sqrt{2}}{3} = 2\sqrt{2}\)
      • \(\arg\frac{z}{w} = -\frac{\pi}{4} - \frac{\pi}{2} = -\frac{3\pi}{4}\)

    3. (c) Describe the transformation that takes the point \(z\) to the point \(\dfrac{z}{w}\), and write \(\dfrac{z}{w}\) in the form \(x + iy\).

      • Dividing by \(w\) turns the point back through \(\arg w = \frac{\pi}{2}\) (clockwise) and divides its distance from the origin by \(|w| = 3\).
      • \(\frac{z}{w} = 2\sqrt{2}\left(\cos\left(-\frac{3\pi}{4}\right) + i\sin\left(-\frac{3\pi}{4}\right)\right) = -2 - 2i\).

  7. The point \(z = 7 + 5i\) is turned through \(\frac{\pi}{2}\) clockwise about the point \(\alpha = 3 + i\), shown on the diagram.

    −112345678910−4−3−2−112345678ReImαz

    1. (a) Find \(z - \alpha\).

      • \(z - \alpha = (7 + 5i) - (3 + i) = 4 + 4i\)

    2. (b) Multiply \(z - \alpha\) by \(\mathrm{cis}\left(-\frac{\pi}{2}\right) = -i\).

      • \((4 + 4i)(-i) = 4 - 4i\), using \(i^2 = -1\).

    3. (c) Hence find the image \(z'\) of \(z\).

      • \(z' = \alpha + (4 - 4i) = 7 - 3i\).
      • Check: \(|z' - \alpha| = |4 - 4i| = |z - \alpha|\).

  8. The point \(z\) is mapped to the point \(w\) by \(w = -3iz + 5\).

    1. (a) Find \(w\) when \(z = 2 + i\).

      • \(w = -3i(2 + i) + 5 = -6i - 3i^2 + 5 = -6i + 3 + 5 = 8 - 6i\)

    2. (b) Describe the two transformations that map the point \(z\) onto the point \(w\), in the order they are applied.

      • \(-3i\) has modulus 3 and argument \(-\frac{\pi}{2}\), so multiplying by it rotates \(\frac{\pi}{2}\) clockwise and enlarges by a scale factor 3 about the origin.
      • Adding 5 then translates every point 5 units to the right.

    3. (c) Find \(z\) when \(w = 2 + 6i\).

      • \(w - 5 = -3 + 6i\)
      • \(z = \dfrac{-3 + 6i}{-3i} = \dfrac{(-3 + 6i)(i)}{3} = \dfrac{-3i - 6}{3} = -2 - i\)

  9. The complex numbers are \(z = 5 + 2i\) and \(w = 4e^{1.3i}\).

    −11234567−1123456ReIm0zw

    1. (a) Find \(|z|\) and \(\arg z\), to 3 significant figures.

      • \(|z| = \sqrt{5^2 + 2^2} = 5.39\)
      • \(\arg z = \tan^{-1}\dfrac{2}{5} = 0.381\) (first quadrant).

    2. (b) Find the angle between the line from \(0\) to \(z\) and the line from \(0\) to \(w\).

      • \(\arg w = 1.3\), so the angle between the lines is \(|1.3 - 0.3805| = 0.919\) radians.

    3. (c) Find the area of the triangle on an Argand diagram with vertices at \(0\), \(z\) and \(w\), to 3 significant figures.

      • \(\text{Area} = \tfrac{1}{2}|z||w|\sin\varphi = \tfrac{1}{2}(5.385)(4)\sin 0.9195 = 8.57\).

In context

  1. Two ropes pull on a post. Each force, in newtons, is written as a complex number: the real part is the force to the right and the imaginary part is the force upwards. The forces are \(z = 120 + 340i\) and \(w = -80 + 60i\).

    1. (a) Find the resultant, \(z + w\), of the two.

      • \(z + w = (120 + 340 i) + (-80 + 60 i) = 40 + 400 i\)
      • Real parts add and imaginary parts add: this is the parallelogram rule.

    2. (b) Find the magnitude of the resultant, to 3 significant figures.

      • \(|z + w| = \sqrt{(40)^2 + (400)^2} = 402\)

    3. (c) Find \(z - w\), the vector from \(w\) to \(z\).

      • \(z - w = (120 + 340 i) - (-80 + 60 i) = 200 + 280 i\)

  2. A radio mast is at the point \(\alpha = -2 + 5i\) on a map, where 1 unit is 1 km. A walker is at the point \(z = 7 - 3i\).

    1. (a) Find \(z - \alpha\) and say what it represents.

      • \(z - \alpha = (7 - 3 i) - (-2 + 5 i) = 9 - 8 i\)
      • It is the vector (displacement) from the radio mast \(\alpha\) to the walker \(z\).

    2. (b) Find how far the walker is from the radio mast, to 3 significant figures.

      • \(|z - \alpha| = \sqrt{(9)^2 + (-8)^2} = 12.0\)

  3. A tile pattern uses a triangle with vertices A(2, 0); B(2, 2); C(0, 4). A program multiplies every vertex by the complex number \(-2 i\).

    1. (a) Describe the effect on the shape of multiplying every vertex by that number.

      • \(|-2i| = 2,\ \arg(-2i) = -\frac{\pi}{2}\)
      • The shape is rotated through \(\frac{\pi}{2}\) clockwise about the origin and enlarged by scale factor \(2\).

    2. (b) Find the vertices of the new shape.

      • \(A' = (2) \times (-2 i) = -4 i\)
      • \(B' = (2 + 2 i) \times (-2 i) = 4 - 4 i\)
      • \(C' = (4 i) \times (-2 i) = 8\)

  4. A Ferris wheel has its hub at \(\alpha = 6 + 10i\) on a plan, in metres, with the ground along the real axis. A seat is level with the hub at \(z = 16 + 10i\). The wheel turns the seat through 0.9 radians anticlockwise about the hub.

    1. (a) Find \(z - \alpha\).

      • \(z - \alpha = (16 + 10i) - (6 + 10i) = 10\)

    2. (b) Find the new position of the seat, with the real and imaginary parts to 3 significant figures.

      • \(z' = \alpha + (z - \alpha)\,\mathrm{cis}\,0.9 = (6 + 10i) + 10(\cos 0.9 + i\sin 0.9)\)
      • \(z' = 12.2 + 17.8i\) (radian mode).

    3. (c) Find how far the seat is from the hub after the turn.

      • \(|z' - \alpha| = |z - \alpha| = 10\) m. A turn does not change the distance from the centre.

Stretch

  1. \(OPQR\) is a square with \(O\) at the origin and \(P\) at the point \(5 + i\). \(R\) is above the line \(OP\).

    −1123456−3−2−11234ReImOP

    1. (a) Find the point \(R\), by rotating \(P\) through \(\frac{\pi}{2}\) anticlockwise about \(O\).

      • \(\mathrm{cis}\left(\tfrac{\pi}{2}\right) = i\)
      • \(R = i(5 + i) = -1 + 5 i\)

    2. (b) Find the point \(Q\).

      • \(Q = P + R = (5 + i) + (-1 + 5 i) = 4 + 6 i\)
      • The corner \(Q\) completes the parallelogram \(OPQR\), which is a square.

  2. \(PQR\) is an equilateral triangle. \(P\) is the point \(p = -2 + i\) and \(Q\) is the point \(q = 3 + 3i\). \(R\) is on the left of the line \(PQ\) as you go from \(P\) to \(Q\). Neither vertex is at the origin, so use \(q - p\).

    −3−2−11234−2−112345ReImPQ

    1. (a) Find \(q - p\), the vector from \(P\) to \(Q\).

      • \(q - p = (3 + 3 i) - (-2 + i) = 5 + 2 i\)

    2. (b) Rotate this vector through \(\frac{\pi}{3}\) anticlockwise about \(P\) by multiplying it by \(\mathrm{cis}\left(\frac{\pi}{3}\right)\). Hence find the vector from \(P\) to \(R\).

      • \((5 + 2 i)\left(\tfrac{1}{2} + \tfrac{\sqrt{3}}{2}i\right) = \frac{5}{2} - \sqrt{3} + \left(1 + \frac{5 \sqrt{3}}{2}\right) i\)

    3. (c) Find the coordinates of \(R\).

      • \(r = p + (\frac{5}{2} - \sqrt{3} + \left(1 + \frac{5 \sqrt{3}}{2}\right) i) = \frac{1}{2} - \sqrt{3} + \left(2 + \frac{5 \sqrt{3}}{2}\right) i\)
      • \(R\) is the point \((\frac{1}{2} - \sqrt{3}, 2 + \frac{5 \sqrt{3}}{2})\).

jmaths.xyz · AI HL · AHL 6.12 Geometry on the Argand diagram