Questions on complex arithmetic and the Argand diagram

Question 1

Given z = 3 + 2i and w = 1 − 4i, find z + w and z − w.

Accepted Answer

z + w = 4 − 2i and z − w = 2 + 6i. B1 for the sum, B1 for the difference. Real parts combine with real, imaginary with imaginary; nothing crosses over.

Question 2

With z = 3 + 2i and w = 1 − 4i, find zw in the form a + bi.

Accepted Answer

zw = 3 − 12i + 2i − 8i² = 3 − 10i + 8 = 11 − 10i. M1 for the expansion, A1 for the answer. The whole calculation is expanding brackets, with i² = −1 the only new step, converting the −8i² into +8.

Question 3

z = 3 − i and w = 4 + 2i. Find z − 2w and z², giving each in the form a + bi.

Accepted Answer

Doubling w gives 8 + 4i, so z − 2w = (3 − i) − (8 + 4i) = −5 − 5i, subtracting real from real and imaginary from imaginary. B1 for −5 − 5i. Squaring, z² = (3 − i)(3 − i) = 9 − 3i − 3i + i² = 9 − 6i − 1 = 8 − 6i. M1 for the expansion, A1 for 8 − 6i. Multiplying by the real number 2 scales both parts at once, and the only step in the square that is not ordinary algebra is replacing i² by −1: dropping that step leaves 9 − 6i, whose real part is 1 too large.

Question 4

The numbers z = 1 + 4i and w = 5 + 2i are plotted on an Argand diagram. Write down the coordinates of the point representing z + w, and name the quadrilateral whose vertices are the points representing 0, z, z + w and w.

Accepted Answer

z + w = 6 + 6i, plotted at (6, 6). B1 for the point. The step from 0 to z is 1 across and 4 up, and so is the step from w to z + w; the step from 0 to w equals the step from z to z + w in the same way. Opposite sides are equal and parallel, so the shape is a parallelogram. B1 for naming it. Addition on the diagram is vector addition, tip to tail. Here |z| = √17 and |w| = √29 are different, so the parallelogram is not a rhombus, and z and w are not multiples of one another, so the four points do not collapse onto a line.

Question 5

Express (2 + 3i)/(1 − i) in the form a + bi.

Accepted Answer

Multiply top and bottom by the conjugate 1 + i. The denominator becomes (1 − i)(1 + i) = 2 and the numerator (2 + 3i)(1 + i) = 2 + 2i + 3i − 3 = −1 + 5i. So the quotient is −1/2 + (5/2)i. M1 for multiplying by the conjugate, A1 for the real denominator, A1 for the quotient.

Question 6

For z = 3 + 4i, find |z|, state where z and its conjugate sit on an Argand diagram, and give the geometric relationship between them.

Accepted Answer

|z| = √(9 + 16) = 5. z sits at the point (3, 4) in the first quadrant; the conjugate 3 − 4i sits at (3, −4). Conjugation is reflection in the real axis, and both points are distance 5 from the origin. B1 for the modulus, B1 for the two positions, B1 for the reflection.

Question 7

Solve z² − 4z + 13 = 0, giving both roots in the form a + bi, and state the relationship between them.

Accepted Answer

The discriminant is 16 − 52 = −36, so z = (4 ± 6i)/2 = 2 ± 3i. The roots are a conjugate pair, as the roots of any quadratic with real coefficients must be when they are not real. M1 for the discriminant, A1 for both roots, B1 for naming them a conjugate pair.

Question 8

Find the real numbers x and y such that (x + yi)(2 − i) = 5.

Accepted Answer

Expanding gives (2x + y) + (2y − x)i = 5. Matching real and imaginary parts gives 2x + y = 5 and 2y − x = 0, so x = 2y and 5y = 5, giving y = 1 and x = 2. Then x + yi = 2 + i, and (2 + i)(2 − i) = 4 + 1 = 5 as a check. M1 for the expansion, A1 for equating real and imaginary parts, A1 for y, A1 for x. One complex equation always carries two real ones.

Question 9

Find the two square roots of −8 + 6i in the form x + yi, where x and y are real.

Accepted Answer

Set (x + yi)² = −8 + 6i. Expanding the left side, x² − y² + 2xyi = −8 + 6i. M1 for squaring x + yi and comparing parts. Real parts give x² − y² = −8, imaginary parts 2xy = 6. A1 for both equations. The second gives y = 3/x, and substituting into the first gives x² − 9/x² = −8, that is x⁴ + 8x² − 9 = 0, a quadratic in x² that factorises as (x² + 9)(x² − 1) = 0. M1 for the quadratic in x². Since x is real, x² = −9 is rejected, leaving x² = 1, so x = ±1, and y = 3/x pairs those with y = ±3 in the same order. The square roots are 1 + 3i and −1 − 3i. A1 for both. Checking, (1 + 3i)² = 1 + 6i − 9 = −8 + 6i. The two roots differ only in sign, since (−z)² = z².

Question 10

The complex numbers z and w satisfy z + 2w = 2 + 5i and 2z − w = 9 − 5i. Find z and w.

Accepted Answer

Multiplying the second equation by 2 gives 4z − 2w = 18 − 10i, and adding the first equation removes w. M1 for eliminating w. That leaves 5z = 20 − 5i. A1. Dividing both parts by 5, z = 4 − i. A1. Then 2w = (2 + 5i) − (4 − i) = −2 + 6i, so w = −1 + 3i. A1. Checking in the second equation, 2(4 − i) − (−1 + 3i) = 8 − 2i + 1 − 3i = 9 − 5i. The elimination is the ordinary one for two linear equations; complex coefficients change nothing about it, and the check is worth doing in the equation that was not used to find w.

Question 11

z = x + yi, where x and y are real, and z* is the conjugate of z. Show that zz* = x² + y², and state what kind of number z + z* is and what kind z − z* is, giving a reason for each.

Accepted Answer

zz* = (x + yi)(x − yi) = x² − xyi + xyi − y²i² = x² + y²: the two cross terms cancel and −y²i² becomes +y². B1 for the product. z + z* = (x + yi) + (x − yi) = 2x, which is real, because the imaginary parts are equal and opposite. B1. z − z* = 2yi, which is purely imaginary, because the real parts are equal and cancel instead. B1. The first result is why multiplying by the conjugate clears a denominator: x² + y² is real, and for z ≠ 0 it is not zero. z − z* is zero exactly when y = 0, that is when z is already real.

Question 12

Given that z = 2 + i is a root of z³ + az² + bz + 10 = 0, where a and b are real, find a, b and the third root.

Accepted Answer

Real coefficients force the conjugate 2 − i to be a root as well, so (z − 2 − i)(z − 2 + i) = z² − 4z + 5 is a factor. Write the cubic as (z² − 4z + 5)(z − r). The constant term is −5r, and matching it to 10 gives r = −2, so the third root is −2. Expanding, z³ − (4 + r)z² + (5 + 4r)z − 5r, so a = −2 and b = −3. Checking the sum of the roots, (2 + i) + (2 − i) + (−2) = 2 = −a. B1 for the conjugate root, M1 A1 for the quadratic factor, M1 for the remaining linear factor, A1 for the third root, A1 for a and b.

Question 13

A polynomial f(z) of degree 5 has real coefficients. Explain why its non-real roots occur in conjugate pairs, and deduce that f(z) = 0 has at least one real root. Then explain why a polynomial of degree 4 with real coefficients need not have any real root, giving an example together with its roots.

Accepted Answer

Take the conjugate of f(z) = 0 term by term. Conjugation leaves each real coefficient unchanged and turns z^k into (z*)^k, so f(z*) = 0 whenever f(z) = 0, and z* is a root whenever z is. B1 for the conjugate root. A non-real root is therefore never alone: its conjugate is a different root, so the non-real roots can be listed in pairs and there is an even number of them. M1 for the pairing. The five roots, counted with multiplicity, cannot all be non-real, since 5 is odd, so at least one root is real. A1. Degree 4 raises no such obstacle, because 4 is even and all four roots may pair up. M1 for the even degree. For example (z² + 1)(z² + 4) = z⁴ + 5z² + 4 has roots i, −i, 2i and −2i, two conjugate pairs and no real root. A1. The count is of roots with multiplicity, and the conclusion is only that the list of real roots is not empty: it names no particular root, and a quintic may have three real roots or five.

Question 14

z = x + yi is a non-zero complex number, where x and y are real. Show that the points representing 0, z and iz on an Argand diagram are the vertices of a right-angled isosceles triangle, and state the geometric effect of multiplying a complex number by i.

Accepted Answer

iz = i(x + yi) = xi + yi² = −y + xi, so the three points are (0, 0), (x, y) and (−y, x). M1 for multiplying by i, A1 for the point (−y, x). The two sides from the origin have lengths √(x² + y²) and √(y² + x²). M1 for both lengths. These are equal, so the triangle is isosceles. A1. The third side joins (x, y) to (−y, x), so its length squared is (x + y)² + (y − x)² = 2x² + 2y², which is the sum of the squares of the other two sides; by the converse of Pythagoras the angle at the origin is a right angle. A1. So multiplying by i turns a number through a quarter turn anticlockwise about the origin and leaves its distance from the origin unchanged. B1. The three points fall together when z = 0, which is why z was taken to be non-zero, and the argument uses the general x + yi rather than one example, so it holds for every such z.

Question 15

The complex number z is not real, and z + 1/z is real. Show that |z| = 1.

Accepted Answer

Since z is not real it is not zero, so 1/z exists. Write z = x + yi with x and y real and y ≠ 0. Multiplying top and bottom by the conjugate, 1/z = (x − yi)/(x² + y²). M1 for the reciprocal through the conjugate, A1 for (x − yi)/(x² + y²). Adding, z + 1/z = x + x/(x² + y²) + [y − y/(x² + y²)]i. A number is real exactly when its imaginary part is zero, so y − y/(x² + y²) = 0. M1 for setting the imaginary part to zero. Factorising, y[1 − 1/(x² + y²)] = 0. A1. z is not real, so y ≠ 0 and the bracket is what vanishes. B1 for the reason y may be divided out. Hence x² + y² = 1, that is |z| = 1. A1. The condition that z is not real carries weight: z = 2 gives z + 1/z = 5/2, which is real while |z| = 2, so dividing by y without checking it is non-zero would prove something false.