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Complex arithmetic and the Argand diagram questions
One new number, i, whose square is minus one, and every quadratic suddenly has its roots. Arithmetic follows the old rules, and a diagram makes the numbers visible.
7 original questions · 23 marks · the complex arithmetic and the argand diagram notes · Complex numbers
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Given z = 3 + 2i and w = 1 − 4i, find z + w and z − w.
Worked 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.With z = 3 + 2i and w = 1 − 4i, find zw in the form a + bi.
Worked 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.Express (2 + 3i)/(1 − i) in the form a + bi.
Worked 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. Realising the denominator is the entire method.For z = 3 + 4i, find |z|, state where z and its conjugate sit on an Argand diagram, and give the geometric relationship between them.
Worked 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.Solve z² − 4z + 13 = 0, giving both roots in the form a + bi, and state the relationship between them.
Worked 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. The ± of the quadratic formula lands on ±6i.Find the real numbers x and y such that (x + yi)(2 − i) = 5.
Worked 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.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.
Worked 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. Reaching for the conjugate first is what keeps this to real arithmetic; substituting z = 2 + i and expanding cubes wastes the structure.
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