"Complex Numbers and Quadratic Equations" — Practice

Question 1

Express z = -1 - i in polar form, giving the principal argument.

Accepted Answer

The modulus is the square root of 1 plus 1, which is root 2. The real part and the imaginary part are both negative, so the point lies in the third quadrant. The acute angle with the real axis satisfies tangent equal to 1, so it is pi over 4. In the third quadrant the principal argument is minus (pi minus the acute angle), which is minus 3 pi over 4. The polar form is therefore root 2 times [cos(-3pi/4) + i sin(-3pi/4)]. Note that an arctangent of the ratio would have returned pi over 4, which is in the first quadrant and wrong by pi.

Question 2

Evaluate root(-4) times root(-25), and explain why the naive answer is wrong.

Accepted Answer

Convert each to i form before multiplying. Root(-4) is 2i and root(-25) is 5i, so the product is 10 i squared, which is minus 10. The naive route multiplies the arguments first to give root(100), which is 10, and this is wrong by a sign. The rule that the root of a product equals the product of the roots is valid only when both numbers are non-negative reals; for negatives the two branches of the square root do not combine consistently, so the identity fails.

Question 3

Show that if |z| = 1 and z is not equal to 1 or minus 1, then (1+z)/(1-z) is purely imaginary.

Accepted Answer

A number is purely imaginary exactly when it equals the negative of its conjugate, so compute the conjugate of the expression. It is (1 + conjugate of z) divided by (1 - conjugate of z). Since the modulus is 1, we have z times conjugate of z equal to 1, so the conjugate of z is 1/z. Substituting gives (1 + 1/z)/(1 - 1/z), and multiplying numerator and denominator by z gives (z+1)/(z-1), which is the negative of the original expression. Therefore the expression equals minus its own conjugate and is purely imaginary.

Question 4

Find the value of i raised to the power 2027, showing the reasoning.

Accepted Answer

Powers of i repeat with period four, cycling through i, minus 1, minus i and 1. So only the remainder of the exponent on division by 4 matters. Dividing 2027 by 4 gives 506 with remainder 3, since 4 times 506 is 2024. The third entry in the cycle, counting i as the first, is minus i. Therefore i to the power 2027 equals minus i.

Question 5

The point 3 + 4i is rotated about the origin through 90 degrees anticlockwise. Find its new position, and then find the position if it is instead rotated 90 degrees anticlockwise about the point 1 + i.

Accepted Answer

Rotation about the origin through 90 degrees anticlockwise is multiplication by i, since i has modulus 1 and argument pi over 2. So the image is i times (3 + 4i), which is 3i plus 4 i squared, that is minus 4 plus 3i. For rotation about 1 + i, first subtract the centre to reduce to the previous case: the point becomes 2 + 3i. Multiply by i to get 2i plus 3 i squared, which is minus 3 plus 2i. Then add the centre back to obtain minus 2 plus 3i.

Question 6

Describe the locus of z satisfying |z - 2i| = 3, and the locus satisfying |z - 1| = |z - i|.

Accepted Answer

The first condition says the distance from z to the point 2i is always 3, so the locus is a circle of radius 3 centred at the point 2i, that is at (0,2) in the plane. The second condition says z is equidistant from the points 1 and i, so the locus is the perpendicular bisector of the segment joining (1,0) and (0,1). That bisector passes through the midpoint (1/2, 1/2) and is perpendicular to the segment, whose slope is minus 1, so the bisector has slope 1 and equation y equals x. In complex terms the locus is the set of points whose real and imaginary parts are equal.

Question 7

If |z - 3| = 2, find the greatest and least possible values of |z|.

Accepted Answer

The condition puts z on a circle of radius 2 centred at the point 3 on the real axis. The modulus of z is its distance from the origin, and the origin lies outside that circle, at a distance 3 from its centre. The greatest distance from the origin to a point of the circle is the distance to the centre plus the radius, giving 3 plus 2, which is 5. The least is the distance to the centre minus the radius, giving 3 minus 2, which is 1. The same bounds follow from the triangle inequality applied to z written as (z - 3) plus 3.

Question 8

Simplify (1 + i) raised to the power 8 using modulus and argument rather than expansion.

Accepted Answer

The number 1 + i has modulus root 2 and argument pi over 4, since it lies in the first quadrant with equal real and imaginary parts. Raising to the eighth power multiplies the argument by 8 and raises the modulus to the eighth power. The modulus becomes root 2 to the eighth, which is 2 to the fourth, that is 16. The argument becomes 8 times pi over 4, which is 2 pi, equivalent to 0. A number with modulus 16 and argument 0 is simply 16. Binomial expansion gives the same answer after considerably more work.

Question 9

If alpha and beta are the roots of 2x squared minus 7x plus 3 equals 0, find the value of alpha squared plus beta squared and of one over alpha plus one over beta, without solving the equation.

Accepted Answer

From the coefficients, the sum of the roots is 7 over 2 and the product is 3 over 2. For the first expression, alpha squared plus beta squared equals the square of the sum minus twice the product, which is 49 over 4 minus 3, that is 49 over 4 minus 12 over 4, giving 37 over 4. For the second, one over alpha plus one over beta equals the sum divided by the product, which is (7 over 2) divided by (3 over 2), giving 7 over 3. Both expressions are symmetric in the roots, which is exactly why they can be rewritten this way.

Question 10

Solve x squared minus 6x plus 25 equals 0 and verify your roots against the coefficient relations.

Accepted Answer

The discriminant is 36 minus 100, which is minus 64, so the roots are complex. Its square root is 8i. The roots are therefore (6 plus or minus 8i) divided by 2, giving 3 + 4i and 3 - 4i, a conjugate pair as real coefficients require. Verifying: the sum is 6, matching minus b over a, and the product is 3 squared plus 4 squared, which is 9 plus 16, that is 25, matching c over a. Nothing about the formula or the relations changed; only the arithmetic moved into the plane.

Question 11

If alpha and beta are the roots of x squared minus 4x plus 1 equals 0, form the quadratic equation whose roots are alpha squared and beta squared.

Accepted Answer

From the coefficients, the sum of the original roots is 4 and the product is 1. The new sum is alpha squared plus beta squared, which equals the square of the old sum minus twice the old product, that is 16 minus 2, giving 14. The new product is alpha squared times beta squared, which is the square of the old product, that is 1. A quadratic is formed as x squared minus the sum times x plus the product, so the required equation is x squared minus 14x plus 1 equals 0.

Question 12

It is claimed that if 1 + 2i is a root of a quadratic then 1 - 2i must also be a root. State when this is true, and construct a quadratic that shows it can fail.

Accepted Answer

It is true only when the coefficients of the quadratic are real. The proof conjugates the whole equation: since conjugation fixes real numbers, the equation reproduces itself with the root replaced by its conjugate, so the conjugate is also a root. If any coefficient is not real, conjugating produces a different equation and the argument fails. For a counterexample, choose the roots first, say 1 + 2i and 3. The sum is 4 + 2i and the product is 3 + 6i, so the equation is x squared minus (4 + 2i)x plus (3 + 6i) equals 0. It has 1 + 2i as a root, its other root is 3, and the two are not conjugates.

Practice — "Complex Numbers and Quadratic Equations"

Sign in to save attempts to your dashboard and track your accuracy over time

Find a verified tutor

Join a live course

Browse coaching centres

About

Categories

Support

Documentation

Get Study Tips & Updates

Download Our App

Download for iOS

Download for Android

Popular Cities