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T HE U NIVERSITY OF SYDNEY SCHOOL OF M ATHEMATICS AND STATISTICS Problem Sheet for Week 3 MATH1901: Differential Calculus (Advanced) Semester 1, 2017 Web Page: sydney.edu.au/science/maths/u/UG/JM/MATH1901/ Lecturer: Daniel Daners Material covered β‘ β‘ β‘ β‘ Roots of complex numbers; Complex exponential function ππ§ = ππ₯ (cos π¦ + π sin π¦); Functions of a complex variable; Sketching the image of a region. Outcomes After completing this tutorial you should β‘ β‘ β‘ β‘ β‘ β‘ find the roots of a complex number; understand the definition of the complex exponential function; manipulate the complex exponential function; solve equations involving the complex exponential function; understand what is meant by a function of a complex variable; sketch the image of a region in the complex plane under a function. Summary of essential material The complex exponential function. exponential function by For any complex number π§ = π₯ + ππ¦, π₯, π¦ β β we define the complex ππ§ βΆ= ππ₯ (cos π¦ + π sin π¦), The sketch below shows the mapping properties of the exponential function. Re π§ = π₯ πβ πβ π ππ₯+ππ¦ = ππ§ π§ = π₯ + ππ¦ Im π§ = π¦ ππ§ π¦ ππ₯ β β βπ It is an invertible map from {π§ β β β£ βπ < Im π§ β€ π} onto β ⧡ {0}. The above mapping properties together with the argument modulus form are useful to determine the images of sets under certain complex functions. Roots of complex numbers. Given π β β, the π-th roots of a complex number πΌ are the solutions to the equation π§π = πΌ. If πΌ = 1 we talk about the roots of unity. By De Moivreβs theorem the roots of unity are equally spaced on the unit circle: ( )π πβ By De Moivreβs theorem we have π2πππβπ = π2πππ = 1, so π2πππβπ , π = 0, β¦ , π β 1, are the roots of unity. If π§ = ππππ is given in polar form, then again by de Moivreβs theorem the π-th roots of π§ are given by 1 β πΌπ = π1βπ πππβπ+2πππβπ π = 0, 1, β¦ , π β 1. Again the π roots lie on a circel. Its radius in π1βπ and they are equally spaced starting from π1βπ πππβπ . Copyright © 2017 The University of Sydney 1 Hints for determining images. Let π βΆ β β β be a function, and let π· β β. The image of π· under π is im(π·) = {π (π§) β£ π§ β π·}. If you need to determine the image of a set under a complex map there are several approaches: β’ Write π§ = π₯ + ππ¦ with π₯, π¦ β β, then do a computation. This is sometimes useful, but a lot of the time inefficient. The method should only be applied as a last step after using some of the techniques below. β’ Use that π§π§Μ = |π§|2 β’ Use geometric properties, in particular |π§1 β π§2 | is the distance between π§1 and π§2 on the complex plane. β’ Write π§ = ππππ in modulus-argument form, in particular if powers of π§ are involved. Questions to complete during the tutorial 1. Express the following complex numbers in Cartesian form: π (a) π2ππβ3 2π π (b) ππ 12 ππ 3 ππ 4 2. Solve the following equations and plot the solutions in the complex plane: β (a) π§5 = 1 (b) π§4 = 8 2(1 + π) 3. Find all solutions of the following equations: (a) ππ§ = π β (c) ππ§ = β1 β π 3 (b) ππ§ = β10 (d) π2π§ = βπ 4. Let π βΆ β β β be the function π§ ξΆβ π§2 . Sketch the following sets, and then sketch their images under the function π . (a) π΄ = {π§ β β β£ Im(π§) = 2} (c) πΆ = {π§ β β β£ |π§| = 1 and Im(π§) β₯ 0} (b) π΅ = {π§ β β β£ Im(π§) = 2 Re(π§)} (d) π· = {π§ β β β£ |π§| = 1} 5. Sketch the following sets and their images under the function π§ ξΆβ ππ§ . (a) π΄ = {π§ β β β£ 0 < Re(π§) < 2 and Im(π§) = π2 } (b) π΅ = {π§ β β β£ Re(π§) = 1 and | Im(π§)| < πβ2} (c) πΆ = {π§ β β β£ Re(π§) < 0 and πβ3 < Im(π§) < π} (d) π· = {π§ = (1 + π)π‘ β£ π‘ β β} 6. Find all solutions of the equation π2π§ β (1 + 3π)ππ§ + π β 2 = 0. 7. (a) Use the definition of the complex exponential function to show that for all π β β and all π β β π β ( πππ )π = 1 + 2 cos π + 2 cos 2π + β― + 2 cos ππ. π=βπ (b) Hence, use the formula for a geometric series to show that 1 + 2 cos π + 2 cos 2π + β― + 2 cos ππ = sin(π + 12 )π sin π2 whenever π β 2πβ€. The expression is called the π-th Dirichlet kernel and appears in the summation of Fourier series. 2 Extra questions for further practice 8. Express the following complex numbers in Cartesian form: (a) πβππ (b) πln 7+2ππ (c) sin(ππ) 9. Solve the following equations and plot the solutions in the complex plane: (a) π§3 = βπ 10. (b) π§6 = β1 (a) Sketch the set π΄ = {π§ β β β£ 1β2 < |π§| < 4, 0 β€ Arg(π§) β€ πβ4}. (b) Sketch the image π΅ of π΄ in the π€-plane under the function π§ ξΆβ 1βπ§. 11. (a) Show that every complex number π§ β β can be expressed as π§ = π€ + 1βπ€ for some π€ β β. (b) Use this substitution to solve the equation π§3 β 3π§ β 1 = 0. 12. Solve, using a completion of squares, the general quadratic equation ππ§2 + ππ§ + π = 0 with π, π, π β β. In other words, prove the standard formula to solve a quadratic equation. Challenge questions (optional) 13. From the definition of πππ = cos π + π sin π we deduce that cos π = πππ + πβππ , 2 sin π = πππ β πβππ . 2π We replace π β β by any complex number π§ β β and define cos π§ βΆ= πππ§ + πβππ§ , 2 sin π§ βΆ= πππ§ β πβππ§ . 2π and call them the complex cosine and sine functions. (a) Show that when π§ is real, cos π§ and sin π§ reduce to the familiar real functions. (b) Show that cos2 π§ + sin2 π§ = 1 for all π§ β β. (c) Show that cos(π§ + π€) = cos π§ cos π€ β sin π§ sin π€ for all π§, π€ β β. (d) Is it true that | sin π§| β€ 1 and | cos π§| β€ 1 for all π§ β β? 14. There is a βcubic formulaβ analogous to the much loved quadratic formula, although it is a lot more complicated. In this question you solve the general cubic equation ππ§3 + ππ§2 + ππ§ + π = 0 with π, π, π, π β β, generalising the method of Question 11. Here is an outline of the strategy: (a) Make a substitution of the form π§ = π’ + πΌ, with πΌ to be determined, to reduce the equation to the form π’3 + ππ’ β π = 0. (b) Now attempt a substitution of the form π’ = π€ + π½βπ€ with a cleverly chosen π½ to reduce the equation to a quadratic in π€3 . (c) You can now solve this quadratic equation for π€, hence back-track to find π§. 3 15. Let π be a given positive integer. A primitive πth root of unity is a solution π§ = πΌ of the equation π§π = 1 with the property that the powers πΌ, πΌ 2 , β― , πΌ πβ1 , πΌ π give all of the πth roots of unity. For example, the 4th root of unity πΌ = π is primitive, since π, π2 = β1, π3 = βπ, π4 = 1 gives us all 4th roots of unity, while the 4th root πΌ = β1 is not primitive, since β1, (β1)2 = 1, (β1)3 = β1, (β1)4 = 1 fails to give us all of the 4th roots of unity. (a) Find all primitive 6th roots of unity. (b) Find all primitive 5th roots of unity. (c) For which values of π, 0 β€ π β€ π β 1, is ππ 2ππ π a primitive πth root of unity? 16. We introduced the complex numbers by saying something along the lines of: βAppend a solution π of the equation π₯2 + 1 = 0 to the real numbers ββ. This is a bit mysterious, and you might ask: βWhat is this magical element π? Where does it live?β. Here is a more formal approach. Let β2 = {(π, π) β£ π, π β β} and define an addition operation and a multiplication operation on β2 by: for all (π, π) β β2 (π, π) + (π, π) = (π + π, π + π) (π, π)(π, π) = (ππ β ππ, ππ + ππ) for all (π, π) β β2 . Let π = (0, 0), π = (1, 0), and π’ = (0, 1). (a) Show that π + (π, π) = (π, π) for all (π, π) β β2 . (b) Show that π(π, π) = (π, π) for all (π, π) β β2 . (c) Show that π’2 + π = π. (d) Explain why β2 with the above operations is really just β in disguise. Thus it is possible to introduce β without ever talking about the βimaginaryβ number π. 4