Chapter 9: Complex Numbers

Transcription

1 Chapter 9: Comple Numbers 9.1 Imaginary Number 9. Comple Number - definition - argand diagram - equality of comple number 9.3 Algebraic operations on comple number - addition and subtraction - multiplication - comple conjugate - division 9.4 Polar form of comple number - modulus and argument - multiplication and division of comple number in polar form 9.5 De Moivre s Theorem - definition - n th power of comple number - n th roots of comple number - proving some trigonometric identities 9.6 Euler s formula - definition - n th power of comple number - n th roots of comple number - relationship between circular and hyperbolic functions 1

2 9.1 Imaginary numbers Consider: 4 This equation has no real solution. To solve the equation, we will introduce an imaginary number. Definition 9.1 (Imaginary Number) The imaginary number i is defined as: i 1 Therefore, using the definition, we will get, 4 4 4( 1) 4i i Eample: Epress the following as imaginary numbers a) 5 b) 8

3 9. Comple Numbers Definition 9. (Comple Numbers) If is a comple number, then it can be epressed in the form : iy, where, y R and i 1. : real part y : imaginary part Or frequently represented as : Re() = and Im() = y Eample: Find the real and imaginary parts of the following comple numbers (a) 1 3i (b) 4i i i 3 3

4 9..1 Argand Diagram We can graph comple numbers using an Argand Diagram. y Re( ) Eample: Sketch the following comple numbers on the same aes. (a) (c) 3 i (b) 3 i 1 3 i (d) 3 i 3 4 4

5 9.. Equality of Two Comple Numbers Given that where 1 a bi and c di 1, C. Two comple numbers are equal iff the real parts and the imaginary parts are respectively equal. So, if 1, then a c and b d. Eample 1: Solve for and y if given Eample : Solve and y. 3 4i (y ) i. for real numbers Eample 3: Solve the following equation for and y where 5

6 9.3 Algebraic Operations on Comple Numbers Addition and subtraction If 1 a bi and c di are two comple numbers, then 1 a c b di Eample: Given. Find a) Z 1 Z b) Z 1 + Z Multiplication If 1 a bi and c di are two comple numbers, and k is a constant, then (i) a bic di (ii) k1 1 ka kbi ac bd ad bci Eample: Given. Find Z 1 Z. 6

7 9.3.3 Comple Conjugate If Note that a bi then the conjugate of is denoted as Division If we are dividing with a comple number, the denominator must be converted to a real number. In order to do that, multiply both the denominator and numerator by comple conjugate of the denominator. 1 1 iy iy 1 iy iy 7

8 Eample 1: Given that 1 1 i, 3 4i. Find 1, and epress it in a bi form. Eample : Given 1 = + i and = 3 4i, find in the form of a + ib. Eample 3: Given Z =. Find the comple conjugate, Write your answer in a + ib form. Eample 4: Given. Find. 8

9 9.4 Polar Form of Comple Numbers P r y O Modulus of, r y. Argument of, arg () = where y tan. From the diagram above, we can see that r cos y r sin Then, can be written as iy r cos irsin r(cos isin ) rcis ( inpolarform) 9

10 Eample 1: Epress i in polar form. Eample : Epress in polar form. Eample 3 Given that 1 = + i and = - + 4i, find such that. Give your answer in the form of a + ib. Hence, find the modulus and argument of. 10

11 9.5 De Moivre s Theorem The n-th Power Of A Comple Number Definition 9.5 ( De Moivre s Theorem) and n R, then If rcos isin n r n cosn isinn Eample 1: a) Write = 1 i in the polar form. Then, using De Moivre s theorem, find 4. b) Use D Moivre s formula to write (-1 i) 1 in the form of a + ib. 11

12 9.5. The n-th Roots of a Comple Number A comple number w is a n-th root of the comple number if w n = or w =. Hence w = 1 = n [ r(cos isin )], 1/ r n k k cos isin n n for k 0,1,,, n 1 Substituting k 0,1,,, n 1 yields the nth roots of the given comple number. Eample 1: Find all the roots for the following equations: (a) (b) 3 i. Eample : Solve and epress them in a + ib form. 1

13 Eample 3: Find all cube roots of. Eample 4: Solve = 0. Sketch the roots on the argand diagram. 13

14 9.5.3 De Moivre s Theorem to Prove Trigonometric Identities De-Moivre s theorem can be used to prove some trigonometric identities. (with the help of Binomial theorem or Pascal triangle.) Eample: Prove that 5 3 cos5 16cos 0cos 5cos and 5 3 sin 5 = 16 sin - 0 sin + 5 sin. Solution: The idea is to write (cos θ + i sin θ) 5 in two different ways. We use both the Pascal triangle and De Moivre s theorem, and compare the results. From Pascal triangle, (cos θ + i sinθ) 5 = cos 5 θ + i 5 cos 4 θ sin θ 10 cos θ sin θ i 10 cos θ sin 3 θ + 5 cosθ sin 4 θ + i sin 5 θ. = (cos 5 θ 10 cos 3 θ sin θ + 5 cos θ sin 4 θ) + i(5 cos 4 θ sin θ 10 cos θ sin 3 θ + sin 5 θ). 14

15 Also, by De Moivre s Theorem, we have (cos θ + i sin θ) 5 = cos 5θ + i sin 5θ. and so cos 5θ + i sin 5θ = (cos 5 θ 10 cos 3 θ sin θ + 5 cos θ sin 4 θ) + i(5 cos 4 θ sin θ 10 cos θ sin 3 θ + sin 5 θ). Equating the real parts gives cos 5θ = cos 5 θ 10 cos 3 θ sin θ + 5 cos θ sin 4 θ. = cos 5 θ 10 cos 3 θ(1 cos θ) + 5 cos θ(1 cos θ) = cos 5 θ 10 cos 3 θ + 10 cos 5 θ + 5 cos θ 10 cos 3 θ + 5 cos 5 θ. = 16 cos 5 θ 0 cos 3 θ + 5 cos θ. (proved) Equating the imaginary parts gives sin 5θ = 5 cos 4 θ sin θ 10 cos θ sin 3 θ + sin 5 θ = 5(1 sin θ) sin θ 10(1 sin θ) sin 3 θ + sin 5 θ = 5(1 sin θ + sin 4 θ) sin θ 10 sin 3 θ + 10 sin 5 θ + sin 5 θ = 5 sin θ 10 sin 3 θ + 5 sin 5 θ 10 sin 3 θ + 10 sin 5 θ + sin 5 θ = 16 sin 5 θ 0 sin 3 θ + 5 sin θ (proved). 15

16 9.6 Eulers s Formula Definition 9.6 Euler s formula states that e i It follows that e in cos isin cosn isinn From the definition, if is a comple number with modulus r and Arg(), ; then r(cos isin ) i re ( ineuler form) Eample: Epress the following comple numbers in the form of (a) 3 + i (b) 4i i re 16

17 9.6.1 The n-th Power Of A Comple Number We know that a comple number can be epress as then i re, i r e 3 3 i 3 r e 4 4 i4 r n r n e e in Eample 1: Given. Find the modulus and argument of 5. Eample : Find in the form of a + ib. Eample 3: Epress the comple number = Then find 1 3i in the form of i re. (a) (b) 3 (c) 7 17

18 9.6. The n-th Roots Of A Comple Number The n-th roots of a comple number can be found using the Euler s formula. Note that: Then, i re ( k ) 1 1 k i r e, k = 0,1 1 1 k i r e, k = 0,1, 1 n r 1 n e k i n, k = 0,1,,,n -1 Eample 1: Find the cube roots of 1 i. Eample : Given. Find all roots of in Euler form. Eample 3: Solve 3 + 8i = 0 and sketch the roots on an Argand diagram. 18

19 9.6.3 Relationship between circular and hyperbolic functions. Euler s formula provides the theoretical link between circular and hyperbolic functions. Since e i cos isin and i e cos isin we deduce that i i i i e e e e cos and sin. i (1) In Chapter 8, we defined the hyperbolic function by cosh e e and sinh e e () Comparing (1) and (), we have so that e coshi tanh i i tan i e i cos i i e e sinhi isin i. 19

20 Also, so that cosi sini i i e e e e cosh i i e e e e sinh i tani i tanh. Using these results, we can evaluate functions such as sin, cos, tan, sinh, cosh and tanh. For eample, to evaluate cos cos( iy) we use the identity cos( A B) cos Acos B sin Asin B and obtain cos cos cosiy sin siniy Using results in (3), this gives cos cos cosh y isin sinh y. (3) 0

21 Eample: Find the values of a) sinh(3 4 i) b) tan 3i 4 c) sin (1 i) 4 (Ans: a) i b) i c) i) 1

22 Pascal s Triangle In general: