Complex number 3. z = cos π ± i sin π (a. = (cos 4π ± I sin 4π ) + (cos ( 4π ) ± I sin ( 4π )) in terms of cos θ, where θ is not a multiple of.

Size: px

Start display at page:

Download "Complex number 3. z = cos π ± i sin π (a. = (cos 4π ± I sin 4π ) + (cos ( 4π ) ± I sin ( 4π )) in terms of cos θ, where θ is not a multiple of."

Antonia Powers
5 years ago
Views:

1 Complex number 3. Given that z + z, find the values of (a) z + z (b) z5 + z 5. z + z z z + 0 z ± 3 i z cos π ± i sin π (a 3 3 (a) z + (cos π ± I sin π z 3 3 ) + (cos π ± I sin π ) + (cos ( π ) ± I sin ( π )) (cos π 3 ±I sinπ 3 ) (cos π 3 ± I sin π 3 ) + (cos π 3 I sin π 3 ) cos π 3 (b) z 5 + z 5 (cos 5π 3 ± i sin 5π 3 ) + (cos 5π 3 i sin 5π 3 ) cos 5π 3. (a) Using demoivre s Theorem to show that sin 5θ a sin 5 θ + bcos θsin 3 θ + cos θ sin θ, where a, b and c are integers to be determined. (b) Express sin 5θ sin θ in terms of cos θ, where θ is not a multiple of. Hence, find the roots of the equation 6x x + 0 in trigonometric form. (a) cos 5θ + i sin 5θ (cos θ + i sin θ) 5 (c + i s) 5, where c cos θ, s sin θ c 5 + 5c (i s) + 0 c 3 (i s) + 0 c (i s) c (i s) + (i s) 5 (c 5 0 c 3 s + 5 c s ) + i(s 5 0 c s c s) Compare imaginary part, we have sin 5θ s 5 0 c s c s sin 5 θ 0 cos θsin 3 θ + 5 cos θ sin θ (b) sin 5θ sin θ sin θ 0 cos θsin θ + 5 cos θ ( cos θ) 0 cos θ( cos θ) + 5 cos θ ( cos θ + cos θ) 0 cos θ + 0 cos θ + 5 cos θ 6 cos θ cos θ + Let x cos θ, 6x x cos θ cos θ + 0 sin 5θ sin θ 0 sin 5θ 0, where θ kπ, where kε. θ kπ 5, where kε. x cos kπ 5, where k 0,,,3,. Since x cos 0 is not a root of 6x x + 0, x cos kπ 5, where k,,3,.

2 3. (a) Find the roots of z 5. (b) Show that one of the roots in (a) is ω cos π 5 + i sin π 5 (c) Show that (i) + ω + ω + ω 3 + ω 0, 5 + i 0+ 5 (ii) + ω + ω 3 5. (a) By de Moirvres Theorem, z 5 z (cis kπ) 5 cis kπ 5, k 0,,, 3,. (b) In (a), k, ω cos π 5 + i sin π 5 Put θ π 5, then 5θ π, 3θ π θ, cos 3θ cos(π θ) cos θ Put x cos π 5, x3 3x x, x 3 x 3x + 0. Since x, dividing the left-hand side of the cubic equation by x, we get x + x 0, x 5 or 5 Rejecting the negative root, we have x cos π 5 5. Using Pythagoras theorem, sin π ( 5 ) Hence ω cos π 5 + i sin π i 0+ 5 (c) (i) Method + ω + ω + ω 3 + ω + cis π 5 + (cis π 5 ) + (cis π 5 )3 + (cis π 5 ) + cis π 5 + cis π 5 + cis 6π 5 + cis 8π 5, by de Moirvres Theorem + cis π 5 + cis π 5 + cis ( π 5 ) + cis ( π 5 ) + [cis π 5 + cis ( π 5 )] + [cis π 5 + cis ( π 5 )] + cos π 5 + cos π 5 + cos π 5 + ( cos π 5 ) + ( 5 ) + [ ( 5 ) ] 0

3 Method + ω + ω + ω 3 + ω + cis π 5 + (cis π 5 ) + (cis π 5 )3 + (cis π 5 ) (cis π 5 )5 cis π 5 (cis π)5 cis π 5 (Geometric series), by de Moirvres Theorem ()5 cis π 5 0 Method 3 Since ω cos π 5 + i sin π 5 is a root of z5, ω 5. + ω + ω + ω 3 + ω ω5 ω ω 0 (c) (ii) + ω + ω (ω ) 3 ω ω6 ω ωω5 ω ω ω +ω i 0+ (+ 5 ) ( ) ω. Show that + i is a root if x The root + i is located on a circle of radius in an Argand diagram and plot all the roots. Method Since complex roots occur in pairs, we consider the polynomial: [x ( + i )][x ( i )] (x + ) (i ) x + x + Since irrational roots occur in pairs, we consider the polynomial: [x + x + ][x x + ] (x + ) ( x) x + 6 Also, note that, x x + (x ) (i ) [x ( + i )][x ( i )] So, we get : x [x + x + ][x x + ] 0 [x ( + i )][x ( i )][x ( + i )][x ( i )] 0 3

4 Roots are: x ± i, ± i ( ± i), ( ± i) x [cos (± π ) + i sin (± π )], [cos (± 3π ) + i sin (± 3π )] (in polar form) Method x x 6 6 (cos π + i sin π) x [cos (kπ + π) + i sin (kπ + π)] x [cos ( kπ+π ) + i sin ( kπ+π )], k 0,,,3 x [cos ( π ) + i sin (π )], [cos (3π ) + i sin (3π )], [cos ( 5π ) + i sin (5π )], [cos (7π ) + i sin (7π )]. Note that : [cos ( 3π ) + i sin (3π)] [ cos (π) + i sin (π )] ( + i) + i is a root of x which is what we want in the first part of the question. Also, other roots can be written in rectangular forms easily. 5. Solve the equation 5z z 3 + z z Since the given equation is symmetric, divide 5z z 3 + z z by z, we get 5 (z + z ) (z + z ) + 0 () Put ω z +, z ω z + z +, eq () becomes 5(ω ) ω + 0 5ω ω 6 0 ω or ω 6 5 When ω, z + z z z + 0 z ± 3 i When ω 6 5, z + z 6 5 5z 6z z 3± i 5

5 6. ABCD is a square with the letters in the anticlockwise order. The points A and B represents + 3i and 6 + i respectively. Find the complex number represented by C and D. Let z A + 3i, z B 6 + i. z B z A (6 + i) ( + 3i) i Rotate z A z B anticlockwisely by π you get z Az D : z D z A (z B z A )e i(π ) z D ( + 3i) ( i) [cos ( π ) + i sin (π )] ( i)[i] + i z D ( + i) + ( + 3i) + 7i z A z B + i Rotate z B z A clockwisely by π, you get you get z Bz C : z C z B (z B z A )e i( π ) z C (6 + i) ( + i) [cos ( π ) i sin (π )] ( + i)[ i] + i z C ( + i) + (6 + i) 8 + 5i 7. The equation z z 3 + kz 8z has imaginary roots. Obtain all the roots of the equation and the value of the real constant k. Since the given has an imaginary root, z ±ai, a > 0 are roots. Hence (z ai)(z + ai) z + a is a factor of z z 3 + kz 8z + 5 z z 3 + kz 8z + 5 (z + a )(z + bz + c) Compare coefficients z 3 term, b Compare coefficients z term, 8 a b 8 a a 3 (take a > 0) Compare constant term, 5 a c 9c c 5 Compare coefficients z term, k a + b 3 7 Hence the equation becomes z z 3 + 7z 8z + 5 (z + 9)(z z + 5) 0 The roots are ±3i, ± i. 8. (a) Let z 3i, find z. (b) Hence solve the equations : (i) w + w 9 + i (ii) w + w 9 + i. (a) z 5 + i (b)(i) w + w 9 + i w + w i (w + ) ( 3i) (w + ) ( 3i) 0 5

6 [(w + ) + ( 3i)][(w + ) ( 3i)] 0 [w 3i][w + + 3i] 0 w 3i or w 3i (ii) w + w 9 + i w + w i (w + ) ( 3i) (w + ) ( 3i) 0 [(w + ) + ( 3i)][(w + ) ( 3i)] 0 [w 3i][w + + 3i] 0 w 3i or w 3i w 3 cis π or w 5 cis(.89) w 3 [cis (kπ + π )] or 5 [cis (kπ.89)] 3 [cis ( kπ+π )] or 3 [cis ( kπ.89 )], k 0,. w 3 [cos π + i sin π ] 3 ( + i) i w 3 [cos 5π + i sin 5π ] 3 ( + i).79.79i w 3 3[cos(.5) + i sin(.5)] i w 3[cos (π.5) + i sin(π.5)] i 9. If z cos θ + i sin θ, show that ( i tan θ) and write +z z in similar form. z (cos θ + i sin θ) cos θ + i sin θ + z ( + cos θ) + i sin θ cos θ + i sinθ cos θ cos θ (cos θ + i sin θ) [cos( θ) + i sin( θ)] (cos θ i sin θ) ( i tan θ) +z cos θ(cos θ+i sin θ) cos θ cos θ z ( cos θ) i sin θ sin θ i sinθ cos θ sin θ (sin θ i cos θ) sin θ [cos ( π θ) i sin (π θ)] sin θ {cos [ (π θ)] + i sin [ (π θ)]} {cos z sin θ{cos[ ( π θ)]+i sin[ (π θ)]} sin θ (π θ) + i sin (π θ)} (sin θ + i cos θ) sin θ ( + i cot θ) 6

7 0. (a) Let z + i, find z n, n,,3,, in Cartesian form. (b) Find z n in polar form. (c) Explain, without plotting, how you can represent z n in Argand diagram. ( ) k n k (a) z n ( ) k ( i) n k ( ) k k i n k { ( ) k ( + i) n k 3 where k,,3. (b) z + i ( + i ) (cos π + i sin π ) z n ( ) n (cos nπ + i sin nπ ), n,,3,, (c) It is easier to write the answers in polar form in (b) and the Argand diagram is easier to draw. If we put z 0 ( + i) 0 z cis π can be plot by rotating anti-clockwisely the vector z0 by an angle π and lengthen the radius by a factor of. Similarly by rotating z anti-clockwisely by an angle π and lengthen the radius of z by a factor of, we can get z, We can get a spiral of points. Yue Kwok Choy /0/07 7

AH Complex Numbers.notebook October 12, 2016

AH Complex Numbers.notebook October 12, 2016 Complex Numbers Complex Numbers Complex Numbers were first introduced in the 16th century by an Italian mathematician called Cardano. He referred to them as ficticious numbers. Given an equation that does