8.3 The Complex Plane; De Moivre s Theorem

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1 Objectives. Plot Points in the complex Plane. Convert a Complex Number Between Rectangular and Polar Form. Find Products and Quotients of Complex Numbers in Polar Form 4. Use De Moivre s Theorem 5. Find Complex Roots 0 May 08 Kidoguchi, Kenneth

2 MTH_ Revisited, The Fundamental Theorem of Algebra By the Fundamental Theorem of Algebra, a second order polynomial has exactly two zeros (aka roots). A second order polynomial of the form: y = f(x) = ax + bx + c, where a, b, & c are real constants: f ( x) x x 0 x b b c a a a b b c a a a 0 May 08 Kidoguchi, Kenneth

3 MTH_ Revisited, The Fundamental Theorem of Algebra Given f(x) = x x +, find the values of x for which f(x) = 0 0 May 08 Kidoguchi, Kenneth

4 ez mz 8. The Complex Plane; De Moivre s Theorem Complex Number Definitions A complex number is any number that can be written in the form: z = a + ib where a and b are real numbers and i, N.B.: EE s use j instead of i! i n the real part of z = a, where a is a real number. the imaginary part of z = b, where b is a real number. The powers of i are: i for n,5,9, for n, 6,0,4, i for n, 7,,5, for n 4,8,,6, Given two complex numbers, z = a + ib and z = a + ib z z a a and b b 0 May 08 5 Kidoguchi, Kenneth

5 b = r sin(q) b = r sin(q) 8. The Complex Plane; De Moivre s Theorem Representation of a Point in the Real Plane vs Complex Plane y Real Plane (x,y) = (a,b) iy Complex Plane z = a + ib q a = r cos(q) x q a = r cos(q) x 0 May 08 6 Kidoguchi, Kenneth

6 Complex Number Definitions If z = a + ib then the complex conjugate of z is: z = z* = a - ib. The product of a number and its complex conjugate is real and nonnegative. A complex number z in trigonometric form is: z = a + ib = r (cos(q) + isin(q)) where: z z r tan(q = b/a = m{z}/ e{z}. Euler's Formula states: e iq = cos(q) + i sin(q) z A complex number z in polar form is: z = r e iq a iba ib a b z 0 May 08 7 Kidoguchi, Kenneth

7 Complex Numbers Sample Hand Computation Given z = - + i, present the analysis to express z in trigonometric and polar form. 0 May 08 8 Kidoguchi, Kenneth

8 Complex Numbers Sample Hand Computation Given z = - + i, present the analysis to express z in trigonometric and polar form. r z ( ) tan( q) / in Quad II z q e i q arctan(-/) cos( q) i sin( q) - iy q 0 x 0 May 08 9 Kidoguchi, Kenneth

9 Complex Numbers Sample Hand Computation Given z = - + i, present the analysis to express z in polar form. If b = arctan(/)+/ & q =arctan(/)+ is b q? 0 May 08 0 Kidoguchi, Kenneth

10 Complex Numbers More Sample Computations Given: z i and z i. Convert z and z to polar form. Evaluate z 6 z. Evaluate z / z present the analysis to: 0 May 08 Kidoguchi, Kenneth

11 Complex Numbers More Sample Computations 0 May 08 Kidoguchi, Kenneth

12 0 May 08 Kidoguchi, Kenneth

13 . z z z 8. The Complex Plane; De Moivre s Theorem Complex Numbers More Sample Computations Given: z i and z i. Convert z and z to polar form. Evaluate z 6 z. Evaluate z / z. r r e i / 6 and tan( q tan( q z e ) ) i / present the analysis to: q 6 q i 6 i i i z z e e e e 6 /6 / 6 9. i/6 e z e z e e i/ i/ i/ i / e i 0 May 08 4 Kidoguchi, Kenneth

14 Complex Numbers Yet Another Sample Computation. Present the analysis to express z = - + i in polar form.. Present the analysis to express z e i in Cartesian form. 9 / 0 May 08 5 Kidoguchi, Kenneth

15 Complex Numbers Yet Another Sample Computation 0 May 08 6 Kidoguchi, Kenneth

16 Identities & Complex Numbers Use e i(a+b) to prove the identities: cos(a+b) = cos(a) cos(b) - sin(a) sin(b) sin(a+b) = sin(a) cos(b) + cos(a) sin(b) 0 May 08 7 Kidoguchi, Kenneth

17 Identities & Complex Numbers 0 May 08 8 Kidoguchi, Kenneth

18 Identities & Complex Numbers Use e i(a+b) to prove the identities: cos(a+b) = cos(a) cos(b) - sin(a) sin(b) sin(a+b) = sin(a) cos(b) + cos(a) sin(b) We have: e i(a+b) = cos(a+b) + i sin(a+b) and: By Uncle Euler e i(a+b) = e ia e ib Elementary Algebra = (cos(a) + i sin(a)) (cos(b) + i sin(b) By Uncle Euler = cos(a) cos(b) + isin(a)cos(b) + icos(a) sin(b) sin(a)sin(b) Given two complex numbers z = a + ib and z = a + ib z = z a = a and b = b Hence: = cos(a) cos(b) sin(a)sin(b) + i(sin(a)cos(b) + cos(a) sin(b)) cos(a+b) = cos(a) cos(b) sin(a) sin(b) sin(a+b) = sin(a) cos(b) + cos(a) sin(b), QED 0 May 08 9 Kidoguchi, Kenneth

19 Complex Algebra Example Solve the simultaneous equations {} and {} for the complex numbers A = a + ib and A = c + id where a, b, c,and d are real numbers. {} A + A = i {} ia - A = {}+{} A ( + i ) = + i i i i A i i i i i 4 i i A i A i A i i A i 0 May 08 0 Kidoguchi, Kenneth

20 More About the Complex Plane iy x (z 0, 0) iy y = z 0 sin(wt) x x = z 0 cos(wt) z = x + iy = z 0 (cos(wt) + i sin(wt) = z 0 e iwt 0 May 08 Kidoguchi, Kenneth

21 Powers of a Complex Number z = r [cos(q) + i sin(q)] = r e iq z = r [cos(q) + i sin(q)] = r e iq r [cos(q) + i sin(q)] z = r [cos(q) + i sin(q)] = r e iq r [cos(q) + i sin(q)] = = = z n = r n [cos(q) + i sin(q)] n = r n e inq r n [cos(nq) + i sin(nq)] DeMoivre's Theorem: If z = r (cos(q) + i sin(q) is a complex number and n is a positive integer then: z n = r n [cos(q) + i sin(q)] n = r n e inq r n [cos(nq) + i sin(nq)] 0 May 08 Kidoguchi, Kenneth

22 DeMoivre's Theorem Example Computation If z i present the analysis to find z. z r z e i /, and tan( q) in quadrant II q / e 4096 i cos(8) isin(8) 0 May 08 Kidoguchi, Kenneth

23 0 May 08 4 Kidoguchi, Kenneth Complex Roots Consider: x 6 = 0 Exactly 6 roots by Gauss,, i i x Note that x 6 = 0 can be rewritten x 6 =,, i i x x 0 = x 6 0 = (x )(x + x +) (x +)(x x +)

24 Roots of a Complex Number Definition of the n th root of a complex number: The complex number u = a + ib is an n th root of z z = u n = (a + ib) n Let: u = s[cos(b) + i sin(b)] and z = r[cos(q) + i sin(q)] where u is an n th root of z so that z = u n. By DeMoivre: z = u n r[cos(q) + i sin(q)] = s n [cos(nb) + i sin(nb)] q k r = s n and nb q k b n Hence, for positive integer n, the complex number z = r[cos(q) + i sin(q)] has exactly n roots given by: n z n q k r cos n q k isin n, k 0,,, n 0 May 08 5 Kidoguchi, Kenneth

25 Roots of a Complex Number - Example Present the analysis to find the 6 th root of. 6 0 k cos 6 0 k isin 6, n 6, k 0,,,,5 k cos k / isink / a + ib iy cos cos cos cos cos cos 0 / sin0 / i i0 / sin / i i / sin / i i / sin / i i0 4 / isin4 / 5 / sin5 / i i i 0 x 0 May 08 6 Kidoguchi, Kenneth

26 Rectangular Form: z = a + ib " a, " b R i complex conjugate z* = z = a - ib r z zz Trigonometric Form: z = r (cos(q) + i sin(q)) a = r cos q b = r sin q r = z tan q = b/a Euler's Equation: e iq = cos(q) + i sin(q) Polar Form z = r e iq 5/6 7/6 Summary 0 May 08 7 Kidoguchi, Kenneth 4/6 z = i 0 iy /6 8/6 0/6 /6 x /6