Chapter 2: Complex numbers - PDF Free Download

Chapter 2: Complex numbers Complex numbers are commonplace in physics and engineering. In particular, complex numbers enable us to simplify equations and/or more easily find solutions to equations. We will explore the damped, driven simple-harmonic oscillator as an example of the use of complex numbers. By the end of this chapter you should be able to... Represent complex numbers in various ways Use complex algebra Complex infinite series Determine functions of complex numbers Use Euler s formula

Real and imaginary parts of a complex number Consider a complex number written as z = x + iy Re z = x What if it is not presented as z = x + iy? We can always simplify to this form with complex algebra. (2 + i) 2 = 4 + 4i 1 = 3 + 4i (1) What if there is a complex denominator? We can multiply in the numerator and denominator by the complex conjugate or conjugate of the denominator. i + 1 i 1 = ( i + 1 i 1 ) ( ) i 1 = i (2) i 1

Real and imaginary parts of a complex number Consider a complex number written as z = x + iy Re z = x Im z = y What if it is not presented as z = x + iy? We can always simplify to this form with complex algebra. (2 + i) 2 = 4 + 4i 1 = 3 + 4i (1) What if there is a complex denominator? We can multiply in the numerator and denominator by the complex conjugate or conjugate of the denominator. i + 1 i 1 = ( i + 1 i 1 ) ( ) i 1 = i (2) i 1

The complex plane Representing a complex number as z = x + iy suggests we can plot them in a complex plane.

The complex plane Representing a complex number as z = x + iy suggests we can plot them in a complex plane. We can also represent in polar form z = re iθ

The complex plane Representing a complex number as z = x + iy suggests we can plot them in a complex plane. We can also represent in polar form z = re iθ Equivalently z = r (cos θ + i sin θ)

The complex plane Representing a complex number as z = x + iy suggests we can plot them in a complex plane. We can also represent in polar form z = re iθ Equivalently z = r (cos θ + i sin θ) The modulus or magnitude is r

Complex algebra We need to work with complex representations z = x + iy = re iθ = r (cos θ + i sin θ). It should become easy to go from one representation to another! For z = x + iy, the complex conjugate z = x iy For z = re iθ, z = re iθ Given z = x + iy, we can find r from r = z = z z = x 2 + y 2 Given z = x + iy, we see cos θ = x/r and sin θ = y/r

Complex power series Previously we had real power series a n x n (3) n=1 We can also consider a power series in z = x + iy a n z n (4) n=1 We need to determine whether a series converges using the ratio test: ρ = lim a n+1z n+1 n a n z n (5) When ρ < 1, the series converges.

Example of convergence test Consider the series, 1 + iz + (iz)2 2! + (iz)3 3! + (iz)4 4! +... (6) We see from this that, ρ = lim a n z n n a n+1z n+1 = lim n iz n + 1 = 0 (7) So indeed this series converges. In fact we will see that it converges to e iz.

Euler s formula We previously had the infinite-series (i.e. Mclaurin series) for cos θ and sin θ, cos θ = 1 θ2 2! + θ4... (8) 4! Using dez dz sin θ = θ θ3 3! + θ5... (9) 5! = ez, we can also represent e z by the series, e z = 1 + z + z2 2! + z3 3! + z4 4! + z5 +... (10) 5! If we take z = iθ, then comparison of these series proves Euler s formula, e iθ = cos θ + i sin θ

Examples using Euler s formula Express z = 2e πi 4 in the form z = x + iy From Euler s formula, 2e iπ 4 = 2 cos π 4 + 2i sin π 4 = 2 + i 2 Express z = ( i 2 1+i ) 6 in the form z = x + iy Using Euler s formula, i = e iπ 2 ( i 2 1 + i ) 6 = ( e iπ 2 Then we use Euler s formula, e iπ 4 and 1 + i = 2e πi 4, then we see ) 6 = ( ) e iπ 6 3πi 4 = e 2 e 3πi 2 = cos 3π 2 + i sin 3π 2 = i

Simple-harmonic oscillator For a pendulum oscillating with small angle θ, or a mass-spring system obeying Hooke s law, we get simple harmonic motion d 2 θ dt 2 + g l θ = 0 d 2 y dt 2 + k m y = 0 y(t) = A cos(ω 0 t) + B sin(ω 0 t) = Ce iω 0t. Note that C is complex, but we usually take A and B to be real. Also ω 0 = k/m. A and B (or C) are determined by conditions at t = 0 (position and velocity)

Damped, driven simple-harmonic oscillator Complex exponentials useful in the damped, driven oscillator We will assume d 2 y dy + 2b dt2 dt + ω2 0y = F D cos(ωt) Y (t) = Ce iωt = Ae i(ωt φ) with the driving force F D e iωt. The A and φ are real and C is complex. Substitute and solve for C, C = F D ω 2 0 ω2 + 2ibω We would like to get C into the form Ae iφ, so we multiply numerator and denominator by complex conjugate of denominator

Damped, driven simple-harmonic oscillator, cont. C = F D ω 2 0 ω2 + 2ibω [ ω 2 0 ω 2 ] 2ibω ω0 2 ω2 2ibω = F [ D ω 2 0 ω 2 2ibω ] (ω0 2 ω2 ) 2 + 4b 2 ω 2 We now need A = C = C C and also an expression for φ. We find F D A = C = (ω0 2 ω2 ) 2 + 4b 2 ω 2 and for φ, tan φ = 2bω ω 2 0 ω2

Damped, driven simple-harmonic oscillator, cont. Then we get Y (t) = Ae i(ωt φ) with the A and φ given on last slide. Then finally, for a driving force F D cos ωt, we take the real part of Y (t), y(t) = A cos (ωt φ) Using complex exponentials was somewhat simpler, because otherwise we would have both sin and cos terms to keep track of, and also solve for A and B (see previously).

LCR circuit LCR circuit completely analogous to damped, driven simple-harmonic oscillator Consider an alternating current source with potential V D = V 0 cos ωt V L + V R + V C = V D L d 2 q dt 2 + R dq dt + q C = V 0 cos ωt Q(t) = Q 0 e i(ωt φ) Solve as a homework problem, but given damped-driven oscillator solution, this should be easy!

Damped, driven simple-harmonic oscillator Is that everything for this problem? No! We only considered particular solutions

Damped, driven simple-harmonic oscillator Is that everything for this problem? No! We only considered particular solutions Particular solution valid for long-time behavior

Damped, driven simple-harmonic oscillator Is that everything for this problem? No! We only considered particular solutions Particular solution valid for long-time behavior We also need homogeneous solutions Homogeneous solutions relevant for no driving force and behavior right after driving force turned on