complex numbers

COMPLEX NUMBERS PART 2

Definition Complex number

A complex number is a number of the form a + bi where a and b are real numbers and

i = √⁠−1  The complex number may also be written as a + ib Set of complex numbers

The set of complex numbers is written as C = {a + bi : a, b ∈ R}

Where do we use them? •  Signal processing (The Scientist and Engineer’s Guide

to Digital Signal Processing by Steven W. Smith) Chapter 30 is on Portal for your perusal. More examples, revision, explanations: http://www.tonmeister.ca/main/textbook/ intro_to_sound_recordingch2.html#x4-370001.6 •  Electrical Circuits http://www.regentsprep.org/Regents/math/algtrig/ATO6/ electricalresouce.htm Chapter on Electronic Circuits on Portal •  High-speed flow over a jet plane

Where do we use them? Fractals: Julia Sets, Mandelbrot set Chapter on Fractals on Portal Documentary: http://www.youtube.com/watch?v=qB8m85p7GsU Visualization of Fractals: http://www.youtube.com/watch?v=8ma6cV6fw24 Another use of complex numbers is in the Fast Fourier Transform (FFT) - one of the most ubiquitous algorithms - used heavily for signal processing. If you have a digital camera, a cell phone, an LCD - FFT is there, bringing complex numbers along. SIMPLE BUT EFFECTIVE EXPLANATION: http://www.picomonster.com/

The polar form of a Complex number A point (a,b) can also be specified by giving the distance, r, of

the point from the origin and the angle, θ , between the line joining the point to the origin and the positive x-axis.

By some simple trigonometry it follows that a = r cos θ and b = r sin θ

The polar form of a complex number is z = r (cos θ + i sin θ) where r is the modulus

and θ is the argument. The modulus r of a complex number z = a + ib is written r = | z | and defined by | z | =

The polar form of a Complex number

REVIEW

•  To add complex numbers in rectangular form, add the real components and add the imaginary components. Subtraction is similar.

•  To multiply complex numbers in polar form, multiply the magnitudes and add the angles. To divide, divide the magnitudes and subtract one angle from the other

EULER’s Formula

Functions like sin (x), cos (x) and ex were

defined for real numbers x in 1201BPS. Using Taylor’s Theorem we can develop a meaningful definition of these functions for an arbitrary complex number z.

Explicit Taylor Polynomials

( ) ( ) ( ) ( ) ( ) ( ) ( )ʹ′ ʹ′ʹ′ ʹ′ʹ′ʹ′

= + + + + +L2 3f, f 0 f 0 f 0 f 0

T f 0 1! 2! 3! !

n n

n x x x x xn

Using this formula Taylor polynomials of functions can often be rather easily computed. Strategy is the following:

1.  Compute several derivatives of the given function. 2.  Evaluate these derivatives at x = 0. 3.  Detect a pattern to find a general formula for f(n)(0).

Taylor Series

Similarly e-iz = cos (z) – i sin (z)

The combination of eiz = cos (z) + i sin (z) is often denoted cis(z)

Let x = iz

Let z = θ (some angle) •  eiθ = cos θ + i sin θ •  e-iθ = cos θ – i sin θ

•  eiθ + e-iθ = cos θ + i sin θ + cos θ – i sin θ = 2 cos θ Dividing both sides by 2

•  eiθ - e-iθ = cos θ + i sin θ –(cos θ – i sin θ) = 2 i sin θ Dividing both sides by 2i

The exponential of a complex number is:

e x+iy = ex eiy = ex (cosy + i siny)

MULTIPLYING AND DIVIDING COMPLEX NUMBERS IN POLAR

FORM Let z1 = r1 e iθ1 and z2 = r2 e iθ2 then Then z1 z2 = (r1 e iθ1 ) (r2 e iθ2 ) = r1 r2 e iθ1 e iθ2

= r1 r2 e iθ1 + iθ2 = r1 r2 e i(θ1 + θ2 )

Cont…

Thus cos (θ1 + θ2) + i sin (θ1 + θ2 ) = e i(θ1 + θ2 ) = e iθ1 e iθ2

= (cos θ1 + i sin θ1) (cos θ2 + i sin θ2)

= (cos θ1 cos θ2 - sin θ1 sin θ2 ) + i(sin θ1 cos θ2 + cos θ1 sin θ2 ) Therefore cos (θ1 + θ2) = cos θ1 cos θ2 - sin θ1 sin θ2 sin (θ1 + θ2 ) = sin θ1 cos θ2 + cos θ1 sin θ2

De Moivre’s Theorem: Trigonometric identities

Equation (e iθ)n = e inθ Expressing both sides in cartesian form, we

have (cosθ + i sinθ)n = cos(nθ) + isin(nθ) This is know an De Moivre’s Theorem

Nth roots of a complex number

Imaginary numbers were originally invented to solve equations such as w2 = - 6

but in fact complex numbers permit solutions of the more general equation wn = z

where z is any complex number and n is a positive integer

w = wn1/n = (wn)1/n = z1/n

Using the polar form of z:

w = (|z|eiθ)1/n = |z|1/neiθ/n That is, w has a modulus equal to the nth

root of the modulus of z, and an argument 1/n times the argument of z.

Nth roots of a complex number

Find all the 6 roots of 1. Another way to write this is, if w6 =1, find all w

1.  Express in polar form, z = 1 = 1ei0 = 1ei(0+2mπ)

2. Find an expression for w in polar form if w6 = 1 then w6 = 1ei(0+2mπ)

And therefore w = (1ei(0+2mπ) )1/6 = 11/6ei(0+2mπ)/6 = 1eim π/3

3. Calculate 6 roots Choose 6 consecutive values of m and

substitute into the expression for w above

We will choose m = -2, -1, 0, 1, 2, 3 but we could have chosen other values Choose m = -2, This gives one sixth root:

w-2 = eiπ(-2)/3 = e(-2/3)π i Choose m = -1, This gives one sixth root:

w-1 = eiπ(-1)/3 = e(-1/3)π i Choose m = 0, This gives one sixth root:

w0 = eiπ(0)/3 = e(0)π i= 1 Choose m = 1, This gives one sixth root:

w+1 = eiπ(1)/3 = e(1/3)π i Choose m = 2, This gives one sixth root:

w+2 = eiπ(2)/3 = e(2/3)π i Choose m = 3, This gives one sixth root:

w+3 = eiπ(3)/3 = e(3/3)π i = eπ i = -1

Note that these six roots are equally spaced around the unit circle each at an angle π/3 = 60o = 360o/6 to the previous one

Find all the 4 roots of z = -1 + i. Another way to write this is, if w4 = -1 + i, find all w

1.  Express -1 + i in polar form, z = -1 + i = Rei(θ+2mπ)

Need to find R and θ R = √2 and tanθ = 1/-1 = -1 θ = tan -1(-1) = -π/4 or -π/4 + π = 3π/4 Draw Argand diagram to find correct θ R = √2 and θ = 3π/4 ∴ w4 = √2ei(3π/4 +2mπ)

Find all the 4 roots of z = -1 + i. Another way to write this is, if w4 = -1 + i, find all w

2.  Find an expression for w in polar form If w4 = -1 + i then w4 = √2ei(3π/4 +2mπ)

and therefore

w = (√2ei(3π/4 +2mπ))1/4 = (√2)1/4ei(3π/4 +2mπ)/4

= (2½)1/4ei(3π/4 +2mπ)/4

= 21/8ei(3π/4 +2mπ)/4

3. Calculate the 4 roots Choose 4 consecutive values of m and substitute into the expression for w.

This time we will choose , m= 0, 1, 2, 3 to give angles in the range 0≤ θ ≤2π.

Choose m = 0, This gives one fourth root:

w0= 21/8ei(3π/4 +2π0)/4

= 21/8ei(3π/4 +0)/4 = 21/8ei3π/16

Choose m = 1, This gives one sixth root: w1 = 21/8ei(3π/4 +2π1)/4

= 21/8ei(3π/4 + 2π)/4 = 21/8ei11π/16

3. Calculate the 4 roots

Choose m = 2, This gives one sixth root: w2 = 21/8ei(3π/4 +2π2)/4

= 21/8ei(3π/4 + 4π)/4 = 21/8ei19π/16

Choose m = 3, This gives one sixth root: w3 = 21/8ei(3π/4 +2π3)/4

= 21/8ei(3π/4 + 6π)/4 = 21/8ei27π/16

Note that these 4 roots all have a magnitude of 21/8

Consecutive roots are equally spaced around the unit circle, each at angle 2π/4 = 90o = π/2 to the previous one

Express the root w3 from the previous example in Cartesian form From z = x + iy = r(cos θ + i sin θ) Therefore w3 = |w3|(cos θ + i sin θ) We know that |w3| = 21/8 and θ3 = 27π/16 Therefore w3 = 21/8(cos27π/16 + isin27π/16)

= 1.0905[0.55557 + i(-0.83147)] = 0.60585 – 0.90672i

Complex solutions to quadratic equations

Let z =x + iy = r eiθ

Where r = and θ = tan-1(y/x) Taking natural log of both sides ln z = ln (x + iy)

= ln (r eiθ ) = lnr + ln (eiθ) = lnr + iθlne = lnr + iθ

= ln ( ) + i tan-1(y/x)

Let us now consider the logarithm of a complex number. If w = ln z

Then, by definition of a logarithm, z = ew

Writing z = x + iy and w = u + iv, we have x + iy = e u + iv = eu eiv

= eu (cos v + i sin v), (Euler’s formula)

EULER’s Formula

Equating real and imaginary parts,

x = eu cos v and y = eu sin v Squaring both these equations and adding gives:

x2 + y2 = e2u (cos2v + sin2v) = e2u ln(e2u) = ln (x2 + y2) 2u lne = ln (x2 + y2)

EULER’s Formula

2u lne = ln (x2 + y2) 2u = ln (x2 + y2) u = ½ ln (x2 + y2)

So that u = ln (x2 + y2) ½ = ln |z|

tan v = y/x + i2nπ, n = 0, ±1, ±2, … So that v = argument(z) + i2nπ, n = 0, ±1, ±2, … Hence ln z = ln |z| + i argument(z) + i2nπ

Example

Evaluate ln(-4 +i4) in the form of x + iy Solution: |-4 + i4| = √(16 + 16) = √32 = 4√2 Argument (-4 + i4) = tan –1(4/(-4)) = tan –1(-1) = π -π/4 = 3π/4 Thus

ln(-4 + i4) = ln(4√2) + i3π/4 + i2nπ n = 0, ±1, ±2, …

Calculus and Complex numbers

The same holds for integration

Definite or indefinite