Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 +...

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1 Complex Numbers Trigonometry Connections to Algebra

Transcript of Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 +...

Page 1: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

11111

Complex Numbers

Trigonometry Connections to Algebra

Page 2: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Overview

• Complex numbers

• Complex numbers in Trig form

• De Moivre’s Theorem

• Polar coordinates

• Parametric coordinates

Page 3: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Defining a Complex Number

• What is -1 ?

We define i as the number that solves x2 = -1

i = -1

i2 = -1

For and non-negative real k, sqrt(-k) = i sqrt(k)

i is called an imaginary number

A complex number is the sum of an imaginary and a real number

Page 4: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Examples, Imaginary Numbers

• Sqrt(-7) =

• Sqrt(-81) =

• Sqrt( -24) =

• -3 sqrt(-16) =

Page 5: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Complex Numbers

If x = ½ [ -6 + sqrt(-16)] we write this as

x = ½ [ -6 + 4 i] = -3 + 2 i

All real numbers are complex numbers since we can write

a = a + 0 i

The form a + b i is called standard form

Page 6: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Adding and Subtracting Complex Numbers

• a + b i + c + d i = a + b + (c + d ) i

Think of a + b i as a + bx, where x = i

• Examples:

(2 + 3 i) + (-5 + 2 i) =

(-5 + 4 i) – (-2 – sqrt(2) i) =

Page 7: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Multiplying Complex Numbers

• Think of as multiplying (a + bx) (c + dx) where x = i, remembering

that (i)2 = -1

Use FOIL

• Examples:

sqrt(-4) (sqrt(-9) =

sqrt(-6) [ 2 + sqrt(-3)] =

(6 – 5 i)(4 + i) =

(2 + 3 i)(2 – 3 i) =

Page 8: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Notes

• It may help to write sqrt(-a) in terms of i before you start

• Just as with reals, (a + b i) (a – b i) = a2 – ( i) 2 b2

which, since (i) = -1 is just

a2 + b2

Page 9: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Example

Show that -3 + 2 i is as solution to x2 + 6x + 13 = 0

Page 10: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Example

• Show that 2 – i sqrt(3) is a solution of x2 – 4x = -7

Page 11: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Powers of i

(i)3 = (i) 2 (i) = - i

(i)6 = (-1)(-1)(-1)= -1

Page 12: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Examples

• i 22 =

• i 28 =

• i 57 =

• i 75 =

Page 13: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Division

Think of (3 - i) / (2 + i ) as [3 – sqrt(x)] / [2 + sqrt(x)]

Remember how you were taught to clear the radical out of the

denominator to simplify these expressions using a conjugate?

Here we use the complex conjugate:

The complex conjugate of [a – b i] is a + b i ]

just as the conjugate of [a – sqrt(3)] is [a + sqrt(3)]

Page 14: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Examples

2/[5-3 i] =

[3 - i] / [ 2 - i] =

[ 6 + sqrt(-36)] / [ 3 + sqrt(-9)] =

Page 15: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Note

We can check our work with inverse operations

Page 16: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Section 10.8

DeMoivre’s Theorem

Page 17: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Complex Numbers in Trig Form

• Let the y axis be imaginary, the x real

• This is called the complex plane

• Rectangular form is z = a + b i

• Trig form is z = r cos + r sin I

– This is also like polar form

r is the magnitude of z (radius of the circle)

is the argument of z

Page 18: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Similar to Unit Circle

• Instead of x and y, we have real and imaginary

• With the unit circle, the radius was 1, here it is r

• On the unit circle x was cos , now it is r cos

• On the unit circle, y was sin , now it is i r sin

Page 19: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Example

• Convert to trig form:

z = -2 -2i

z = 6 + 2i

Page 20: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Alternate Notation

Z = r (cos + i sin ) = r cis

Page 21: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Example: converting from Trig to Rectangular

• Z = 12 cis(/6)

r =12, = /6

z = 12 [ cos /6 + i sin /6] = 12/2 [ sqrt(3) + i ]

= 6 [ sqrt(3) + i ]

• Z = 13 cis [arctan(5/12)]

cos = 12/13, sin = 5/13

z = 13 [ cos + i sin ] = 12 [ 12/13 + 5/13 i]

= 12 + 5 i

Page 22: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Why are we bothering?

• Some of our calculations become much easier

Page 23: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Products

• Z1 = 3 + 3 i , Z2 = 0 + 2i

The product, Z1 Z2 = (3 + 3 i )(0 + 2i) = -6 + 6 i

• The magnitude of the product is sqrt (72) = 6 sqrt(2)

and the angle is arctan(-1) = 135 deg

• The angle of Z1 is 45 deg, the length is 3 sqrt(2)

• The angle of Z2 = 90 deg, the length is 2

• Note that the length of the product is the product of the

lengths, and the angle is the sum of the angles of the factors

Page 24: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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We can do the product in more general form

• Z1 Z2 = r1(cos a + i sin b) r2(cos b + i sin b)

= r1 r2 [(cos a – i sin a ) ( cos b + i sin b)]

= r1 r2 [cos a cos b + i sin a cos b +

i sin b cos a – sin a sin b)

= r1 r2 [(cos a cos b - sin a sin b) +

i ( sin a cos b + sin b cos a)]

= r1 r2 [ cos (a + b) + i sin (a + b)]

• Similarly Z1/Z2 = r1/r2 [ cos (a-b) + i sin ( a-b)]

Page 25: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Example

• Z1 = -3 + i sqrt(3), Z2 = sqrt(3) + i

– Write Z1 and Z2 in trig form and compute Z1 Z2

– Compute Z1/Z2

– Verify by using rectangular form

Page 26: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Solution

• Z1 = -3 + i sqrt(3), Z2 = sqrt(3) + i

– Write Z1 and Z2 in trig form and compute Z1 Z2

Z1 in in Q2, r = 2sqrt(3), = 150

Z2 in Q1, r = 2, = 30

So Z1= 2sqrt(3)(cos 150 + i sin 150)

Z2= 2(cos 30 + i sin 30)

Z1 Z2 = 2 sqrt(3) cos 150 + i sin150) 2 (cos30 + i sin 30)

= 4 sqrt(3)[cos (150 + 30) + i sin (150 + 30)]

= 4 sqrt(3)[ -1 + 0] = -4 sqrt(3)

Page 27: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Example Cont

– Compute Z1/Z2

2 sqrt(3)[cos 150 + i sin 150] / [2(cos 30 + i sin 30)]

= sqrt(3) [cos (150 – 30) + i sin (150 – 30)]

= sqrt(3)[cos 120 + i sin 120]

= sqrt(3) [ -1/2 + sqrt(3)/2 i ]

= sqrt(3)/2 + 3/2 i

Page 28: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Continued

– Verify by using rectangular form

Z 1 Z 2 = (-3 + sqrt(3) i) (3 + i )

= -3 sqrt(3) – 3 i + 3 i + sqrt(3) i 2

= -4 sqrt(3)

Page 29: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Examples

What Quadrant? Write in trig form

7 – 7 i

2 – 2 sqrt(3) i

5 sqrt(7) – 5 sqrt(7)

-6 + 6 sqrt(3) i

Page 30: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

Review

Rectangular to trig form: x is real, y is imaginary

Calculate the length (r, moduli, |z|, etc.)

Determine the angle [Usually arctan (y/x)]

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Page 31: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Example

• Graph and write in rectangular form

12 cis (/6)

5 sqrt(3) cis (7 /6)

Page 32: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Example

• Find the moduli (length) and angle of the product

Z1 = 1 + sqrt(3) i, Z2 = 3 + sqrt(3) i

Page 33: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

Review: Products and Quotients in Trig Form

z1 = r1(cos a + i sin a), z2 = r2 ( cos b + i sin b)

z1 z2 = r1 r2 [ cos (a+b) + i sin (a+b)]

z1 /z2 = (r1/r2) [ cos (a-b) + i sin (a-b)]

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Page 34: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

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Example

Give the moduli, r, and angles, then compute the product and

verify r1r2 = r and the sum of the angles is the product’s angle

z1 = 1 + sqrt(3) i, z2 = 3 + 3 i

Page 35: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

Solution

z1 = 1 + sqrt(3) i, z2 = 3 + sqrt(3) i

r1 = sqrt( 1 + 3) = 2; angle = arctan [sqrt(3)] = 60, Q1

r2 = sqrt(9 + 3) = 2 sqrt(3); angle = arctan[sqrt(3)/3] =30, Q1

Rectangular: (1 + sqrt(3) i) [ 3 + sqrt(3)i] =

3 - 3 + i [3 sqrt(3) + sqrt(3)] = i 4 sqrt(3)

angle = 90

r1 r2 = 4 sqrt(3)

Sum of angles 60 +30 = 90

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Page 36: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

Example

Compute z1/z2 in trig form and check by using rectangular

coordinates

z1 = -sqrt(3) + i, z2 = 3

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Page 37: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

Solution

r1 = sqrt(3 + 1) = 2; angle = arctan(-1/sqrt(3) = 150 deg

r2 = 3; angle = 0

z1/z2 = 2/3 [ cos 150 + i sin(150)]

Rectangular: (-sqrt(3) + i)/( 3) = -sqrt(3)/3 + i /3

r = sqrt[3/9 + 1/9} = sqrt(4/9) = 2/3

angle = arctan( -sqrt(3)/3) = 150 deg

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Page 38: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

DeMoivre’s Theorem

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Page 39: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

The Theorem

We know z1 z2 = r1 r2 [cos (a+b) + i sin(a+b) ]

What if z1 and z2 are the same? we get

r2 [ cos 2a + i sin 2a ]

If we multiply it by the vector again, we get

r3 [ cos 3a + i sin 3a ]

So zn = rn [ cos na + i sin na ]

So what???

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Page 40: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

It is easy to compute power of complex numbers!

Find z9 if z = - ½ - ½ i

r = sqrt(2)/2, angle = 5/4

z9= [sqrt(2)/2)] [cos 45/4 + i sin 45/4 ]

= sqrt(2)/32 [cos 5/4 + i sin 5/4 ]

= sqrt(2)/32 [ -sqrt(2)/2 – i (sqrt(2)/2]

= -1/32 – 1/32 i

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Page 41: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

Solutions to Polynomial Equations

Show z = -2 – 2 i is a solution to

z4 - 3z3 – 38 z2 - 128z – 144 = 0

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Page 42: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

Solution

z4 - 3z3 – 38 z2 - 128z – 144 = 0

z = 2 sqrt(2) cis 225

z4 = [2 sqrt(2)] 4 cis (4 x 225) = 64 cis 900 = 64 cis 180 = 64 + i 0

z3 = [2 sqrt(2)] 3 cis (3 x 225) = 16 sqrt(2) cis 315 =

= 16 sqrt(2) [ sqrt(2)/2 – i sqrt(2)/2] = 16 – 16i

z2 = [2 sqrt(2)] 2 cis (2 x 225) = 8 cis (450) = 8[ 0 + i ]

We have 64 – 3(16 – 16 i) – 38(8i) – 128(2 – 2 i) – 144=

64 – 48 + 256 – 144 + i( 48 – 304 + 256) i = 0

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Page 43: Complex Numbers Products • Z1 = 3 + 3 i, Z2 = 0 + 2i The product, Z1 Z2 = (3 + 3 i)(0 + 2i) = -6 + 6 i • The magnitude of the product is sqrt (72) = 6 sqrt(2) and the angle is

nths Roots

Similar to De Moivre’s:

The nth root of z (z1/n) =

r1/n [ cos ( /n + 2/n) + i sin ( /n + 2/n) for n = 1 to n-1

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