Download i, Z2

Complex Numbers
Trigonometry Connections to Algebra
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Overview
• Complex numbers
• Complex numbers in Trig form
• De Moivre’s Theorem
• Polar coordinates
• Parametric coordinates
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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
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Examples, Imaginary Numbers
• Sqrt(-7) =
• Sqrt(-81) =
• Sqrt( -24) =
• -3 sqrt(-16) =
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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+0i
The form a + b i is called standard form
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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) =
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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) =
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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
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Example
Show that -3 + 2 i is as solution to x2 + 6x + 13 = 0
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Example
• Show that 2 – i sqrt(3) is a solution of x2 – 4x = -7
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Powers of i
(i)3 = (i) 2 (i) = - i
(i)6 = (-1)(-1)(-1)= -1
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Examples
• i 22 =
• i 28 =
• i 57 =
• i 75 =
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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)]
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Examples
2/[5-3 i] =
[3 - i] / [ 2 - i] =
[ 6 + sqrt(-36)] / [ 3 + sqrt(-9)] =
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Note
We can check our work with inverse operations
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Section 10.8
DeMoivre’s Theorem
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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
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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 
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Example
• Convert to trig form:
z = -2 -2i
z = 6 + 2i
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Alternate Notation
Z = r (cos  + i sin ) = r cis 
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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
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Why are we bothering?
• Some of our calculations become much easier
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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
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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)]
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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
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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)
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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
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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)
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Examples
What Quadrant? Write in trig form
7–7i
2 – 2 sqrt(3) i
5 sqrt(7) – 5 sqrt(7)
-6 + 6 sqrt(3) i
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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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Example
• Graph and write in rectangular form
12 cis (/6)
5 sqrt(3) cis (7 /6)
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Example
• Find the moduli (length) and angle of the product
Z1 = 1 + sqrt(3) i, Z2 = 3 + sqrt(3) i
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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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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
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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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Example
Compute z1/z2 in trig form and check by using rectangular
coordinates
z1 = -sqrt(3) + i, z2 = 3
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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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DeMoivre’s Theorem
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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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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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Solutions to Polynomial Equations
Show z = -2 – 2 i is a solution to
z4 - 3z3 – 38 z2 - 128z – 144 = 0
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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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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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