complex numbers 1 2

8/13/2019 Complex Numbers 1 2

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Chapter 13: Complex NumbersSections 13.1 & 13.2

() Chapter 13: Complex Numbers 1 / 19
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Denitions Complex numbers and complex plane

1. Complex numbers

Complex numbers are of the form

z = x + iy , x , y R , i 2 = 1.

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1. Complex numbers


z = x + iy , x , y R , i 2 = 1.In the above denition, x is the real part of z and y is the imaginarypart of z .



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1. Complex numbers


z = x + iy , x , y R , i 2 = 1.In the above denition, x is the real part of z and y is the imaginarypart of z .

The complex numberz = x + iy may be

represented in thecomplex plane as thepoint with cartesiancoordinates (x , y ).

y

x 0

z=3+2i

1

1

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Denitions Complex conjugate

Complex conjugate

The complex conjugate of z = x + iy is dened as

z = x iy .


D i i C l j
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Complex conjugate


z = x iy .As a consequence of the above denition, we have

e (z ) = z + z

2 , m (z ) =

z z 2i

, z z = x 2 + y 2. (1)


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Complex conjugate


z = x iy .As a consequence of the above denition, we have

e (z ) = z + z

2 , m (z ) =

z z 2i

, z z = x 2 + y 2. (1)

If z 1 and z 2 are two complex numbers, then

z 1 + z 2 = z 1 + z 2, z 1z 2 = z 1 z 2. (2)


Denitions Modulus of a complex number
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Modulus of a complex number

The absolute value or modulus of z = x + iy is

|z

|= z z =

x 2 + y 2.

It is a positive number.


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Modulus of a complex number

The absolute value or modulus of z = x + iy is

|z

|= z z =

x 2 + y 2.

It is a positive number.

Examples: Evaluate the following

|i ||2 3i |


Algebra of complex numbers
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2. Algebra of complex numbers

You should use the same rules of algebra as for real numbers, butremember that i 2 = 1.Examples:




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# 13.1.1: Find powers of i and 1/ i .


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g p




Assume z 1 = 2 + 3 i and z 2 = 1 7i . Calculate z 1z 2 and (z 1 + z 2)2.


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g p




Assume z 1 = 2 + 3 i and z 2 = 1 7i . Calculate z 1z 2 and (z 1 + z 2)2.Get used to writing a complex number in the form

z = ( real part ) + i (imaginary part ),

no matter how complicated this expression might be.


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g p

Algebra of complex numbers (continued)

Remember that multiplying a complex number by its complexconjugate gives a real number.


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Examples: Assume z 1 = 2 + 3 i and z 2 =

1

7i .


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1

7i .

Find z 1z 2

.


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1

7i .

Find z 1z 2

.

Find z 1z 2

.


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1

7i .

Find z 1z 2

.

Find z 1z 2

.

Find

m 1z 13 .


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1

7i .

Find z 1z 2

.

Find z 1z 2

.

Find

m 1z 13 .

# 13.2.27: Solve z 2 (8 5i )z + 40 20i = 0.


Polar coordinates form of complex numbers Denitions
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3. Polar coordinates form of complex numbers

In polar coordinates,

x = r cos(), y = r sin(),

where

r = x 2 + y 2 = |z |,is the modulus of z .

y

y

x x 0

z=x+iy


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3. Polar coordinates form of complex numbers

In polar coordinates,

x = r cos(), y = r sin(),

where

r = x 2 + y 2 = |z |,is the modulus of z .

y

y

x x 0

z=x+iy

The angle is called the argument of z . It is dened for all z = 0 ,and is given by

arg( z ) = =arctan y x if x 0arctan y x + if x < 0 and y 0arctan y x if x < 0 and y < 0

2n.


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Principal value Arg (z )

Because arg(z ) is multi-valued, it is convenient to agree on aparticular choice of arg(z ), in order to have a single-valued function.


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Because arg(z ) is multi-valued, it is convenient to agree on aparticular choice of arg(z ), in order to have a single-valued function.The principal value of arg(z ), Arg (z ), is such that

tan( Arg (z )) = y x , with < Arg (z ) .


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Because arg(z ) is multi-valued, it is convenient to agree on aparticular choice of arg(z ), in order to have a single-valued function.The principal value of arg(z ), Arg (z ), is such that

tan( Arg (z )) = y x , with < Arg (z ) .

Note that Arg (z ) jumpsby 2 when onecrosses the negative realaxis from above.

y

x 0

Arg( z )=

1

1

Arg( z )> -


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Principal value Arg (z ) (continued)

Examples:Find the principal value of the argument of z = 1 i .

y

x 0 1

1


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Principal value Arg (z ) (continued)

Examples:Find the principal value of the argument of z = 1 i .Find the principal value of the argument of z = 10.

y

x 0 1

1


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Polar and cartesian forms of a complex number

You need to be able to go back and forth between the polar andcartesian representations of a complex number.

z = x + iy = |z |cos() + i |z |sin().


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z = x + iy = |z |cos() + i |z |sin().

In particular, you need to know the values of the sine and cosine of multiples of / 6 and / 4.


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z = x + iy = |z |cos() + i |z |sin().


Convert cos6

+ i sin6

to cartesian coordinates.


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z = x + iy = |z |cos() + i |z |sin().


Convert cos6

+ i sin6

to cartesian coordinates.

What is the cartesian form of the complex number such that |z | = 3and Arg (z ) = / 4?


Polar coordinates form of complex numbers Eulers formula

l f l
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Eulers formula

Eulers formula reads

exp(i ) = cos( ) + i sin(), R .



E l f l
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Eulers formula


exp(i ) = cos( ) + i sin(), R .

As a consequence, every complex number z = 0 can be written asz = |z |(cos() + i sin()) = |z |exp(i ).



E l f l
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Eulers formula


exp(i ) = cos( ) + i sin(), R .

As a consequence, every complex number z = 0 can be written asz = |z |(cos() + i sin()) = |z |exp(i ).

This formula is extremely useful for calculating powers and roots of complex numbers, or for multiplying and dividing complex numbers inpolar form.


Polar coordinates form of complex numbers Integer powers of a complex number

I t f l b
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Integer powers of a complex number

To nd the n-th power of a complex number z = 0, proceed as follows



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To nd the n-th power of a complex number z = 0, proceed as follows1 Write z in exponential form,z = |z |exp(i ) .



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2 Then take the n-th power of each side of the above equationz n = |z |n exp(in) = |z |n (cos(n) + i sin(n)) .



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2 Then take the n-th power of each side of the above equationz n = |z |n exp(in) = |z |n (cos(n) + i sin(n)) .

3 In particular, if z is on the unit circle (|z | = 1), we have(cos() + i sin()) n = cos( n) + i sin(n).

This is De Moivres formula.



Integer powers of a complex number (continued)
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Examples of application:Trigonometric formulas

cos(2) = cos 2() sin2(),sin(2) = 2 sin( )cos().

(3)



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Examples of application:Trigonometric formulas

cos(2) = cos 2() sin2(),sin(2) = 2 sin( )cos().

(3)

Find cos(3) and sin(3) in terms of cos() and sin().


Polar coordinates form of complex numbers Product and ratio of two complex numbers

Product of two complex numbers
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The product of z 1 = r

1 exp (i

1) and z

2 = r

2 exp (i

2) is

z 1 z 2 = ( r 1 exp (i 1)) ( r 2 exp (i 2))= r 1r 2 exp (i (1 + 2)) . (4)



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1 exp (i

1) and z

2 = r

2 exp (i

2) is


As a consequence,arg( z 1 z 2) = arg( z 1) + arg( z 2), |z 1z 2| = |z 1| |z 2|.



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1 exp (i

1) and z

2 = r

2 exp (i

2) is


As a consequence,arg( z 1 z 2) = arg( z 1) + arg( z 2), |z 1z 2| = |z 1| |z 2|.

We can use Equation (4) to show that

cos(1 + 2) = cos ( 1)cos(2) sin(1)sin(2) ,sin(1 + 2) = sin ( 1)cos(2) + cos ( 1)sin(2) .

(5)



Ratio of two complex numbers
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Similarly, the ratio z 1z 2

is given by

z 1z 2

= r 1 exp (i 1)r 2 exp (i 2)

= r 1r 2

exp(i (1 2)) .



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is given by

z 1z 2

= r 1 exp (i 1)r 2 exp (i 2)

= r 1r 2

exp(i (1 2)) .As a consequence,

argz 1z 2

= arg( z 1) arg( z 2),z 1z 2

= |z 1||z 2|

.



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is given by

z 1z 2

= r 1 exp (i 1)r 2 exp (i 2)

= r 1r 2

exp(i (1 2)) .As a consequence,

argz 1z 2

= arg( z 1) arg( z 2),z 1z 2

= |z 1||z 2|

.

Example: Assume z 1 = 2 + 3 i and z 2 = 1 7i . Findz 1z 2

.


Polar coordinates form of complex numbers Roots of a complex number

Roots of a complex number
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p

To nd the n-th roots of a complex number z = 0, proceed as follows



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p

To nd the n-th roots of a complex number z = 0, proceed as follows1 Write z in exponential form,

z = r exp (i ( + 2p )) ,

with r = |z | and p Z .



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p


z = r exp (i ( + 2p )) ,

with r = |z | and p Z .2 Then take the n-th root (or the 1 / n-th power)

n z = z 1/ n = r 1/ n exp i + 2p n

= n r exp i + 2p n

.



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p


z = r exp (i ( + 2p )) ,

with r = |z | and p Z .2 Then take the n-th root (or the 1 / n-th power)

n z = z 1/ n = r 1/ n exp i + 2p n

= n r exp i + 2p n

.

3

There are thus n roots of z , given by

z k = n r cos + 2k n + i sin + 2k

n, k = 0 , , n 1.



Roots of a complex number (continued)
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The principal value of n z is the n-th root of z obtained by taking = Arg (z ) and k = 0.



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The principal value of n z is the n-th root of z obtained by taking = Arg (z ) and k = 0.The n-th roots of unity are given by

n

1 = cos2k

n + i sin2k

n = k

, k = 0 , , n 1where = cos(2 / n) + i sin(2/ n).



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The principal value of n z is the n-th root of z obtained by taking = Arg (z ) and k = 0.The n-th roots of unity are given by

n

1 = cos2k

n + i sin2k

n = k

, k = 0 , , n 1where = cos(2 / n) + i sin(2/ n).In particular, if w 1 is any n-th root of z = 0, then the n-th roots of z are given by

w 1, w 1, w 12, , w 1n 1.



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Examples:Find the three cubic roots of 1.



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Find the four values of 4 i .



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Find the four values of 4 i .

Give a representation in the complex plane of the principal value of the

eighth root of z = 3 + 4 i .


Polar coordinates form of complex numbers Triangle inequality

Triangle inequality
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|z 1 + z 2| |z 1|+ |z 2|.This is called the triangle inequality. Geometrically, it says that thelength of any side of a triangle cannot be larger than the sum of thelengths of the other two sides.


Polar coordinates form of complex numbers Triangle inequality

Triangle inequality
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|z 1 + z 2| |z 1|+ |z 2|.This is called the triangle inequality. Geometrically, it says that thelength of any side of a triangle cannot be larger than the sum of thelengths of the other two sides.

More generally, if z 1, z 2, . . . , z n are n complex numbers, then

|z 1 + z 2 + + z n | |z 1|+ |z 2|+ + |z n | .

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