Natural Sciences Notes > Mathematics for Natural Sciences Notes

This is an extract of our **Complex Numbers** document, which
we sell as part of our **Mathematics for Natural Sciences Notes** collection written by the top tier of
Cambridge University students.

* The following is a more accessble plain text extract of the PDF sample above, taken from our Mathematics for Natural Sciences Notes.
Due to the challenges of extracting text from PDFs, it will have odd formatting:
*

Mathematics for NST Part IA Cambridge University, 2012-2013

Notes for Complex Numbers The fundamental theorem of algebra: Gauss (1799) proved in his 'Fundamental Theorem of Algebra' that the field of complex numbers is algebraically closed. That is, it encloses and completes the field of numbers as a whole. A direct result of his proof is that the following n

P( z)=an z + an-1 z Has

n-1

n th order polynomial:

+...+ a1 z +a 0=0, an [?] 0

n possibly-repeated complex roots for all possible complex coefficients a0 , ... an .

Equivalently, the equation as at least one complex root. The imaginary part of any root may also be zero such that the root is purely real. Non-rigorous proof of the fundamental theorem of algebra: The fundamental theorem of algebra can be 'proven' as follows:

*

Generate a locus of points in the complex plane corresponding to given

P( z)

for a

|z|=R

*

Continuously deform the locus of points from large

*

Crossing the origin indicates the presence of a value

R to

R=0

|z|=R at which the value of

the polynomial is zero. Thus each time the locus crosses the origin as it is deformed, the presence of a root is confirmed

a0 [?] 0 , since a0 =0 gives a trivial root

Assuming that that

P( z) is simply divided by

*

n times

P( z) is almost a single circle of radius a1 z

a0

Thus, during the continual deformation of one to the other, the locus must cross the origin

*

n-1 :

P( z) is almost a circle centred on the origin, which

R , the locus of

At very small centred on

*

to form a polynomial of order

R , the locus of

For large

wraps around the origin

*

z

z=0 in which case

Hence

n times

P(z)

must have

n complex roots

The imaginary exponential: The most important result in complex number analysis, which underpins the algebraic rules used to manipulate such numbers, concerns the exponential of an imaginary number:

e

ith

Mathematics for NST Part IA Cambridge University, 2012-2013 The value of this exponential is derived using the Taylor Series for

e x =1+x +

x :

x 2 x3 x 4 x 5

+ + + +...

2! 3 ! 4 ! 5 !

[?] e ith =1+ith-

(

e

[?] e ith = 1-

th2 i th3 th 4 ith 5

- + +

+...

2! 3 ! 4 ! 5 !

th2 th 4

th3 th 5

+ + ... +i th- + +...

2! 4 !

3! 5 !

) (

)

ith

[?] e =cos th+i sin th

[?] r e ith =r ( cos th+i sinth )

Algebraic laws derived from the imaginary exponential: The algebraic laws which allow for the manipulation of complex numbers, including de Moivre's Theorem, are derived from the definition of the imaginary exponential above: i th1

Let

z 1=r 1 e =r 1 ( cos th1 +isin th1 )

Let

z 2=r 2 ei th =r 2(cos th2 +isin th2 ) 2

Multiplication:

z 1 z 2=( r 1 e i th )( r 2 ei th ) =r 1 r 2 e 1

2 i (th 1+th2 )

[?] z 1 z 2=r 1 r 2 (cos ( th 1+ th2 ) +isin ( th1 +th 2) )

th

( 1+th 2)

[?] r 1 cis th1 x r 2 cisth 2=r 1 r 2 cis

Division:

z 1 r 1 e i th r 1 i ( th -th )

=

= e z2 r2 ei th r 2 1

1 2

2 [?]

z1 r 1

= (cos ( th1-th 2 ) +i sin ( th1 -th2 ) ) z2 r 2

Mathematics for NST Part IA Cambridge University, 2012-2013

th

( 1-th2 ) r cisth 1 r 1

[?] 1

= cis

r 2 cisth 2 r 2 De Moivre's Theorem: n

( r e ith ) =r n e nith

n

[?] [ r ( cos th+i sin th ) ] =r n ( cos ( nth ) +i sin ( nth )) nth

() n

[?] ( r cis th ) =r n cis

Manipulating numbers in the complex plane: Im Re

z 2(c , d ) z 1(a , b) z 3(a+c , b+d ) The manipulation of complex numbers may be represented by geometric operations in the complex plane: Addition/subtraction:

z 1=a+ib

Addition or subtraction of

z 2=c +id

*

to give

z3

to or from

:

Corresponds to a shift in co-ordinates of

(+- a ,+- b) in the Argand diagram

*

Geometrically equivalent to the

addition/subtraction of position vectors

z2

r1 r2

z 1 and

from the origin of the Argand diagram.

Mathematics for NST Part IA Cambridge University, 2012-2013

z2 Im Re

th2 z1 z3 r2 r1

th1

th1 +th2

Mathematics for NST Part IA Cambridge University, 2012-2013 Multiplication:

z 1=r 1 e

Multiplication of give

*

i th1

with

i th2

z 2=r 2 e

to

z3 : Corresponds to a rotation by angle

+-th 2 about the origin, coupled with a change in

distance from the origin by factor

r2

:

[?] z 1 z 2=r 1 r 2 ei th ei th =r 1 r 2 e i(th +th ) 1

*

If

|z 2|=1

2 1

2 , the complex number is

simply rotated about the origin by angle

th2

Curves in the Argand Plane: Since the real and imaginary axes of the Argand plane are geometrically equivalent to the

x and

y

axes of a Cartesian plane, the locus of points

|z +a+ib|=r in the Argand

plane are equivalent to the circle in the Cartesian plane given by the equation

( x+a )2 + ( y +b )2=r 2 :

|z +a+ib|=r , z=x + yi

[?]|x +a+i ( y +b )|=r

[?] ( x +a )2 + ( y +b )2=r 2 Transformations of circles in the Argand Plane:

R

I Translation: Complex number

u+ vi is added to

z , such that:

|z|=r With centre at to:

|z +u+iv|=r

(0,0) is transformed

Mathematics for NST Part IA Cambridge University, 2012-2013

Dilation:

R

I For a dilation in the real direction, the real part of real number

z

is multiplied by some

u , such that:

|z|=|x + yi|=r -|ux + yi|=r

R

I For a dilation in the imaginary direction, the imaginary part of multiplied by some real number

z

is

v ,

*Buy the full version of these notes or essay plans and more in our Mathematics for Natural Sciences Notes.*