Complex Numbers Chapter 3

1
Complex Numbers (Chapter 3)

• Ideas from Further Pure 1
• What complex numbers are
• The idea of real and imaginary parts
• if z 4 5j, Re(z) 4, Im(z) 5
• How to add, subtract, multiply and (especially)
  divide complex numbers
• The Argand diagram
• Modulus (easy) and argument (often wrong)

2
Complex Numbers

• Loci in the Argand diagram
• E.g.

z is closer (or as close) to 2 than to -2j
the direction from 2 to z is p/4 a half-line

• Equations a polynomial equation of degree n has
  precisely n (real or complex) roots (some may be
  repeats)
• The complex roots of polynomial equations with
  real coefficients occur in conjugate pairs

3
De Moivres theorem

• Polar form if z x yj has modulus r and
  argument ?,

• Multiplication and division

and
and
may need to add or subtract 2p to give principal
argument p lt ? p
4
De Moivres theorem

• As a consequence of these two,

IN F.BOOK
Uses of de Moivres theorem 1. cos n? and sin n?
in terms of powers
by de M
useful abbreviation
by Binomial

• Then equate real and imaginary parts
• May need to use

5
De Moivres theorem

• 2. cosn? and sinn? in terms of multiple angles

?
?
and
KNOW
?
?
and
KNOW
6
De Moivres theorem

• Example
• Express cos3? in terms of cos 3? and cos ?

do this right!
dont forget 2s
?
use integration
7
Complex exponents

• Use Summing series
• Example

Call this C, define S
Then C jS
This is a G.P. with a 1, r
which we have to simplify to find C
Sum to infinity
8
Complex exponents
is

• Complex conjugate of

C jS
Now
so C jS
C is the real part C
9
Complex roots

• We want to find the nth roots of a complex number
  w
• Suppose

Then

• On the Argand diagram
• roots lie on a circle, radius
• they are separated by 2p/n so form an n-sided
  polygon inscribed in the circle

10
Complex roots

• Example Find the cube roots of 8 8j

Cube roots have modulus
Arguments
so principal arguments are
so roots are
11
Complex loci

• There is one more technique in FP2

The vector from 2 to z is p/4 ahead of the vector
from -5 to z
so locus is arc of circle, endpoints -5 and 2
12
Questions Winter 06
13
Examiners Report

• (i) Found difficult! Modulus much better than
  argument
• (ii) This was done well, but sometimes proofs
  lacked sufficient detail
• (iii) Not done well some candidates appeared not
  to have been taught this
• Link with part (ii) not recognised
• Some had sums to n terms but went on to let n
  tend to infinity

14
Questions Summer 06
15
Examiners Report

• (a) (i) Most could write this down
• (ii) This exact question is in the book!
• A lot got tangled up, or left out the powers
  of 2
• (b) (i) Little trouble
• (ii) Efficiently done, although some gave
  arguments outside the range
• (iii) Many did not see connection with (ii)
• and started again