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Complex Numbers Mathematics Skills Guide

This is one of a series of guides designed to help you increase your confidence in handling mathematics. This guide contains both theory and exercises which cover:1. 2. 3. 4. 5.

Introduction Argand diagram The four rules Modulus-Argument form Roots

There are often different ways of doing things in mathematics and the methods suggested in the guides may not be the ones you were taught. If you are successful and happy with the methods you use it may not be necessary for you to change them. If you have problems or need help in any part of the work then there are a number of ways you can get help. For students at the University of Hull  Ask your lecturers.  You can contact a maths Skills Adviser from the Skills Team on the email shown below.  Access more maths Skills Guides and resources at the website below. 

Look at one of the many textbooks in the library.

Web: www.hull.ac.uk/skills Email: [email protected]

1. What are Complex Numbers? 2 You should know that the solution of the quadratic equation ax  bx  c  0 is given

by x 

 b  b2  4ac . The term b 2  4ac discriminates between the three 2a

possibilities:- if b2  4ac  0 the equation has 2 distinct real roots if b2  4ac  0 the equation has two equal real roots if b2  4ac  0 the equation has no real roots. 2 Solving the equation x  2x  17  0 using the formula gives 2  4  68 2   64 2  8  1 x    1  4 1 2 2 2

So far these roots have been ignored and described as not real but by using j to represent 1 we can write the solution to the above equation as x  1  4 j . Using this technique means that we can solve ALL quadratic equations. Mathematicians use i to represent 1 while many others use j , as we shall here. Note as j   1 then j 2   1 , j 3  j 2  j  1  j   j , j 4  j 2  j 2   1  1  1 etc. Numbers such as z  a  jb are called complex numbers with real part a and imaginary part b . NOTE b is the imaginary part NOT jb . 2. Argand diagram Complex numbers can be shown on co-ordinate axes - called an Argand diagram with the real numbers on the x -axis and the imaginary numbers on the y -axis, x  jy being shown as the point ( x, y ). y

In the diagram P( x, y ) represents the complex number z  x  jy This implies that if a  jb  c  jd then (a, b) and (c, d ) must be the same point which can only be true if a  c and b  d .

O

P(x, y)

x

This leads to a very important statement:For two complex numbers to be equal both the real and imaginary parts must be equal.

1

3. The Four Rules Addition, subtraction, multiplication and division of complex numbers are defined so that they comply with normal algebra. Addition: ( a  jb)  ( c  jd )  (a  c)  j(b  d ) Subtraction: ( a  jb)  (c  jd )  ( a  c)  j(b  d ) Multiplication:

(a  jb )(c  jd )  ac  jad  jbc  j 2 bd  ( ac  bd )  j (ad  bc )

2 2 Note ( a  jb)( a  jb)  a 2  j 2b 2  a 2  b 2 ie factors of a  b are ( a  jb)( a  jb) Also a( c  jd )  ac  jad

Conjugates If z  a  jb then a  jb is called the conjugate of z and denoted by z * .

z  z*  a  jb  a  jb   2a and zz*  a  jb a  jb   a 2  j 2 b 2  a 2  b 2 ie both z  z * and zz * are real. When solving quadratic equations with real coefficients the roots are always in conjugate pairs. For instance consider the quadratic equation 2x 2  6x  5  0 The solution is x 

6

 6 2  4  2  5 4



3 1  6  4  6  4 j    j 4 4 2 2

3 1 3 1  j and x2    j are of the form a  jb, a  jb which are 2 2 2 2 conjugates. It can be proved that this is always the case as long as the coefficients a, b and c are real.

The roots x1  

2 Note that, given equation is ax  bx  c  0 ,

 b  b 2  4ac then it can be shown that the 2a b c sum of the roots is  and the product . a a This is true for both real and complex roots. with solution x 

2

Division: To divide two complex numbers multiply both the numerator and the denominator by the conjugate of the denominator. This makes the denominator real as zz * is real for all complex numbers z . Divide ( 2  3 j) by (1  2 j ) The conjugate of 1  2 j is 1  2 j hence 2 3j (2  3 j )(1  2 j ) 2  4 j  3 j  6 j2 2  7 j 6 4 7j       1 2 j (1  2 j)(1  2 j) 1 4 5 5 1 4 j2

Examples 1. Given that z  1  j ,   1  5 j (i) Express the following in the form a  jb a) 3z  2 * ; b) z 2 ; c)

z ; d)  3 

(ii) If (m  jn) z   find the values of m and n (i) a) 3 z  2 *  3(1 j)  2(1 5 j)  3  3 j  2  10 j  1 7 j b) z 2  (1  j ) 2  1  2 j  j 2  1  2 j  1  2 j c)

1 j (1  j)(1  5 j) 1  6 j  5 j2  4 6 j 2 3 j        1  5j (1  5 j)(1  5 j) 1  25 26 13 13  z

d)  3    1 5 j 3   1 5 j15 j 2  (1  5 j)(1  10 j  25 j2 )  (1  5 j)( 24  10 j)   24  10 j  120 j  50 j 2   74  110 j or, using the binomial theorem,

 3  1 5 j 3  1 3 5 j  3 5 j 2   5 j 3  1 15 j  75 j 2  125 j 3   74 110 j (ii) (m  jn)z   (m  jn)z   

Equate real parts Equate imaginary parts Solving the equations together gives Alternative method write m  jn 

 z

( m  jn)(1  j)  1  5 j ( m  n)  j( n  m)  1 5 j m n 1 nm5 m  2, n  3

and use division method as in (i) (c)

2. Find the square root of 5  12 j Let the square root of Squaring Equating real parts

5  12 j be (a  jb) with a and b real

5  12 j  a2  2 jab  b2 5  a2  b2 (i)

3

Equating imaginary parts From (ii) a 

6 sub into (i) b

12  2ab

(ii)

2

36 6  b2 5     b2  2 b   b

Multiply through by b 2

5b2  36  b4

Rearrange & factorise

b4  5b2  36  0  (b 2  9)(b 2  4)  0  b2  9 or 4 giving b  3 j or  2

but b is real (stated above) using a  b6 gives

hence b  2 a  3 when b  2 & a  3 when b  2

so the square roots of 5  12 j are (3  2 j ) and (3  2 j ) Exercise 1 1. Given z1  1  j, z2  2  j, z3  3 j, z4  1  j find

i  3z1  z 2 vi

(z1 *)  z4 z2

x 

z2 2 z1  z3

ii  z 3  3z 2 vii  xi

iii  2z 4  z3  z 2

z33

z1( z2*) z3 z4

viii   z2 z3 2  xii 

iv  ix 

z1 z4

v 

3( z2 *) z3

z12 z2 2   z1z2 2

m and n where m  jnz1  z4

2. Find the square roots of (i) 3  4 j , (ii) 21  20 j

4. Modulus - Argument form a) If we represent the complex number z  x  jy by the point ( x, y) in the Argand diagram, then the length OP is called the modulus of z and is written as r  z 

2

x y

P(x, y) y r

2

Note by putting y  0 , this is consistent with the definition of x for real numbers.

 O

x

The angle between the positive direction of the x -axis and OP (  radians) is called the argument of z . The angle should always be given such that       and is written as arg z or arg ( z ). x y and sin  = This gives cos  = r r y  x  r cos , y  r sin and tan  x 4

hence z  x  jy  r cos  jr sin  r(cos  j sin ) which is called the modulus argument form. [Note you can also have the Euler form z  re j . ]

Multiplication and division are a lot easier if the Complex numbers are in mod-arg form but addition and subtraction are easier when they are in Cartesian form. b) Multiplication & division using mod-arg form a) Multiplication Given z1  r1(cos1  j sin1 ),

z2  r2 (cos 2  j sin 2 )

then z1 z2  r1 (cos1  j sin1 ) r2 (cos2  j sin2 ) multiplying out gives

z1 z2  1r r2 (cos1 cos2  j2 sin1 sin2  j sin1 cos2  j cos1 sin2 )  r1 r2   cos1 cos 2  sin1 sin2   j(sin1 cos2  cos1 sin2 )  r1r2  cos1   2   j sin1  2 

This uses the trig identities for cos 1  2  and sin1  2  that you should know. In the Euler form z1  r1e j 1 , z 2  r2 e j 2 , giving

z1 z2  r1 e j1  r2 e j2  r1 r2 e j1 e j2  r1 r2 e j1 2 

j    From z1 z2  r1 r2  cos1 2   j sin1  2  or from z1z 2  r1 r2e 1 2

we have and

| z1 z2 |  r1r2  | z1 |  | z2 | arg(z1z2 )  1   2  arg z1  argz2

but you do need to check that the angle is in the correct range    1  2   ! 7 3 For example if 1  2  then, in the range, we would write arg( z1 z2 ) = 2 2 Division: In the same way we can show that

z1 z2



z1 z2

z  arg 1   arg( z1 )  arg( z 2 )  z2  again it is important to check that the angle is in the correct range. And

5

n b) Value of z

z  r(cos  j sin  )

Given Then, from above,

z2  r2 (cos 2  j sin 2 )

And

z 3  r (cos  j sin  )  r 2(cos 2  j sin 2 )  r3(cos 3  j sin 3 )

De Moivre developed this further and proved that

z n  r n cosn   j sinn   for all values of n . This is fairly obvious for positive integer values of n , but it can be proved for negative values (see Example 2(a) below) and, with care, for fractional values. Examples 1. Given z   2  j 2   1 j 3 (a) Express z and  in modulus-argument form and then (b) find (in modulus-argument form) (i)  z (ii) and hence







 , (iii)   z



11 , cos  11 in surd form (c) find the values of sin 12 12

1 (a) There are at least two methods for this Draw a diagram: Method 1 – the safest & best! y z  2 j 2 P From the diagram, using  2 , 2  triangle PON, by Pythagoras, OP = 2 and, by trigonometry O N angle PON= 4

x

hence angle POx = 34 this gives similarly Method 2



Let then

cos  =

4

  1  j 3  2 cos 3  i sin3

equate real parts and imaginary parts Squaring and adding gives hence



z   2  j 2  2 cos 3  j sin 3



4



z  r(cos  j sin )

 2  j 2  r(cos  j sin )  2  r cos  2  r sin r2

2 , sin = - 2 2 2

6

1  cos = 1 , sin = giving   34 2 2





so z   2  j 2  2 cos 34  j sin 34 as before Note from section 4 we could have said r  z

x 2  y2 and tan 

y but you really x

need the diagram to be sure you have the correct value for  . y (b) (i)  z   z  2  2  4 arg( z )  arg   arg z  3    13 . 4

3

 

12

But the argument of a complex number cannot be greater than  hence from the diagram arg( z )  11 12



11  j sin 11  z  4 cos 12 12

x  



(ii)

 z



 z



2 1 2

 3  5   arg   arg  arg z    3 4 12  z   5  5    1 cos  j sin  hence z 12 12   (iii)

 

 2  2 |  | 4; arg  2  2 arg 

2 3

2 2    j sin  hence  2  4  cos 3 3  (c) but

11    11  j sin from part (i)  z  4  cos  12 12  



z  2j 2

 1  j 3   

=  2 (1  3 )  j 2 (1  3 ) comparing (A) & (B)



2 2 3 j

(A)



2 2 3



(B)



  j sin 11 =  2 (1  3 )  j 2 (1 3 )  z  4 cos 11 12 12

equating real and imaginary parts:

 11 real: 4 cos  12   11   cos   12 

  11     2 (1  3 ) imaginary: 4 sin 12   2 (1  3 )    2( 1  3) 2 (1  3 )   11  ; sin  4 4  12  7

Example 2. Use De Moivre’s Theorem to, simplify 1 cos5  j sin5 (b) cos3  j sin34 (c) (a) 6 cos   j sin  cos   j sin  3 3





(a) 1 cos   j sin   j cos   sin   cos  j sin  cos  j sin 

cos   j sin 

 cos

2

2

  sin









cos  j sin

 cos2   j2 sin2  

cos  j sin  cos  j sin 1

If De Moivre’s theorem works for negative indices we’d get 1  cos     j sin     cos   j sin  cos   j sin But this is the same as above, so De Moivre’s theorem works for negative indices. (b) By De Moivre’s Theorem cos 3  j sin3  4  cos12  j sin12 (c) Using De Moivre’s theorem cos 5  j sin 5 cos 5  j sin5 cos 5  j sin5   6 cos 2  j sin2 cos 63  j sin 63 cos 3  j sin3





1  cos 5  j sin5 cos 2  j sin2   cos5  j sin5  cos  2   j sin  2    cos 3  j sin3

Exercise 2 1. Express each of the followingin modulus argument form giving your answersas fractions of  except in part ( v) :

i  1

j ii   1 j 3

iii   1 iv 

12  2 j

v 3  4 j vi 

j

2. Using the answers from question 1, express the following



in mod  arg form i  1  j   1  j 3

  ii  112 j 23j

 

3. Using your answers from question 2, express cos 512 as a fraction 4.UseDe Moivre' s Theorem to simplifythe following

i  (cos2 

j sin2 )(cos 5  j sin5 )

( ii ) (cos3  j sin3 )4 ( iv)

( iii) (cos 2  j sin2 )(cos 3  jsin3 ) 2

(cos 5  jsin 5 )(cos 2  j sin 2 (cos 3  j sin 3 )

2

(v )

1 3 cos   j sin3 8

 vi  cos3

 j sin3

6



( vii) cos 4  j sin 4

3  cos 3

 j sin 3

2

5. Given the complex number a  jb , where a and b are real numbers, find z 2 and 1 in terms of z

2

a and b . Verify that z2  z and 1  1 . z z

6. (i) By writing z  a  jb , (with a and b real), solve the equation z 2  2  2 3 j (ii) Express z and 2  2 3 j in modulus-argument form and verify that

arg( 2  2 3 j )  2  arg z , 2  2 3 j  z 2

5. Roots of Unity and Roots of Complex Numbers If z 3  1 then

z3  1  0

factorising

( z  1)( z2  z  1)  0

hence

2 z  1 or z  z  1  0

2 Solving z  z  1  0 leads to

The 3 cube roots of unity are

But  z1 2   

 1 j 3   2 

2

 2

z

 1 1 4 2

1 j 3 and  z2  2    2    

1  3 2

1  j 3 2

z1  1, z2 

 2 2 j 3  1 j 3 1 2 j 3  ( j 3 ) 2   4 4 2





, z3 

1 j 3 2

1  j 3 2

 z2

1 2 j 3( j 3) 2  2 2 j 3  1 j 3   4 4 2

 z1

2 3 The roots are often written as 1,  and  where   1

And as z   is a solution of z2  z  1  0 then 2    1  0 2

also z   is a solution so 1  2  4  0  1  2    0 as 3  1

In Modulus-Argument form: Let cos  j sin be a cube root of 1 then

cos 

j sin 3  1

by De Moivre' s Theorem this gives

cos 3  j sin3   1

equating real and imaginary parts gives

cos 3  1, sin 3  0

cos3  1  3  2 m    2 m  3  2k      3 (where m, n & k are integers). n sin 3  0  3  n      3  giving

 

2 , 0, 2 in the range -      3 3

9

The values, outside the range       , are repeats of the values you already have. For example   4 is equivalent to   2  etc. 3

3

So the 3 cube roots of unity are

 

 

 

 

 cos  2  j sin  2    1 3 3   2   cos 0   j sin 0   1,



 cos 2  j sin 2     1  3 3   2  which is the same as before (mathematics is consistent)!

j 3

 

2

j 3 2

 

y



If the 3 roots are put on an Argand Diagram the lines from the origin to the 3 points representing the complex roots are at angles of 2 or 120o 3

x

to each other. Points representing  and 2 are reflections of each other in the x -axis as they are conjugates. Note the labels  and 2 can be interchanged.

1 2 1

4

4

If z  1 then z  1  0 which factorises into

( z 2  1)( z 2  1)  0 giving ( z  1)( z  1)( z  j)( z  j)  0 so the 4 fourth roots of 1 are 1, -1, j ,  j and on the Argand diagram we have the points representing the 4 fourth roots at (1, 0), (-1, 0), (0, 1) and (0, -1) and the lines from the origin to the points are at angles of 90o to each other.

x

-1

1

-1

Examples 2 1. If  is a complex cube root of 1, simplify (1   )(1 ) 2 3 If  is a complex cube root of 1 then 1      0 and   1 2

Hence 1    and 1   2 2

2

3

giving (1   )(1 )  () ( )    1 Or: multiply out: (1  2 )(1 )  1   2  3 2

3

 (1     )    0  1 1. 7 2. If  is any complex eighth root of 1 show that    is real In mod-arg form 1  cos0  j sin0 or cos(2n )  j sin(2n ) ( n an integer)

If  is an eigth root of unity then 8 = 1 = cos(2n )  j sin(2n ) by De Moivre’s Theorem

  cos

2n  8

j sin 28n for n  0 to 7

Or, to get the argument correct, take n as –3 to 4, but for all values of n :

ω  cos 2 n8  j sin2 n8



ω 7  cos 148n  j sin148n

giving ω  ω 7  cos 2n8  j sin 2n8  cos 148n   j sin 14n8  10









But cos14 n  cos 2 n  2 n  cos  2n  cos 2 n 8 8 8 8 14 2 2 2     n n n n and sin  sin2 n   sin    sin



8

7

hence    

8

cos 28n 





j sin 28n 

8



cos 28n 

8

j sin 28n

 2cos28n which is real.

3. Find the 5 fifth roots of unity and put them on an Argand diagram. Let z  cos   j sin be a fifth root of 1 then z 5  1

or cos   j sin  5  1  cos 5  j sin5  1  cos 5  1 and sin 5  0  5  2n where n is an integer Solving for  in the range -  <    to give distinct values of cos + j...