Transcript complex number - Runnymede Mathematics Department

Complex numbers
What is
There is no number which squares to
make -1, so there is no ‘real’ answer!
1 ?
Mathematicians have realised that by defining the imaginary number i 
many previously unsolvable problems could be understood and explored.
If
1,
i   1 , what is:
i 3  i
i 2  1
i4  1
 100  10i
 4  2i
 3  3i
A number with both a real part and an imaginary one is called a complex number
Eg
Complex numbers are often
referred to as z, whereas real
numbers are often referred to as x
z  2 3i
The imaginary part
of z, called Im z is 3
The real part of z,
called Re z is 2
A complex number in the
form z  x  iy is said
to be in Cartesian form
Manipulation with complex numbers
Techniques used with real numbers can still be applied with complex numbers:
z = 5 – 3i,
WB1
w = 2 + 2i
Express in the form a + bi, where a and b are real constants,
(a) z2
(b)
z
w
a) z  5  3i   25  30i  9  16  30i
2
2
Expand & simplify as usual,
remembering that i2 = 1
z 5  3i 2  2i 10  10i  6i  6 4  16i  1  2i




b)
2
8
4

4
w 2  2i 2  2i
An equivalent complex number with a real
denominator can be found by multiplying by
the complex conjugate of the denominator
If z  x  iy then
its complex conjugate
is z*  x  iy
Modulus and argument
The complex number z  x  iy
can be represented on an Argand
diagram by the coordinates x, y 
Eg z1  1  3i Eg
x2  y2
2
Eg
z1 
Eg
z2 
22  12 
Eg
z3 
22  22  2 2
3
5
4
4
Remember the definition of arg z

arg z 3
1
2
z2
2
3
2

 3  tan 1     arg z 3 
2

z1 1, 3
z1
3
z 
12 
z3
Eg z 3  2  2i
z2  2  i
The modulus of z,
z 3  2,2
Im
Re
z2 2,1
The principal argument arg z  
is the angle from the positive real axis
to z x, y in the range      


 1 tan
1
 3 3

 arg z1 

3
1  1 

tan
2
   0.463...
2
 arg z2  0.46 (2dp)
WB2 The complex numbers z1 and z2 are given by
z1  2  8i
z2  1 i
z1 2  8i 1 i


z2
1 i
1 i
2  2i  8i  8

1 1
 6  10i

 3  5i
2
Find, showing your working,
(a)
z1
in the form a + bi, where a and b are real,
z2
z1
z2
(b) the value of
The modulus of
z  x  iy is z 
(c) the value of arg
 3,5
  3  5i 
 32  52 
x2  y2
z1
, giving your answer in radians to 2 decimal places.
z2
Im
The principal argument arg z  
is the angle from the positive real axis
to z x, y in the range      


34
Re

5
  tan 1   1.03 arg z      2.11
3
WB3
z = 2 – 3i
(a) Show that z2 = −5 −12i.
z 2  2  3i   4  6i  6i  9  5  12i
Find, showing your working,
(b) the value of z2,
z 2   5  12i 
2
 52   122  13
(c) the value of arg (z2), giving your answer in radians to 2 decimal places.
Im
 12 
  tan 1   1.176...
5

 5,12
Re
arg z       1.97
(d) Show z and z2 on a single Argand diagram.
Im
Re
z
z2
Complex roots
In C1, you saw quadratic equations that had no roots.
Eg x  4 x  13  0
2
x
4
 4 2  4  1 13
2 1
Quadratic formula
 b  b 2  4ac
If ax  bx  c  0 then x 
2a
2
We can obtain
complex roots
though
4  6i
4   36
 2 3i


2
2
We get no real answers because
the discriminant is less than zero
We could also obtain these roots
by completing the square:
This tells us the curve
x 2  4 x  13  0
y  x 2  4 x  13
 x  2  4  13  0
will have no
intersections with
the x-axis
 x  2  9
2
2
 x  2  3i
 x  2  3i
WB4
z1 = − 2 + i
(a) Find the modulus of z1
2i 
 22  12
 5
(b) Find, in radians, the argument of z1 , giving your answer to 2 decimal places.
Im
 2,1
 1
  tan    0.463...
2
1
arg z      2.68

Re
The solutions to the quadratic equation z2 − 10z + 28 = 0 are z2 and z3
(c) Find z2 and z3 , giving your answers in the form p  iq, where p and q are integers.
z 2  10 z  28  0
  z  5   25  28  0
2
 z  5   3
(d) Show, on an Argand diagram, the points
representing your complex numbers
Im
2
 z  5  i 3
 z  5i 3
Re
WB6
f x   x 3  x 2  44 x  150


Given that f x   x  3 x  ax  b , where a and b are real constants,
2
(a) find the value of a and the value of b.
x  3x2  ax  b  x 3  ax 2  bx  3 x 2  3ax  3b
 x3  a  3x 2  b  3a x  3b
Comparing coefficients of x2  a  2
Comparing coefficients of x0  b  50
(b) Find the three roots of f(x) = 0.


f x   x  3 x 2  2 x  50  0
either x  3  0  x  3
or x  2 x  50  0
2
(c) Find the sum of the
three roots of f (x) = 0.
1 7i   1 7i    3
So sum of the three roots is -1
  x  1  1 50  0
2
  x  1  49
 x 1  7i
2
 x  1 7i
Problem solving with roots
In C2 you met the Factor Theorem:
If a is a root of f(x)
then ( x  a ) is a factor
Eg Given that x = 3 is a root of the equation x  ax  b  0 ,
(a) write down a factor of the equation,
2
(b) Given that x = -2 is the other root, find the values of a and b
 x  2 is the other factor
 x  3x  2 is the equation factorised
 x 2  x  6  0 expanding
 a  1, b  6
In FP1 you apply this method to complex roots…
 x  3
Problem solving with complex roots
We have seen that complex
roots come in pairs:
Eg x  4 x  13  0
2
 x  2  3i
This leads to the logical conclusion that if a complex number
z  x  iy is a root of an equation, then so is its conjugate z*  x  iy
We can use this fact to find real quadratic factors of equations:
WB5 Given that 2 – 4i is a root of the equation z2 + pz + q = 0,
where p and q are real constants,
(a) write down the other root of the equation,
2 4i
(b) find the value of p and the value of q.
z  2  4i z  2  4i   0
 z 2  2  4i z  2  4i z  2  4i 2  4i   0
 z 2  2 z  4iz  2 z  4iz  4  8i  8i  16i 2  0
 z 2  4 z  20  0  p  4, q  20
Factor theorem:
If a is a root of f(x)
then ( x  a ) is a factor
WB7 Given that 2 and 5 + 2i are roots of the equation
x 3  12 x 2  cx  d  0
c, d  R
(a) write down the other complex root of the equation.
5  2i
(b) Find the value of c and the value of d.
x  5  2i x  5  2i 
 x 2  5  2i x  5  2i x  5  2i 5  2i 
 x 2  5 x  2ix  5 x  2ix  25  10i  10i  10i 2
 x 2  10 x  35
x
2

 10 x  35 x  2
(c) Show the three roots of this
equation on a single Argand diagram.
Im
 x 3  2 x 2  10 x 2  20 x  35 x  70
 x 3  12 x 2  55 x  70
 c  5, d  70
Re
Problem solving by equating real & imaginary parts
Eg Given that 3  5i  a  ib 1 i 
where a and b are real, find their values
Equating real parts:
a  ib 1 i   a  ai  bi  b
 a  b  a  bi
 a  b  3 (1)
Equating imaginary parts:
 a  b  5 (2)
(1)  (2)  2a  8  a  4
Sub in (2)  4  b  5  b  1
WB8 Given that z = x + iy, find the value of x and the value of y such that
z + 3iz* = −1 + 13i
where z* is the complex conjugate of z.
z  x  iy  z*  x  iy
then z  3iz*  x  iy  3ix  iy 
 x  iy  3ix  3 y
 x  3 y    y  3 xi
Equating real parts:  x  3 y  1 (1)
Equating imaginary parts:  y  3 x  13
(1)  3  3 x  9 y  3
(2)
(3)
(3)  (2)  8 y  16  y  2
Sub in (1)  x  6  1  x  5
Eg Find the square roots of 3 – 4i in the form a + ib, where a and b are real
a  ib 2  a 2  2abi  b2


 a 2  b2  2ab i
Equating real parts:  a 2  b 2  3 (1)
Equating imaginary parts:  2ab  4 (2)
(2)  a 
2
b
sub in (1) 
4
b2
 b2  3
 4  b 4  3b 2
 b 4  3b 2  4  0



 b2  1 b2  4  0
 b  1 as b real
sub in (2)  a  2
Square roots are -2 + i and 2 - i
Eg Find the roots of x4 + 9 = 0  x 2  3i
a  ib 2  a 2  2abi  b2


 a 2  b2  2ab i
Equating real parts:  a 2  b 2  0 (1)
Equating imaginary parts:  2ab  3 (2)
(2)  a   23b
sub in (1) 
9
4b 2
 b2  0
 9  4b 4  0
 b4 
9
4
b
3
2
sub in (2)  a  
Roots are
3
2
3
2
i
3
2
3
2
,
3
2
3
2
3
2
i
3
2
,  233  i
2
3
2
,  233  i
2
3
2
Modulus-argument form of a complex number
r z  x y
2
z  x  iy and
If
then
2
Im
zx, y 
  arg z
z  r cos  ir sin 
r
 z  r cos  i sin  
known as the modulus-argument
form of a complex number
z  2  i in the form
z  r cos  i sin  
Eg express
From previously,

r  5 and   
6
so
z  5 cos 6   i sin  6 
y  r sin 

x  r cos
Re
z  2 cos 34   i sin  34 
in the form z  x  iy
Eg express
x  r cos
 2 cos 34 
 1
y  r sin 
 2 sin  34 
 1
 z  1 i
The modulus & argument of a product
Eg if
It can be shown that:
It can also be shown that:
z1 z 2  z1 z 2
argz1z2   argz1   argz2 
z1  2  i and z 2  1 3i
z1 z 2  z1 z 2  22  12  12  3 2
Eg if
z1  2  i and z 2  1 3i
  tan 1 21   0.46...
Im
 5  10  5 2
z1

This is easier than
evaluating z1z2 and then
finding the modulus…
z1z2  2  i 1 3i   2  6i  i  3  5  5i
z1z2  5  5i  5   5  50  5 2
2
2

z2
Re
arg z1    0.46...
  tan 1  31   0.32...
arg z2  2   
 1.24...
argz1z2   0.46...   1.24...
  4
The modulus & argument of a quotient
Eg if
It can be shown that:
It can also be shown that:
z1
z1

z2
z2
 z1 
arg   argz1   arg z 2 
 z2 
z1  2  i and z 2  1 3i
Eg if
From previously,
z1
z1
5
1



z2
z2
10
2
arg z1  0.46...
arg z2  1.24...
This is much easier than
z
evaluating z21 and then
finding the modulus…
arg
z1 2  i 1  3i 2  6i  i  3



  101  107 i
z2 1 3i 1  3i
1 9
z1
z2
  101  107 i 
 101 2  107 2 
z1  2  i and z 2  1 3i
50
100

1
2
   0.46...   1.24...
z1
z2
 1.71...
WB9
z = – 24 – 7i
(a) Show z on an Argand diagram.
Im
(b) Calculate arg z, giving your answer
in radians to 2 decimal places.

 24,7
a  ℝ, b  ℝ.
It is given that w = a + bi,
Given also that w  4 and arg w 
5
6
(c) find the values of a and b
w  4cos 56  i sin
5
6
  2
3  2i
(d) find the value of zw
z1 z 2  z1 z 2
z
zw  z w  25  4  100
 242   72
w  4 given
 25
 7 
  tan    0.283...
 24 
1
Re
arg z     
 2.86
Modulus-argument form
z  r cos  i sin  
where r  z
and   arg z
Complex numbers
Manipulation with
complex numbers
z 5  3i 2  2i 10  10i  6i  6 4  16i
 21  2i




8
w 2  2i 2  2i
44
Modulus and argument
z  52   32  34
arg z  
Complex roots
w  2  2i
z  5  3i
Using:
Im
  tan 1 35 

Re
Also z1 z 2  z1 z 2
z1
z2

z1
z2
argz1z2   argz1   argz2 
arg
z
   arg z   argz 
z1
z2
1
2
w is a root of z  az  b . Find the values of a and b
2
z  2  2i z  2  2i   z 2  z2  2i   2  2i z  2  2i 2  2i 
 z 2  2 z  2iz  2 z  2iz  4  4i  4i  4
 z2  4z  8
Equating real & imaginary parts
wz*  p  qi
Find the values of p and q
wz*  2  2i 5  3i   4  16i
Equating real parts:  p  4
Equating imaginary parts:  q  16