Download File - PROJECT MATHS REVISION

Lucan Community College
Mathematics Department
Complex Numbers
5th Year Maths – February 7 2010 - Mr Duffy
Please Read and study the following notes on (1) Division of Complex
Nos. and (2) The equality of Complex Nos. Write a summary of this into
your hardbacks for Tuesday’s class.
 (1) Division of Complex Numbers
Reminder from Friday’s Class:
The conjugate of the complex number (a + bi) is (a – bi) ie. Change teh sign of
the imaginary part ONLY.
For any two complex conjugates (a + bi) and (a – bi): (a + bi)(a – bi) = a + b
[conjugate product rule – 1st no squared + 2nd no. squared.]
To divide two complex numbers, multiply both the numerator and
denominator by the conjugate of the denominator.
Example. Divide
Solution:
·
=
(3  7i)(5  2i)
52  22
Multiply, using FOIL in the numerator; use conjugate product rule in
denominator
=
15  6i  35i  14i 2
25  4
=
15  14  41i
29
Substitute –1 for i and collect like terms.
=
then separate the real and imaginary parts; both over the denominator
=
 (2) Equality of Complex Numbers
If we are told that 2 complex numbers are equal, we can
let the real parts equal to each other
and separately,
let the imaginary parts equal to each other, therefore creating two equations.
Example 1
If a  bi  c  di
Then we can say that a  c and b  d
Please note, that when equating complex numbers, we never use the i part of the
questions; we just use the numbers in front of the is (the coefficients).
Example 2:
If 5  i  x  (5  2 y )i , find the value of x and y.
Step 1: If necessary, write both sides of the equation in the form of a real number plus or
minus an imaginary number.
On the LHS of the equation we clearly have 5 as our real number and -1 as our imaginary
number.
On the RHS of the equation we have x as our real number as it is not being multiplied or
divided by any imaginary number.
The imaginary number on the RHS is everything inside the bracket, as it is all being
multiplied by the imaginary number. So, the imaginary part on the RHS is (5-2y)
So, we get
5  x and
1  5  2y
Step 2: Working on the second equation we get
1  5  2y
2y  5 1
2y = 6
6
y
2
y = 3 and x = 5
Example 3:
Find a and b if 3(a  bi)  4(ai  b)  6  3i
Solution
Multiply out the brackets.
3a  3bi  4ai  4b  6  3i
Tidy up the left hand side so that the real numbers are together and the imaginary
numbers are together.
{NB the real numbers are all the terms which do not have any i’s in them.}
3a  4b  3bi  4ai  6  3i
[3a  4b]  (3b  4a)i  6  3i put reals in bracket and factorise the i out of 2nd set of terms
3a  4b  6
3b  4a  3
We now have two equations which we need to solve simultaneously in order to find a and
b.
3a  4b  6
 4a  3b  3
Multiply the top equation by +4 and the bottom one by 3
12a  16b  24
 12a  9b  9
25b  15
15
b
25
3
b
5
Subbing this vale for b into any of the original equations we can find our value for a.
3a  4b  6
3
3a  4( )  6
5
3a  4(0.6)  6
3a  2.4  6
3a  6  2.4
3a  3.6
3 .6
a
3
a  1 .2
6
a
5
Now try the following sample questions on your own – finish for homework.