Transcript COMPLEX NUMBER ctd

F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• Algebra, from Arabic al-jebr meaning “reunion of
broken parts, is the branch of Maths concerned with
the study of the rules of operations and relations and
the constructions and concepts arising from them,
including terms, polynomials, equations, ...
• When letters and numbers are used together, the
mathematics is called generalised arithmetic or
algebra. Some of the rules used in algebra include the
commutative law (a+b = b+a), the associative law
((a.b).c = a.(b.c)), the distributive law (a.(b-c)=ab-ac), ...
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• Refers to distributing operational sign outside the
brackets to the terms inside the brackets
• Only applies to specific combinations of operations in
specific order
• Multiplication is distributive over addition or
subtraction, i.e.
• a(b+c) = ab + ac
• a(b – c) = ab – ac
• Division is distributive over addition or subtraction, i.e.
• 1/a (b + c) = b/a + c/a
• 1/a (b – c) = b/a – c/a
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• Example:
• 6 (4m – 45n) = 6x4m – 6x45n = 24m – 270n
• Work out the following and simplify where
possible
• ½ (46p – 64r)
• 32(21y + 15x)
• 10(0.7a + 2.3b)
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• In the expression 35, 5 is the exponent or the index of 3.
3 is then called the base. Both the base and the index
can be numbers or letters. Functions of the type 3x
where the index is a variable are called exponential
functions.
• X m x Xn = Xm + n i.e. when multiplying exponential
functions with the same base, add the indices
• Xm /Xn = Xm – n i.e. When dividing exponential
functions with the same base, subtract the indices
i.e. If an exponential function is raised
• (Xm)n = Xmn
to another power, multiply the indices
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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X0 = 1
(X  0)
X –m = 1/Xm (X  0)
X1/n = n  X
Xm/n = n  Xm or ( n X)m
• Without using a calculator, evaluate:
• (a) 1003/2;
• (b) 32 -2/5
• Simplify 27x/3 / 9x/2
• Find the value of 16 -3/4
• Show that 42-x = 16/22x
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• Mathematical statement separated by equal
sign, e.g. 17X = 51
• Can be extracted from given data, from
various functions (e.g. quadratic,
trigonometric, exponential or logarithmic)
• Solving an equation is to find a value of the
variable or unknown that makes the
statement true; e.g. 17X = 51  X = 51/17 or
X = 3.
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• 3x is an exponential function with base 3 and
variable x. Infinite set of values for x can result
in table of values that can be used to draw
function y = 3x in the Cartesian plan.
• But 3x = 81 is an exponential equation. Such
equations can be solved by expressing both
RHS and LHS in terms of the same base, i.e. 3x
= 34  x = 4.
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• 1. Solve 22x + 3 + 1 = 9 x 2x  22x x 23 + 1 = 9 x 2x
 (2x)2 x 8 + 1 = 9 x 2x; let p = 2x p2 x 8 + 1 = 9 x p
 8p2 + 1 = 9p  8p2 – 9p +1 = 0 (quadratic equation)
 8p2 – 8p – p + 1 = 0  8p(p – 1) – 1(p-1) = 0
 (8p – 1) (p – 1) = 0  p = 1/8 or p = 1  2x = 1/8 or
2x = 1 2x = 2-3 or 2x = 20  x = -3 or x = 0.
• 2. Solve the simultaneous equations 3x x 9y = 1 and 22x x
4y = 1/8 3x x 32y = 30 and 22x x 22y = 2 -3
 x + 2y = 0 and 2x + 2y = -3
 x = - 2y and 2(-2y) + 2y = - 3 or -2y = -3
 y = 3/2 or 1.5 and x = - 3.
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• Logarithmic function is the inverse of exponential
function (i.e. If y = ax, then x is the logarithm of y
to the base a and is written loga y = x; e.g. log10
100 = 2 because 10 2 = 100.
• Most common bases in logarithm are 10 resulting
in log y = x or lg y = x without indicating base and
e resulting in natural or naperian logarithm
denoted ln y = x. ex is an exponential function
with base e. Other bases can be used.
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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Loga(PQ)=loga(P)+loga(Q); ln(PQ) = ln(P)+ln (Q)
Loga(P/Q)=loga(P)–loga(Q); ln(P/Q)= ln(P)–ln(Q)
Loga (Pn) = n loga (P);
ln (Pn) = n ln (P)
Loga (1) = 0 (for a  0); ln (1) = 0
Loga (a) = 1; ln(e) = 1 and ln (ex) = eln(x) = x
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• Find log27. Let log27 = x  2x = 7 log (2x) = log 7
 x log 2 = log 7 x = log 7/log2 = 2.8
• €500 is invested at 10% p.a. compound interest.
After how many years will it have amounted to
€1500? C = P (1 + r/100)n is the formula to find
the capital C with r as %age compound interest
p.a. and P being the Principal and n the number
of years. Stop at 1 d.p.
• 1500 = 500 (1 + 10%)n  1500/500 = (110%)n 
log3 = n log (110%)  n = log3 log (110/100) 
n = log3 / (log110 – log 100) = 11.5 years (1 d.p)
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• This is a function whose highest power of the variable is 2;
e.g. ax2 + bx + c
• graph of a quadratic function , y = f(x) = ax2 + bx + c has a
characteristic shape called parabola.
• a˃0
max.
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X
• min.
a˂0
• Maximum and minimum values of y = f(x) = ax2 + bx + c are
given by f (- b/2a) and –b/2a is the x –coordinate of the
point where f (x) has its turning point and axis of symmetry.
• Find the minimum or maximum values of the following
functions and the values of x where this occurs: (a) x2 – 6x –
1; (b) x2 + 2x – 3; (c) (1 – x)(x + 2); (d) x2 + 2bx + c
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• A quadratic function that is or can be equated
to a constant is a quadratic equation and the
general form of a quadratic equation is ax2 +
bx + c = 0.
• To solve a quadratic equation is to find its
roots, usually 2, though not always.
• Quadratic equations can be solved by
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• Factorisation, i.e. equation with roots α and
reduces to (x – α)(x – = 0 and α +β = - b/a; αβ = c
• e.g. solve x2 – 4x – 5 = 0 by factorisation
• Completing the square, i.e. turning LHS into
perfect square. Ensure that coefficient of x2 is 1.
• e.g. solve 5x2 + 2x – 2 = 0 by completing the
square
• Quadratic formula, i.e.
for ax2 + bx + c = 0. In this formula,
is called Discriminant (D).
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• if D ˃ 0, the equation has two real and separate roots;
if D ˂ 0, the equation has no real roots; if D = 0, the
equation has two real and equal roots, i.e, it has one
real root.
a ˃ 0 D˃0
a˂0
X
a˃0 D = 0
X
a˃0 D˂0
X
a<0
a<0
• e.g. solve x2 – 4x – 5 = 0 using the quadratic formula.
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• Two quadratic equations, a linear and a quadratic
equation or two or more linear equations can be
solved simultaneously, meaning they have a/or
common solution(s). Subject/substitution is the
most common method used in such cases.
• e.g. the straight line y = 2x + 1 cuts the curve y =
x2 + 3x in points A and B. Find the coordinates of
A and B correct to 3 d.p.
• e.g. find the x- coordinate(s) of the point(s) of
intersection of the graphs of the following
functions: y = x2 – 2x – 3 and 2 y – 4 x2 + 6 = 8x.
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• ASSIGNMENT2 [To be completed]
• Find the values of k if the equation
x2 + (k – 2)x + 10 – k = 0 has equal roots
• Find the values of p if the line x + y = p is a tangent to
the circle x2 + y2 = 8
• A rectangular enclosure is made against a straight wall
using three lengths of fencing, two of length x metres.
The total length of fencing (excluding the wall) is 50m.
• (a) show that the area enclosed is given by 50x – 2x2
• (b) hence find the maximum possible area which can
be enclosed and the value of x for this area.
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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Series in Maths are best understood when related to Sequences
A Sequence is a set of things (usually numbers) that are in order.
It can be Infinite or Finite
Examples:
{1, 2, 3, 4 ,...} is a simple sequence (and it is an infinite sequence)
{1, 3, 5, 7} is the sequence of the first 4 odd numbers (and is a finite
sequence)
{1, 2, 4, 8, 16, 32, ...} is an infinite sequence where every term
doubles
{a, b, c, d, e} is the sequence of the first 5 letters alphabetically
{f, r, e, d} is the sequence of letters in the name "fred"
{0, 1, 0, 1, 0, 1, ...} is the sequence of alternating 0s and 1s (yes they
are in order, it is an alternating order in this case)
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• When we say the terms are "in order", we are free to define what
order that is! They could go forwards, backwards ... or they could
alternate ... or any type of order you want!
• A Sequence usually has a Rule, which is a way to find the value of
each term; e.g.
• the sequence {3, 5, 7, 9, ...} starts at 3 and jumps 2 every time.
• The rule can be generalised as a formula for nth term, where n could
be any term number we want.
• So instead of saying "starts at 3 and jumps 2 every time" we write
this: 2n+1
• Now we can calculate, for example, the 100th term: 2 × 100 + 1 =
201
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• To make it easier to use rules, we often use this special style: Tn is the term
n is the term number
• Example: to mention the "5th term" you just write: x5
• So a rule for {3, 5, 7, 9, ...} can be written as an equation like this:
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And to calculate the 10th term we can write: x10 = 2n+1 = 2×10+1 = 21
Can you calculate x50 doing this?
Example2: Calculate the first 4 terms of this sequence: {an} = { (-1/n)n }
Calculations:
a1 = (-1/1)1 = -1
a2 = (-1/2)2 = 1/4
a3 = (-1/3)3 = -1/27
a4 = (-1/4)4 = 1/256
Answer:
{an} = { -1, 1/4, -1/27, 1/256, ... }
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• Arithmetic Sequences also known as Arithmetic Progression (AP)
• In an Arithmetic sequence the difference between one term and
the next is a constant.
• Example: 1, 4, 7, 10, 13, 16, 19, 22, 25, ...
• This sequence has a difference of 3 between each number. Its Rule
is Tn = 3n-2
• In General you could write an arithmetic sequence like this: {a, a+d,
a+2d, a+3d, ... } where:
• a is the first term, and d is the difference between the terms (called
the "common difference")
• And you can make the rule by:
• (We use "n-1" because d is not used in the 1st term).
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• Example: (a)what is the 15th term of the AP -5,
-2, 1, ...? (b) which term is 28?
• Answer: in this AP, a= - 5 and d = 3 
• (a) T15 = a + (15 – 1)d = -5 + 14(3) = 37
• (b) Tn = a + (n-1)d  28 = -5 + (n – 1) (3) 28
+ 5 = 3n – 3 33 + 3 = 3n  n = 12, i.e. 28 is
the 12th term in the given AP.
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• Work out the following:
• (a) Find a formula in n terms of n for the nth term of
the AP 15, 9, 3, ... and hence find the 30th term
• (b) The nth term of an AP is given by Tn = 2n + 9. Find (i)
the first term; (ii) the common difference
• (c) If the 5th term of an AP is – 4 and the 10th term is
16, find the first term and the common difference
• (d)The first term of an AP is – 4 and the 15th is double
the 5th term. Find the 12th term.
• (e) the sum of three consecutive terms of an AP is 18
and their product is 120. find these terms
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• Arithmetic mean (AM) refers to the middle
term between two consecutive terms of an
AP; e.g. If a, b, c are three consecutive terms
of and AP, b is the arithmetic mean between a
and c.
• This is because b – a = c – b 2b = a + c 
• E.g. The AM between -7 and 3 = (- 7 + 3)/2 = -2
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• Geometric Sequences
• In a Geometric sequence, also known as Geometric
Progression (GP) each term is found by multiplying the
previous term by a constant. Example: 2, 4, 8, 16, 32, 64,
128, 256, ...
• This sequence has a factor of 2 between each number. Its
Rule is xn = 2n
• In General you could write a GP like this: {a, ar, ar2, ar3, ... }
where: a is the first term, and r is the factor between the
terms (called the "common ratio"). Note: r should not be 0.
• And the rule is:
(We use "n-1" because ar0 is the
1st term)
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• Example1 : state the common ratio of (a) 12, 4, 4/3, ...; (b) ½, 1/3, 2/9, ...; (c) 32/3, 16, 24,
...
• Answer:
• To find r, divide a term by the preceeding one,
i.e. (a) -4/12 = - 1/3; (b) 1/3
½ = 2/3;
(c) 16 32/3 =3/2.
• Example 2: find (a) the 12th term of the GP
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• Work out the following:
• Find (a) the 12th term of the GP 128, 64, 32, ... (b)
a formula for the nth term
• The 4th and 8th terms of a GP are 3 and 1/27
respectively. Find the possible values of a and r.
• A store finds that it is selling 10% less of an article
every week. In the first week it sold 500. in which
week will it be first selling less than 200?
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• Geometric mean (GM) refers to the middle term
between of three consecutive terms of a GP; i.e.
if a, b and c are consecutive terms of a GP, then b
is the geometric mean of a and c.
• This is because b/a = c/b b2 = ac or
• E.g. The GM of 2 and 32 is 8 because 2x32 =8
• Below are some other sequences, but our focus
will be more on AP and GP that we will use for
Arithmetic Series and Geometric Series.
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• Some other sequences include
– Triangular Numbers
• 1, 3, 6, 10, 15, 21, 28, 36, 45, ...
• The triangular number sequence is generated from a pattern of
dots which form a triangle.
• By adding another row of dots and counting all the dots we can
find the next number of the sequence:
• But it is easier to use this Rule:
• Example: the 5th Triangular Number is x5 = 5(5+1)/2 = 15, and the
sixth is x6 = 6(6+1)/2 = 21
– Square Numbers
• 1, 4, 9, 16, 25, 36, 49, 64, 81, ...
• The next number is made by squaring where it is in the pattern.
Rule is
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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– Cube Numbers
• 1, 8, 27, 64, 125, 216, 343, 512, 729, ...
• The next number is made by cubing where it is in the
pattern. Rule is
– Fibonacci Sequence
• This is the Fibonacci Sequence
• 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...
• The next number is found by adding the two numbers
before it together:
• The 2 is found by adding the two numbers before it (1+1)
• The 21 is found by adding the two numbers before it (8+13)
etc...
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• Rule is
• That rule is interesting because it depends on
the values of the previous two terms.
• Rules like that are called recursive formulas.
• Example: term "6" would be calculated like this:
• x6 = x6-1 + x6-2 = x5 + x4 = 5 + 3 = 8
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• A series is defined as the sum of the terms of
a sequence. Finite sequences and series have
defined first and last terms, whereas infinite
sequences and series continue indefinitely.
• Arithmetic series are given by adding up terms
of an AP; whilst geometric series result from
adding up terms of a GP
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• E.g. What is the sum of the first 20 natural numbers?
The first 20 natural numbers are an AP 1, 2, 3, ...,20.
Let the sum be
S20 = 1 + 2 + 3 + ... + 20; but also
S20 = 20 + 19 + 18 + ... + 1. Adding both gives
2 S20 = 21 + 21 + 21 + ... + 21 i.e. 20 (first term + last
term)  2S20 = 20x21 = 420 
S20 = ½ x20x(1st + last term).
• General formula for arithmetic series generated from
an AP with n terms and whose first term is a whilst last
term is L is S n = ½ n (a + L); but the first and last terms
are not always known, hence the general formula is
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• Example 1: the integers from 12 to 55 are added
up. What is their sum?
• Answer: there are (55 -12) + 1 = 44 integers. 
S44 = 44/2 (12 + 55) = 1,474
• Example 2: the sum of the first 8 terms of an AP is
12 and the sum of the first 16 is 56. find the AP.
• Answer: S8 = 8/2[2a + 7d] = 12 8a + 28d = 12;
S16 = 16/2(2a + 15d) = 56  16a + 120d = 56  a
= (12-28d)/8 16/8(12 – 28d) + 120d = 56  24
– 56d +120d = 56 64d = 32  d = ½ a = (12
– 28/2)/8 = - ¼  AP is – ¼, ¼ , ¾ , ...
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• Work out the following:
• A spot of light is made to travel across a screen in a
straight line and the total distance travelled is 115.5
cm. The distances travelled in successive seconds are in
AP. It travels 1.5 cm in the first second and 9.5 in the
last. How long did it take to cover the total distance?
• The sum of the first 10 terms of an AP is 80 and the
sum of the next 12 terms is 624. what is the AP?
• The sum of the first 10 terms of an AP is 7/2 times the
sum of the first 4 terms. (a) find the ratio of the 10th
term to the 4th term. (b) given that the 5th term is 2,
find the sum of the first 10 terms.
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• If Sn denotes the sum of the first n terms of a
GP, then Sn = a + ar + ar2 + ... + arn-1
• Multiplying throughout by r, rSn = ar + ar2 + ar3
+ ... + arn-1 + arn . Subtracting (ii) from (i) yields
Sn – rSn = a – arn or Sn (1 – r) = a (1 – rn ) which
gives Sn = a(1 – rn )/(1 – r).
• Hence formula :
for r  1
or
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• Example 1: Find the sum of the first 10 terms of 1/8, ¼,
½, ...
• Answer: a = 1/8; r = 2; n = 10 S10 = [1/8 (210 – 2)]/(2 –
1) = 1023/8.
• Example 2: what is the least number of terms of the GP
2, 3, 9/2, ... To be added to give a total greater than
30?
• Answer: a = 2; r = 3/2. Sn = [2(3/2)n - 1]/(2 – 1) =
4(1.5n - 1)and this must be ˃ 30. Hence (1.5n – 1) > 7.5
or 1.5n > 8.5 [using log, n = 5.3 the least value of n
(which must be an integer) satisfying this inequality is
6. So 6 terms must be added.
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• Work out the following:
• Find first 6 terms of 64, 16, 4, ...
• The 3rd and 6th terms of a GP are 108 and – 32
respectively. Find (a) the common ratio; (b)
the first term; (c) the sum of the first 6 terms
• The first term of a GP is 64 and the common
ratio is ½. How many terms must be added to
obtain a total of 127½?
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• The sum to infinity of the GP a, ar, ar2 , ... is
provided – 1 < r < 1.
• A real series is something of the form: a1 + a2 + a3 + ... with ai
ϵ R. It is often
Written as
ai
i=1
using the notation invented by the Swiss mathematician
Leonhard Euler (1707 – 1783)
where the
sign stands for "summation".
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• A complex number can be visually represented as a pair of numbers
(a,b) forming a vector on a diagram called an Argand diagram,
representing the complex plane. Re is the real axis, Im is the
imaginary axis, and i is the imaginary unit, satisfying i2 = −1.
• A complex number is a number that can be put in the form a + bi,
where a and b are real numbers and i is called the imaginary unit,
where i2 = −1. In this expression, a is called the real part and b the
imaginary part of the complex number. The complex number a + bi
can be identified with the point (a, b). A complex number whose
real part is zero is said to be purely imaginary, whereas a complex
number whose imaginary part is zero is a real number. In this way
the complex numbers contain the ordinary real numbers while
extending them in order to solve problems that cannot be solved
with only real numbers.
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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Im
b
a+bi
0
a
Re
The set of all complex numbers is denoted by ℂ.
The real number a of the complex number z = a + bi is called the real part of z, and the
real number b is often called the imaginary part. By this convention the imaginary part
is a real number – not including the imaginary unit: hence b, not bi, is the imaginary
part. The real part is denoted by Re(z) or ℜ(z), and the imaginary part b is denoted by
Im(z) or ℑ(z). For example,
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• Anyway, this new number was called "i", standing for "imaginary",
because "everybody knew" that i wasn't "real". (That's why you
couldn't take the square root of a negative number before: you only
had "real" numbers; that is, numbers without the "i" in them.) The
imaginary is defined to be: i = (- 1). Then:
• When dealing with imaginaries, YOU MUST ALWAYS DO THE i-PART
FIRST!
• Example 1: Simplify (–9).
• Answer:
• Example 2: Simplify (–25).
• Answer:
• Work out the following: (a) Simplify (–18); (b) Simplify (–6).
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• In your computations, you will deal with i just as you would with x,
except for the fact that x2 is just x2, but i2 is –1:
• Simplify 2i + 3i.
• 2i + 3i = (2 + 3)i = 5i
• Simplify 16i – 5i.
• 16i – 5i = (16 – 5)i = 11i
• Multiply and simplify (3i)(4i).
• (3i)(4i) = (3·4)(i·i) = (12)(i2) = (12)(–1) = –12
• Multiply and simplify (i)(2i)(–3i).
• (i)(2i)(–3i) = (2 · –3)(i · i · i) = (–6)(i2 · i)
• =(–6)(–1 · i) = (–6)(–i) = 6i
• Note this last problem. Within it, you can see that , because i2 = –1.
Continuing, we get:
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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This pattern of powers, signs, 1's, and i's is a cycle:
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
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• In other words, to calculate any high power of i, you can convert it
to a lower power by taking the closest multiple of 4 that's no bigger
than the exponent and subtracting this multiple from the exponent.
For example, a common trick question on tests is something along
the lines of "Simplify i99", the idea being that you'll try to multiply i
ninety-nine times and you'll run out of time, and the teachers will
get a good giggle at your expense in the faculty lounge. Here's how
the shortcut works:
• i99 = i96+3 = i(4×24)+3 = i3 = –i
• That is, i99 = i3, because you can just lop off the i96. (Ninety-six is a
multiple of four, so i96 is just 1, which you can ignore.) In other
words, you can divide the exponent by 4 (using long division),
discard the answer, and use only the remainder. This will give you
the part of the exponent that you care about. Here are a few more
examples:
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•
•
•
•
•
•
•
•
Examples:
Simplify i17.
i17 = i16 + 1 = i4 · 4 + 1 = i1 = i
Simplify i120.
i120 = i4 · 30 = i4· 30 + 0 = i0 = 1
Simplify i64,002.
i64,002 = i64,000 + 2 = i4 · 16,000 + 2 = i2 = –1
It's time to move on to complex numbers. The standard
format for complex numbers is "a + bi"; i.e., real-part
first and i-part last.
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•
•
•
•
•
•
•
•
•
•
•
Complex numbers are “binomials" of a sort, and are added, subtracted, and
multiplied in a similar way. (Division, which is further down the page, is a bit
different.) First, though, you'll probably be asked to demonstrate that you
understand the definition of complex numbers.
Solve 3 – 4i = x + yi
Finding the answer to this involves nothing more than knowing that two complex
numbers can be equal only if their real and imaginary parts are equal. In other
words, 3 = x and –4 = y.
To simplify complex-valued expressions, you combine "like" terms and apply the
various other methods you learned for working with polynomials.
Simplify (2 + 3i) + (1 – 6i).
(2 + 3i) + (1 – 6i) = (2 + 1) + (3i – 6i) = 3 + (–3i) = 3 – 3i
Simplify (5 – 2i) – (–4 – i).
(5 – 2i) – (–4 – i)
= (5 – 2i) – 1(–4 – i) = 5 – 2i – 1(–4) – 1(–i)
= 5 – 2i + 4 + i = (5 + 4) + (–2i + i)
= (9) + (–1i) = 9 – i
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• You may find it helpful to insert the "1" in front of the second set of
parentheses (highlighted in red above) so you can better keep track
of the "minus" being multiplied through the parentheses.
• Simplify (2 – i)(3 + 4i).
• (2 – i)(3 + 4i) = (2)(3) + (2)(4i) + (–i)(3) + (–i)(4i)
• = 6 + 8i – 3i – 4i2 = 6 + 5i – 4(–1)
• = 6 + 5i + 4 = 10 + 5i
• For the last example above, FOILing works for this kind of
multiplication, if you learned that method. But whatever method
you use, remember that multiplying and adding with complexes
works just like multiplying and adding polynomials, except that,
while x2 is just x2, i2 is –1. You can use the exact same techniques for
simplifying complex-number expressions as you do for polynomial
expressions, but you can simplify even further with complexes
because i2 reduces to the number –1.
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• The complex conjugate of the complex number z
= x + yi is denoted * or z and defined to be x − yi.
Geometrically, z is the "reflection" of z about the
real axis. In particular, conjugating twice gives the
original complex number: z = z.
• The real and imaginary parts of a complex
number can be extracted using the conjugate:
Re(z) = ½ (z + z) and Im(z) = ½i (z – z )
• Conjugation distributes over the standard
arithmetic operations:
• z + w = z + w; zw = z . w; z/w = z/w
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• The reciprocal of a nonzero complex number z
= x + yi is given by
• 1/z = z/zz = z/(x2 + y2)
• Adding and multiplying complexes isn't too
bad. It's when you work with fractions (that is,
with division) that things turn ugly. The issue
with complex numbers is that imaginaries
should not be left in the denominator. So how
do you handle this?
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• Suppose you have the following exercise:
• Simplify 3/2i
• This is pretty "simple", but they want me to get rid of
that i underneath, in the denominator. The 2 in the
denominator is fine, but the i has got to go. To do this, I
will use the fact that i2 = –1. If I multiply the fraction,
top and bottom, by i, then the i underneath will vanish
in a puff of negativity:
• So the answer is
• This was simple enough, but what if they give you
something more complicated?
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•
•
Simplify
If I multiply this fraction, top and bottom, by i, I'll get:
•
Since I still have an i underneath, this didn't help much. So how do I handle this
simplification? I use something called "conjugates". The conjugate of a complex
number a + bi is the same number, but with the opposite sign in the middle: a – bi.
When you multiply conjugates, you are, in effect, multiplying to create something
in the pattern of a difference of squares:
•
Note that the i's disappeared, and the final result was a sum of squares. This is
what the conjugate is for, and here's how it is used:
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So the answer is
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• In the last step, note how the fraction was
split into two pieces. This is because,
technically speaking, a complex number is in
two parts, the real part and the i part. They
aren't supposed to "share" the denominator.
To be sure your answer is completely correct,
split the complex-valued fraction into its two
separate terms.
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• Remember that the Quadratic formula solves "ax2 + bx + c = 0" for the
values of x. Also remember that this means that you are trying to find the
x-intercepts of the graph. When the Formula gives you a negative inside
the square root, you can now simplify that zero by using complex
numbers. The answer you come up with is a valid "zero" or "root" or
"solution" for "ax2 + bx + c = 0", because, if you plug it back into the
quadratic, you'll get zero after you simplify. But you cannot graph a
complex number on the x,y-plane. So this "solution to the equation" is not
an x-intercept. You can make this connection between the Quadratic
Formula, complex numbers, and graphing:
• x2 – 2x – 3
x2 – 6x + 9
x2 + 3x + 3
•
•
•
a positive number inside
square square root
two real solutions
two distinct x-intercepts
zero inside the square
root
one (repeated) real solution
one (repeated) x-intercept
F. Ndererehe, BA Educational Sciences UNR, MA Educational Sciences - UNR
a negative number inside the
root
two complex solutions
no x-intercepts
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• As an aside, you can graph complexes, but not in the x,yplane. You need the "complex" plane. For the complex plane,
the x-axis is where you plot the real part, and the y-axis is
where you graph the imaginary part. For instance, you would
plot the complex number 3 – 2i like this:
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• This leads to an interesting fact: When you learned about regular
("real") numbers, you also learned about their order (this is what
you show on the number line). But x,y-points don't come in any
particular order. For instance, you can't say that (4, 5) "comes after"
(4, 3) in the way that you can say that 5 comes after 3. All you can
do is compare "size", and, for complex numbers, "size" means "how
far from the origin". To do this, you use the Distance formula, and
compare which complexes are closer to or further from the origin.
This "size" concept is called "the modulus". For instance, looking at
our complex number plotted above, its modulus is computed by
using the Distance Formula:
• Note that all points at this distance from the origin have the same
modulus. All the points on the circle with radius (13) are viewed as
being complex numbers having the same "size" as 3 – 2i.
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•
•
•
•
•
Work out the following:
Find c1 + c2 if c1 = 5 + 3i and c2 = 3 + 5i
Simplify i69
Simplify the following (a) 8/4i; (b) 6/(3 +2i)
Which of the following complex numbers are
at a distance of 5 units from the origin?
(a) 3 – 4i; (b) 4 + 3i; (c) 1 + 5i; (d) 0 + 5i; (e) 5 + 0i
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