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A practical tutorial on precalculus for high school students who have completed Algebra 1.;useful for self-learning and also as review for starting calculus course...many advanc d concepts are introduced.in a simple manner; The tutorial has several simple examples and exercise problems to practise...Several applications are given in each topic, including exercise using graphing uitlity for polar equations.
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Adventures in Precalculus - 1
Dr N K Srinivasan
What is Precalculus?
If you scan through a 'Precalculus' book, you will note that it
is a course with a jumble of topics--more like a bridge course
from Algebra 2 to regular Calculus.
But a careful study will show you something: It teaches you
several mathematical functions and how to manipulate them and use
them for 'modeling' real-world problems.
Well, you have to learn many mathematical functions because these
functions are the 'ingredients' for using Calculus.
While learning these functions, you will use a bag of clever
tricks,call them 'tools of the trade" if you like.
This article teaches you some of the clever tricks you will
use---so that 'precalculus' becomes easy and interesting to study
and develops your 'calculus muscles' to handle tougher math
problems later.!
Note: Get a graphing utility and use as required.
1 Completing the square:
Take a simple example: Solve x2 +6x -16 =0
You can, of course, use the quadratic
formula ,if you remember it !
Is there a simple way to solve?
Recall : (x+a)2 = (x+a)(x+a) = x2 +
2ax + a2
Keep this equation in mind. We will proceed from right to left
while "completing the square".
Take the example and rewrite this equation as follows:
x2 + 6x = 16
Take the coefficient of x, which is 6
in this example; divide this by 2 and
square the number: that is 6/2=3 and 3
x3 =9
Add this number both sides: x2 + 6x +9 = 16
+9
You can factor the left side using the relation given earlier:
(x + 3)2 = 25
Take the square root on both sides:
x+3 = +/- 5
So the roots are: x+3 =5 or x=2
and x+3 = -5 or x=-
8
This method is far easier than using quadratic
formula.
Try the following problems before you proceed:
1 solve x2 + 4x -32 =0 [Ans ; x=
4, x=-8
2 Solve x2 - 8x -9=0 [Ans:
x=9;x=-1]
3 Solve 3x2 +6x -24 =0 [Ans x=2
;x=-4]
Example 2
This method--"completing the square has
other uses too. We will see one
application:
Solve x2 + y2 +2x +6y -6 =0
This problem appears 'tricky'...well --
this is a 'standard' equation for a
circle.
Recall that the equation for a circle
with centre C(0,0) at the origin is :
x2 + y2 = r2
where r is the radius.
The equation changes when the center is shifted to C'(h,k) position.
The equation becomes:
(x-h)2 + (y-k)2 = r2
Now the given equation can be factored by "completing the
square" for both x and y . Rewrite the expression:
(x2 + 2x + ) + ( y2+6y + ) = -6
Add 1 to the first bracket and 9 to the second bracket and the same numbers on the right side too.
(x2 + 2x +1) + (y2 + 6y + 9) = -6 +1+9
Factoring for x and y we get:
(x+1)2 + (y+3)2 = 4
The problem is solved : this is a circle with centre at C(-1,-3)
and radius 2.
Problem to solve:
1 find the center and radius of the circle: x2 + y2 -4x +10y-7=0
[Ans
center C(2.-5) radius 6]
2 Find the center and radius of this circle: x2 +x + y2 -6y +1/4 =0
[Ans
center (-1/2,3), radius 3]
Example 3 Find the vertex of the parabola whose
equation is:
x2 -6x-y +10 =0
Let us rewrite this first: (x2 -6x )
+10 = y
The coefficient of x is 6; so let
us put 9 inside the bracket and
subtract 9 outside:
( x2 -6x +9) -9
+10 =y
(x -3)2 = y -
1
This is a parabola, with vertex at
[3,1]
Try this:
1 Find the vertex of the parabola :
2x2 -8x -y + 7 =0
We can extend this method to find the equation of an ellipse from the
equation in standard form:Read this from your text book.
2 Function of a function
Suppose you have a function: y= f(x) = 2x +3
You have another function : z = g(x) = x2 +1
you can generate another function f
[ g(x)]
Write this as: f[ g(x)] = f [x2 +1]
Then write the function f(x) replacing
x by x2 + 1.
We get f[g(x)] = 2(x2 +1) + 3
You also get f[f(x)] by writing ;
f [ 2x+3] = 2 (2x+3) + 3 = 4x + 9
Try writing g[g (x)]
Example: Given f(x) = 1/x + 3. g(x)= x -1. write f[g(x)] and
g[f(x)]
Are they same? no.
Complex formulae can be created by using this method:
Example 1: The electrical resistance of a wire increases with
temperature as follows:
R = a + bT + c T2
Now the temperature of an electrical heater varies with time t as follows:
T = d + 4 t
Find the rate of heating by using the relation : Heat produced H
= I2 R as a function of time: R =
f(T) = a + bT + c T.T
T (t) = d + 4t
So f [ T (t)] = f [ d+4t] = a + b
[d+4t] + c [ d+4t][d+4t]
Now you see the usefulness of function
of function.
Here is another example:
example 2: The atmospheric air pressure
decreases almost linearly [like a
straight line] as we ascend to about
10000 feet.The pressure P = P0 - 2 (h)
where h is height in feet.
A baloon ascends at the rate of 50 feet per minute. h = 50 t
where t is time in minutes from the ground level.
Write the equation for change in pressure with respect to time
for a traveller going up in the balloon.
3 Inverse functions
Take the function y = f(x) = 2x + 3
You can find the inverse function by switching x and y:
Replace x by y abd y by x: x= 2y + 3
Now solve for Y : 2y = x - 3
y= (x-3)/2
Note that this function is an inverse function of f(x) . Let us
write this as f-1 (x)
Note that the inverse function is not the reciprocal of
f(x); it is not 1/f(x).
The inverse function is the reflection to the original
function across the line y=x. Y=x is the diagonal line
that passes through the origin.
Draw the lines y= 2x +3 and the line y=(x-3)/2 in a
graph paper. You will see the two lines as mirror
images of each other.
Example 2 Find the inverse function for the parabola: y
= 4 x.x
Switching x and y ; x= 4 y.y
or y= sqrt(x)/2
Plot again ,using your graphing utility, the two
curves: y= 4xx and y= sqrt(x)/2 .Do they form
reflection across the diagonal line y=x? Check this.
What if we find f[ f-1 (x)] ?
Take the example of f(x)= 2x+3 , and f-1
(x) = (x-3)/2
f[ (x-3)/2] = 2[(x-3)/2] +
3 = x-3 +3 =x
Are you surprised with the result?
We took x, transformed it into
f(x) ,then took its inverse and again
revert it back by f(x) ,getting x
again.
So, you can always check your result of
inverse function by forming f [ f-1
(x)]----if you get x, you are right.
Exponential function and logarithmic function are inverses of each
other.
y =f(x)= e x
Switch y and x: x = ey
Solve for y: ln x = y= f-1 (x)
So the two functions are inverses. Plot the two functions using
your graphing utility; you will find that they are curves reflected
over y=x.
Try these problems:
1 If f(x) = sqrt(2x+3), find its inverse function.
2 If f(x) = 2ex
find its inverse function.
Application
We can use inverse functions to find the conversion formula for
temperature in Centigrade scale to Fahrenheit and the reverse
one:
If x is the temp in centigrade and y the temp in Fahrenheit, we
have the relation as follows:
y = 32 + (9/5) x
Let us find the inverse function; switch x and y in the above
equation:
x = 32 + (9/5) y
Solve for y: y = (x-32) (5/9)
This formula gives the conversion fromula when x is in Fraherheit
and y the centigrade.!
Example: If 212 F is the boiling point of water, find the
temperature in centigrade scale:
deg C = (212-32) (5/9) = (180 x 5 )/9 = 100 deg C
So, the boiling point of water is 212 deg F or 100 deg C.
3 Shifting and scaling
Many functions can be shifted around and scaled up or down
[multiplied or divided] by suitable manipulation.
Consider a parabola with vertex at the origin V(0,0):
y = x2
Suppose you want to shift the vertex to V(2,0) , change x to x-2.
y = (x-2)2
You want to shift the vertex to V(2,1), change y to y-1.
y-1 = (x-2)2
Consider a circle with center at the origin C(0,0) and radius 2 units>
The equation for the cirlce is : x2 + y2 = 4
Suppose you want to shift the center of the circle to P(-2,1) with
the same radius,
(x+2)2 + (y-1)2 = 4
Another transformation you do is to enlarge or shrink by multiplying or dividng x or y.
y = k x.x is a parabola that rises very fast along the y axis--if k is greater than 1.
Take y = 2 x2
Compare this parabola with y = x2
We call this process "scaling"
Again consider the graph of y = k .
You can choose k=1, k=2 k=1/2 Draw the graphs and study the
shapes you get.
1 Write the equation for a ellipse with centre at (-2,2)
2 Write the equation for a cirlce with center at (2,-2) and radius
3 units.
4 parametric equations
When we relate x to y through the function , y = f(x) , we have a
single graph of y versus x.
Often it is convenient to use another variable which is
intermediate and common to both:
Let us call this intermediate variable t; then x = f(t) and y=
g(t)
We use two functions now, both functions of t.
Consider a sphere whose surface area is related to radius: A = f
(r) = 4 .pi.r.r
the volume is related to radius: V = g(r)= 4.pi.r.r.r/3
Here the radius is the common parameter.
We can take the ratio of surface area to volume: A/V = f(r)/g(r)=
3/r
So writing parametric equations help us to see new relations.Thus
if you want to study cooling of a sphere [like our Earth] we want
to study the ratio of surface area to volume.As r increases, the
ratio of surface area to volume decreases, so the cooling rate
could get lower. If your house-top is like a hemispherical dome,
it would cool slower if the radius of the dome increases. Did the
cathedral builders think of this factor...may be!
5 Solving exponential equations
Let us start with a simple equation to solve for x using
exponents:
Solve : 2x = 8
Let us write the right side number '8'
in terms of exponent of base 2.
8=2 .2.2 = 23
Equate the powers or indices both sides, since they have
the same base:
x =3 is the answer.
Try the following:
1 Solve 3x = 81 [Ans: x=4]
2 Solve 4x = 256 [Ans : x=4]
3 Solve 5x = 125 [Ans : x=
3]
4 solve 3 (2x+1) = 243 [Ans: x=2]
---------------------------------------
--------
Method of substitution
Solve : e2x + 5 ex + 6 = 0
This equation suggests that it can be
converted into a simple algebraic
equation without exponential function
by 'substituting" exponential function
by a variable u.
Let u = ex
Then the equation becomes:
u2 - 5u + 6 = 0
Solving we get (u -3)(u-2) =0
So, u = 3 and u = 2
Therefore: u = ex = 3 ex = 2
The answer is x = ln3 or ln2
Try the following:
Solve: e2x -2 ex -15 =0
[ans: x= ln 5 or ln -3
Discard the second one!There is no log
function for a negative number]
Taking log on both sides:
Many equations can be solved by
converting to log function on BOTH
SIDES:
1 Solve e2x = 9
Take natural log on both sides: ln (e2x
) = ln 9
2x =ln 9
x = (1/2) ln9
2 Solve e3k =2 [Ans k
= ln2/3]
Exponential Growth
WE use exponential function to "model" growth of
populations. "Population" is a general word; it may mean the
human population or population of rats, population of birds,
population of houses or population of cell phones.
To write an equation for growth of population, we start with an
initial time t=0 when the population is "n".Then the population
at a later time t is given by N(t):
N(t) = n e kt
where k is a constant, called 'growth constant" and t is the time
measured from the start time.
Let us take an example. Joe is a realtor who wants to model the
growth of houses in the new neighborhood called Chico town. Chico
had a population of 1200 homes in the year 2000. So n=1200.
Joe finds that by the year 2005, that is ,five years later, the
number of houses has grown to 3000.
So N(t) is 3000 where t is five years.
Now we can write the growth equation: 3000 = 1200 e k.5
Joe has to calculate the growth constant first. What is
the value of 'k'?
The trick is to take log on both sides which you have
learned in the previous section....Taking the natural
logarithm 'ln':
ln 3000 = ln 1200 + 5k
5k = ln 3000 - ln 1200
= ln (3000/1200) = ln 2.5
Using the calculator: ln 2.5 =0.916
k = 0.184
So the growth equation becomes: N(t) = 1200 e0.18 t
Now Joe can "predict" the future population by plugging t , provided
the growth follows the same trend;that is there is no major
catastrophe [like earth quake] or economic depression which will
change the trend.
Let us predict the population for the year 2015. t=15
N(t) = 1200 e 0.18 .15
= 1200 e2.7
= 1200 x 14.9
= 17855 or roughly
18000
So, Joe can expect about 18000 homes in Chico town by the
year 2015.
You see that exponential growth models can be used to
predict future growth, once you find the growth constant.
[ All models are constructed to help you to predict the
future.]
Doubling Time
It is easy to understand the rate of growth if we calculate the time it takes
for the population to double itself.This time , we denote by t', would depend
on the growth constant.Higher the growth constant k ,shorter the doubling
time.
To find this time let us reverse the procedure: take N(t) = 2n
Find t for a given k : 2n = n ekt'
Dividing by n, we get 2 = ekt'
Taking log on both sides, we get ln 2 = kt'
or doubling time t' = ln 2/k
Larger the k value or growth rate, shorter the doubling time!
Now ln 2 =0.69 or take it as 0.7
Doubling time t' = 0.7/k
Let us find the doubling time for Joe's model: k = 0.18
Doubling time t' = 0.7/0.18 = 3.88 or nearly 4 years.
The population of homes in Chico town was 3000 in the year 2005. Four
years later, that is in 2009, it could be 6000 homes! in the year 2013 ,it
could be 12000; in the year 2017, it could be 24000.
Try this problem:
1 The number of toucon birds in Amazon forest in an area of 1000 acres
was 150 in the year 2004. When the bird counting was done in 2007, it
was 210.
1 What is the growth constant?
2 Find the doubling time.
3 Predict the toucon population in years 2010 and 2013.
[Hint: 210 = 150 e3k k=0.114 doubling time t'=
6.2 years ]
2 India's human population was 300 million in 1945. It rose to 1000 million
or 1 billion in the year 1990. Find the growth rate in percent and also
predict the likely population for the year 2015. Also find the doubling time.
{Ans k = 0.019 or 1.9% doubling time t'= 37 years.
Note: The population growth rate has decreased in India; so the growth
constant now: k = 0.013 or 1.3% .
3 If the bacteria population in a soup triples every three hours, how long it
will take for initial population of 20 bacterias to grow to 1000.
[Hint tripling time t" = ln 3/k Find k since t" = 3 hours.ln3=1.1 k=0.37 t=10.5
hours]
Exponential Decay
There are processes in which population may decrease---say fish
population in lake or deer population in a forest [because food is limited or
there are predators in the environment---things conservationists and
ecologists worry about.] We can model this too by writing an exponential
function with (-k) t:
N(t) = n e-kt
'k' is called a "decay constant". Find k as before;
"Half life" is the time it takes for the population to reduce to half its
initial value. As for doubling time, Half life = ln 2/k = 0.7/k
Radioactive decay of an isotope is a classic example of exponential decay.
Radium , discovered by Marie Curie and her husband Curie in Paris, has a
half life of nearly 1600 years.
So, the decay constant for radium is 0.7/1600 = 0.00043. Not bad, the half
life is a long one.
1 One of the isotopes of the trans-uranic element Californium Cu250
has a half life of 13 years. Find its decay constant . Calculate how long it
will take for 500 grams of this isotope to decay to 75 grams:
k = 0.7/13 = 0.054 N(t) = n e -0.054 t
If N(t) = 75, n= 500, we get 0.054 t = ln 500/75 = ln 6.66 = 1.89
t= 1.89/0.054 = 35 years
Try this:
1 In a reserve forest, the deer population was 250 in the year
1950. It decreased to 100 by the year 2000. Find the decay
constant and the half-life for this population.When it will
decrease to 50 deers, that is without conservation efforts.
[Ans k = (1/50)ln 0.4 =0.008 N(t) = 250 e-0.008t
Half life = 0.7/.008 = 87.5 years The population
of deer will decrease to 50 by the year
2087.]
2 A radioactive isotope used for
radiation therapy in a clinic has a
half life of 110 days. If 1 gram of
this isotope was bought, how much will
be left after nearly 360 days [1
year.].
Can you make a quick guess without
calculation:
Half life is 110 days. So after 110
days, you will have 0.5 grams;after 220
days, you will have .25 grams ; after
330 days 0.125 grams...So after 360
days, you will have less than .125
grams of the radioactive isotope.
Logarithmic Function
Logarithmic functions are inverse functions of exponential. So
the two are closely related.
Suppose you write y = 2x .
The same function can be written as : log2 y = x
When we write log y , we mean that the
base is 10.
When we write ln y , we mean that the
base is 'e'.
Keep this in mind.
There are only two rules that you have
to keep in mind for all manipulations
of log functions.
Rule 1 log (xy) = log x + log y
For example : log (2x) = log 2 +
log x
log ( x z) = log
x + log z.
We get the second rule from the
first:
log ( x2 )= log ( x.x)
= log x + log x
= 2 log x
Now log ( x3 ) = log
(x.x.x) = log x + log x+log x
= 3 log x
Generalizing this, we can write the
second rule:
Rule 2: log (xn ) = n log x
This is a very useful rule in manipulation of log equations.
We get rule 3 from the last two rules: log (x/y) = log (xy-1)
log (x/y) = log x + log (Y-1 )
Applying the rule 2, we get Rule 3:
Rule 3 : log (x/y) = log x - log y
Let us keep these three rules in mind>
WE can expand or condense some equations and log functions using
the three rules:
1 Expand log (2 x3 )
Do this in small steps: log ( 2 x3 ) =
log 2 + log (x3 )
=
log 2 + 3 log x
2 Expand log ( 3 y2 /4) Ans: log 3 +
2 log y - log 4
3 Expand log ( x y/z2 ) Ans: log x +
log y - 2log z
Application:
1 Volume of a cylinder v= (pi)r2 h
where r is the radius and h, the
height.
Expand log v [ans: log
v = log pi +2log r+log h]
2 The electrical resistance of a wire R
= R' l / (pi r2 )
Expand R
Condenzing a log expression is the
reverse process and easy to do.
Example: Condense: log3 + 2log r -3
log h [Ans: log(3r2 /h3 )
1 Condense: log (.5) + log m + 2 log
v [ans: log (0.5mv.v)
2 Condense: log 4 -log3 +log pi + 3log
r
3 Condense: 2 ln x - (1/2) ln (1+x)
Solving log equations
1 Solve for x log 10 3 = x
3 log 10 = x
x = 3
2 Solve log3 81 = x 3x = 81
x=4
3 Solve logx 256 = 4 x4 =256= 4 x
4 x 4 x 4 x=4
Changing the base This is an important method you must keep in mind.
Take log to the base 5 of x : log5 x = ln x /ln 5 = log x
/log 5
This convenient "change of base" is useful in solving many
problems with 'difficult' bases.
1 Solve 5x =13 log5 13 = x
Change the base from 5 to e: ln 13/ln
5 = x
Use the calculator to find ln 13 and ln
5 ans x =1.59
2 Solve log 3x = 0.25
10 0.25 = 3x x= 0.593
3 Solve 2log x -log 72 + log 2 =0
2 log x = log 72 - log 2
Condense both sides: log (x2 ) = log
(72/2)
= log 36
So, X = +6 OR - 6.
4 Solve 3 log (x) - log (27) = 0
ans x=3
5 Solve ln 2x + ln 5 = 3
[Hint ln 10x = 3; 10x=e3]
6 Solve ln 3 - 2ln x = ln 12
ans x = +/- (1/2)
7 Solve log8 x = 4/3
8 Condense: 3 log(x+2y) - log y
+(1/2) ln ( x+3z)
---------------------------------------
---------
Complex Numbers
How do you find the square root of a negative number say, -4.
Let us split this up: -4 = 4 . -1
The square root of 4 is +2 or -2. Let us take the +2 root.
To find the square root of -1 , sqrt(-1) , we invent a new
quantity called 'i' ,an imaginary number. Clever trick.
√ -1 = i
Now we shall build up some properties of this imaginary quantity
'i'.
Firstly sqrt(-4) = sqrt(4.-1) = (2).sqrt(-1) = 2i
This means that i.i or i2 = -1
Keep this in mind: "i squared is equal
to -1".
Now let us find i3 = i.i.i = -1.i = -i
What is i4 = i.i.i.i = (-1)(-1) = 1
which is a real number.
What is i5 = i again. and so on.
1 Find the value of i9
i9 = i4 . i4 .i = (1)(1)i = i
2 Find the value of i15 [Ans -
i]
You may think that this whole thing or
scheme is crazy. But let us extend
further by inventing a "complex
number".
Mathematicians invent many new abstract
things like this for
solving "problems".
A complex number z is like this : z = a
+ bi
where a and b are real numbers.
For example z = 3 + 4i is a complex
number. Here 3 is the real part and 4i
is the imaginary part.
Now z1 = 3i --> this is also a
complex number with real part zero;
only the imaginary part is given.
I use z and z1, z2,z3.....to write
complex numbers.
Arithmetic Operations
Let us invent the usual operations of addition, subtraction,
multiplication and division for complex numbers...This is like
learning new rules in a new country.
Addition
This is simple. Add the real parts and add the imaginary parts
separately and form the added complex number:
z1 = 3 + 4i z2= 2 - 2i
z1 + z2 = (3+2) + (4-2)i= 5 + 2i
Try the following :
1 z1 = 4i and z2= 2-3i z1+z2 = 2 +i
2 z1 = -2 + 3i z2= 4 + 3i z1+z2 =
3 z1= 1-3i z2= 2-2i z1+z2 =
4 z1= 2-3i z2 = 1-i z1 + z2=
Subtraction
Subtract the real parts and separately subtract the imaginary
parts and form the new complex number.
1 z1= 4+ 3i z2= 2 + i z1 - z2 = (4- 2) + (3-1)i = 2 +2i
2 z1= 3+2i z2= -i z1 - z2 =
3 z1= 1 - 2i z2 = 1 -i z1- z2 = -i
Multiplication
This operation gets complicated.
First we invent a new number called "complex conjugate" of a
given complex number.
If z = a + bi , its complex conjugate z' = a - bi
For example, z= 3+4i, its complex conjugate is z' = 3 -4i
Again if z= 2-3i , its complex conjugate is z' = 2 + 3i
1 Write the complex conjugates of the following complex numbers:
z = 1+i z'=
z= -1-i z'=
z= 2 -3i z' =
You may wonder why we invented the complex conjugate for any
complex number. You will see shortly.
Suppose z1 = 3+4i and z2= 2 +i
z1.z2 = (3+4i)(2+i)
Using FOIL method, you can work out the multiplication:
z1.z2 = (6 + 3i + 8i + 4i.i)
z1.z2 = 6 + 11i + 4(-1) = 6 -4 + 11i = 2 + 11i
Try the following:
1 z1= 2-3i z2= 1-3i z1.z2
=
[Ans: -7 -9i]
2 z1= 4i z2= 2+3i z1.z2=
3 z1= 1+2i z2= 2-4i z1.z2 =
Now you know how to multiply tow complex numbers, using
FOIL method, let us try this:
multiply z and its complex conjugate z';
z= 2+3i z'= 2- 3i
z.z' = (2+3i) (2-3i) = ( 4 + 6i -6i-9i.i) = 4 +9 = 13
The answer is simple: if z = a+bi , then z.z' = a2 +b2
The result is a real number. This result z.z'= |z|
2, .This is the square of the magnitude
of the complex number.
[Does this sound like Pythagorian
theorem applied to complex
numbers...yes...the real part is the x
value, imaginary part is the vertical
or y value .Then the complex number is
the point a+bi. The hypotenuse is the
value sqrt(a.a + b.b).]
Example 1 Find the magnitude of the
complex number:
z= 3+4i z' = 3-4i
z.z' =3.3 + 4.4 = 25
mag of z = |z| = sqrt(25) = 5
Example 2 > Find the magnitude of the
complex number:z= 2-i
z=2-i z'= 2+i z.z' = 2.2 +
1.1= 5
|z| = sqrt(5).
Try the following:
1 find |z| , given z= 2+2i [Ans |
z| = sqrt(8)
2 Find |z| for z= 3-4i {ans
|z| = 5)
3 Find |z| for z= 2-5i [ ans
|z| = sqrt(29)]
Now we are ready to learn division of
complex numbers.
Division
Suppose z1= 2+i and z2= 3+4i
then z1 / z2 = (2+i)/ (3+4i)
Take the complex conjugate of the denominator: z2' = 3-4i
Multiply both numerator and denominator by z2'
z1/z2 = (2+i)(3-4i) /(3+4i)(3-4i)
The denominator now becomes: (3+4i)(3-4i)= 3.3+4.4= 25
Numerator: (2+i)(3-4i) = 6 +3i -8i +4 = 10-5i
Divide the numerator by 25: z1/z2 = (10/25) - (5/25)i = 2/5 -
(1/5) i
Example 1 Find z1/z2 if z1= 3-i and
z2= 2-3i
z1/ z2 = (3-i)(2+3i)/(2-3i) (2+3i)
= (3-i)(2+3i)/(4+9)
= (1/13) [ 6 -2i+9i+3]
= (1/13) [9 - 7i]
= (9/13) - (7/13)i
Try the following:
1 z1= 2+i z2= 3-2i z1/z2= ?
2 z= 4i z1= 1+2i z/z1=?
3 z= 2+3i z1 = 3i z/z1=?
Now we have learned to do addition,
subtraction, multiplication and
division of complex numbers.
One major thing you have to learn :
exponents of complex numbers.
Exponents of complex numbers
1 z= 2+i find z2 = z.z = (2+i)(2+i)= 4 +2i +
2i +i.i = 3+3i
2 Find z3 and z4. [ans:
(3+3i)(2+i)=
3 If z3 = z2 .z3 , then z3 = z5
4 Find z = Z5/z2 = z3
Application
1 One important application is in solving quadratic equations:
Find the root of x.x +3x +3 =0
Applying quadratic formula: x= [-3 +/- sqrt(3.3 - 4.3)]/2
The discriminant b.b-4ac in quadratic formula for ax2 +bx+c =0 is
negative.
x = -(3/2) +/- sqrt(-3)/2
For square root of -3, write sqrt(3)i.
x = -3/2 + sqrt(3)/2 i
and = -3/2 - sqrt(3)/2 i
The result :we get two complex number roots!
Try the following:
1 solve x.x + x +1 =0
2 Solve x.x + 2x + 3 =0
2 Second application: There are several applications in electricity and
magnetism and in other topics in physics and engineering.
For an alternating current [AC], the current passing through a
coil is found to be:
I = E /Z where I is the current, E the voltage and
Z, the impedance.
If E = 8(cos20 + i sin 20), z= R + i Xl = 6+ 3i,
find I.
I = 8(cos 20+ i sin 20) (6-3i)/ (6+3i)(6-
3i)
= (8/45) [6.cos20 + 6sin20 i -3 cos20i
+3 sin20]
From trig tables we get: cos 20 = 0.94 sin
20 = 0.34
I = (8/45) [ (6x0.94 +3 x 0.34) + i (6
x0.34 - 3 x 0.94)]
= (8/45) (6.66 - 0.78i)
= 1.198 - 0.1404i
Advanced topic in complex numbers
Here are a few topics that you can skip during first level of
learning this subject -precalc.
Euler found that eix = cos x + i sin x , a complex
quantity.
It follows that if x= pi , eipi = cos pi + i sin pi
cos pi = -1 and sin pi =0
So we get e i pi = -1
or e pi.i +1 =0
This is one of the famous equations of Euler combining pi and e,
both transcendental numbers.
-------------------------------------------------
Polar Coordinates
You are used to representing points and lines in two-dimensional
graph paper using x and y axes....the horizontal axis is the x
axis and the vertical axix is the y axis. The two axes are
prependicular to each other.So we call it 'Rectangular
coordinate " system or "Cartesian Coordinate" system after the
French mathematician Rene Descartes.[1596-1650]
Can we use a different system which uses only one length and an
angle....
Take a point P with coordinates x and y.
Join P with the origin O (0,0) . You get a line OP. Call this
distance r.
Find the angle OP makes with x axis. Call it θ (theta).
Then you can see that the point can be represented by these
two "coordinates" --namely r and theta, instead of x and y.
These two are called "Polar Coordinates".
It is easy to switch from rectangular coordinates x and y to
polar coordinates, r and theta.
Using trig functions: x = r cos (theta)
y = r sin (theta)
How to find r and theta ,given x and y?
r = sqrt ( x2 + y2)
theta = y/x
Take the point with P (3,4)
Convert to Polar coordinates: r=
sqrt( 9 + 16) = 5
tan ( θ )
= 4/3
theta =
53.12 deg
1 Convert to polar coordinates: P(1, -
2)
r = sqrt( 1+4) =
θ = arctan (-2/1) = arctan (-2)
2 Convert to polar coordinates: Q(1,3)
r=
theta=
3 Convert from polar to rectangular:
r=2 theta= 30
[Ans: x = r cos(theta) y= r
sin(theta)]
4 Convert from polar to rectangular:
r=3 theta= 45
x= y=
Complex Plane
Now you know about complex numbers and also polar coordinates, we
can work on "complex planes".
Take a complex number z= 3+ 4i
We can represent this number in regular x-y coordinate system,
provided you make a slight change. We will call the vertical axis
or y axis as "imaginary axis" (Im ) and then 4i is the same as y
coordinate in the x-y plot.
This plane with x axis and Im
axis of iy is called complex plane [aka
Argand plane].
So a complex number z = 3+2i is
represented by a point P(3,2) in the x-
iy plane as shown in the figure. What
is "r" now abd the angle theta in the
figure. You can draw this figure first
and represent r and theta.
The distance from the origin to P, OP =
r = sqrt( x2 +y@ )
tan (θ ) =
y/x
So, you can rewrite z= x+iy into 'polar
coordinates"
z = x + iy = r cos (θ) + i sin
(θ)
or z = r { cos (θ) + i sin
(θ) }
We can use a short form for this: z =
r cis(θ)
Example 1
1 Convert the complex number z= 3+4i ,
into polar form, using trig functions:
r = sqrt(9+16) = 5
theta = tan -1 (4/3) =
53.12 deg
So, z= 3+ 4i = 5 cis (53.12) = 5
(cos 53.12 + i sin 53.12)
Try this:
1 Convert to polar coordinates in the
complex plane:
z=2+i {Ans:
z= √3 [cis 0.01]
2 z= 3 - 2i
{ ans: z =√13 [cis 33.66]}
Euler Relation
Leonhard Euler [1707-1783]made
synthesis: he related exponential
function to trig functions and thus
made a revolution in math in his
period.This relation is so useful that
many things in precalc and calculus
become easy and straight-forward.
This relation is just this:
exp ( i θ ) = cos (theta) + i sin (theta) =
cis (theta)
With this relation, we can write a complex number z in
another way:
z = r
exp(i θ)
Thus z= 3+4i = 5 cis(theta)= 5 ei 53
We can do lot of manipulations with Euler relation:
1 Find z2 for z= 3+4i
z = 5 exp( 53o i) z.z = 25 exp
(106i)
2 Find z3 for z= 3+4i
z.z.z = (5.5.5) [ exp( 159i)]
So, we can generalize: zn = rn [exp
(i.n. theta)
[This equation is "de Moivre's theorem
found in all precalc text books.]
You will find that complex numbers are
easy to work with in polar coordinates
and with Euler relation, using
exponential functions!
Complex division:
If z1= r1 exp(i π/2) and z2= r2
exp(iπ/4)
then z1/z2 = (r1/r2) exp [i (π/2 -
π/4)]
Example 1 If z= -1 +√3i find z3
First convert z into polar form: r= 2 θ =120 = 2π/3
z= 2. exp(i (2π/3)
Using De Moivre theorem,
z.z.z = 8 exp(i 2π )= 8 (cos 2pi + i sin 2pi)= 8
Example 2
Find the cube root of z= 8 cis (60 deg)= 8( cos 60deg + i
sin 60)
z(1/3) = 2 [cos( 20 deg) + i sin
(20 deg)]
Try the following:
1 Find the product of two complex
numbers:
z1=5 cis(90 deg) z2= 3 cis(45 deg)
Ans z1.z2=
15cis(135)=15[(-0.707)+0.707i]
2 Find 3cis(305deg) / 9 cis(65deg)
Ans : (1/3) cis (305-65)deg)= (1/3)
[cos 240 + sin 240]
---------------------------------------
----------
Polar Equations
You are familiar with equations in x and y, the rectangular
coordinate system.
We can write equations using polar coordinates ,that is, r and
theta. This becomes easy to generate several nice graphic figures
as we will see now.
Let us begin with a straight line equation: y = mx + c
or y - mx -c =0
Let us susbstitute for x and Y : x = rcos (theta) y= r sin
(theta)
r sin (θ) - m r cos (θ) = c
Rewriting : r = c/ [(sin (θ) - m cos
(θ)]
This equation in the form r = f(theta): r ,a function of theta
is a "polar equation".
You can try this with a graphing utility and draw a straight
line.
How would you write a polar equation for a circle?
The answer is simple: r=constant
r=3 is a circle with 3 units as radius and theta changes from o
to 360 degrees.
A Spiral to draw:
Consider this Polar equation in which the radius increases with
theta , as theta changes from o to 360 degrees:
r = k θ where k is a constant ; Let k= 1/10
Let us write r and theta for a few values of theta:
theta [degree] 0 10 30 90 120 160
180 240 360
r [cm] 0 1 3 9 12
16 18 24 36
Can you visualize the figure you will get? The answer is " a
spiral" with expanding radius.
This is also called "Archimedes Spiral. "
Now keep your graphing utility ready to generate these figures
given by the following "Polar Equations":
1 r = 1+ cos (theta) [Cardiod--a heart
shaped figure]
2 r = 3 cos (2theta) [ a "rose" with 4
petals]
3 r = 3 cos (3 theta) [ a flower with
three petals]
4 r2 = cos (2theta) [ a
'lemniscate" - a figure of eight type]
{ Challenge: try writing these
equations in rectangular form: it will
be tedious!!
Rewrite r = 2sin (theta) in rectangular
form:
Multiply both sides by r: r.r = 2 r
sin(theta)
x.x + y.y =
2 y
y.y - 2y +x.x
=0
Completing the square for y : y.y - 2y
+1 +x.x =1
(y-1)2 +
x.x =1
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-------------------------
---------------------------------------
----------------------
Hope you enjoyed the adventures in
Precalc with simple examples and only a
few exercise problems.You can dig up
more applications from the text book
and increase your adventure sessions.
You can always return to this
'adventure tutorial" for a quick
review.
If this tutorial has helped you learn
the basic things easily [besides your
teacher and text books] or supplemented
your class work, I feel amply rewarded
for my effort .
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