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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

Adventures in Precalculus

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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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Page 1: Adventures in Precalculus

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.

Page 2: Adventures in Precalculus

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

Page 3: Adventures in Precalculus

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.

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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

Page 5: Adventures in Precalculus

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: 

Page 6: Adventures in Precalculus

       (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

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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

Page 8: Adventures in Precalculus

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: 

Page 9: Adventures in Precalculus

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

Page 10: Adventures in Precalculus

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

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              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.

Page 12: Adventures in Precalculus

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.

Page 13: Adventures in Precalculus

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

Page 14: Adventures in Precalculus

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)

Page 15: Adventures in Precalculus

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)

Page 16: Adventures in Precalculus

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

Page 17: Adventures in Precalculus

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.

Page 18: Adventures in Precalculus

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,

Page 19: Adventures in Precalculus

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]

Page 20: Adventures in Precalculus

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.

Page 21: Adventures in Precalculus

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

Page 22: Adventures in Precalculus

                                    

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

Page 23: Adventures in Precalculus

                                      

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.

Page 24: Adventures in Precalculus

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

Page 25: Adventures in Precalculus

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.]

Page 26: Adventures in Precalculus

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.

Page 27: Adventures in Precalculus

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.

Page 28: Adventures in Precalculus

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

Page 29: Adventures in Precalculus

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     

Page 30: Adventures in Precalculus

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

Page 31: Adventures in Precalculus

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.

Page 32: Adventures in Precalculus

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

Page 33: Adventures in Precalculus

        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:

Page 34: Adventures in Precalculus

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:

Page 35: Adventures in Precalculus

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 )

Page 36: Adventures in Precalculus

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

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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

Page 38: Adventures in Precalculus

             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]

Page 39: Adventures in Precalculus

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

Page 40: Adventures in Precalculus

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

Page 41: Adventures in Precalculus

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.

Page 42: Adventures in Precalculus

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: 

Page 43: Adventures in Precalculus

         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. 

Page 44: Adventures in Precalculus

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

                                          

Page 45: Adventures in Precalculus

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

Page 46: Adventures in Precalculus

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

Page 47: Adventures in Precalculus

     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)

Page 48: Adventures in Precalculus

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

Page 49: Adventures in Precalculus

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=? 

Page 50: Adventures in Precalculus

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   

Page 51: Adventures in Precalculus

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

Page 52: Adventures in Precalculus

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)]

Page 53: Adventures in Precalculus

         = (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  

Page 54: Adventures in Precalculus

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.

Page 55: Adventures in Precalculus

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

Page 56: Adventures in Precalculus

                              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= 

Page 57: Adventures in Precalculus

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.

Page 58: Adventures in Precalculus

      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@ )

Page 59: Adventures in Precalculus

                            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

Page 60: Adventures in Precalculus

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]

Page 61: Adventures in Precalculus

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)

Page 62: Adventures in Precalculus

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)]

Page 63: Adventures in Precalculus

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)]       

Page 64: Adventures in Precalculus

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:

Page 65: Adventures in Precalculus

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.

Page 66: Adventures in Precalculus

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:

Page 67: Adventures in Precalculus

              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]

Page 68: Adventures in Precalculus

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

Page 69: Adventures in Precalculus

Completing the square for y :  y.y - 2y

+1 +x.x =1

                                (y-1)2 +

x.x =1

--------------------------------------------------------------------------

-------------------------

                         

---------------------------------------

----------------------

Page 70: Adventures in Precalculus

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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