foundations of 4U maths book - S.K.Patel (1 Viewer)

*Pooja*

Member
Joined
Feb 29, 2004
Messages
244
Location
on the other side of the moon
hey ppl. does anyone have the solutions to exercise 4O in this textbook. its the revision exercise for complex numbers and some of them are way too hard for me.

the stupid topic happens to be tested in my half yrlies.

pm me if u have the answers pls. thanks.
 

nike33

Member
Joined
Feb 18, 2004
Messages
219
post the questions up (dont have the book) and ill give em a go (as im sure will other people :))
 

aud

Member
Joined
Jan 21, 2004
Messages
417
Location
Cherrybrook -> Forest Lodge
Gender
Male
HSC
2004
Originally posted by *Pooja*
hey ppl. does anyone have the solutions to exercise 4O in this textbook. its the revision exercise for complex numbers and some of them are way too hard for me.

the stupid topic happens to be tested in my half yrlies.

pm me if u have the answers pls. thanks.
I have Maths Extension 2 2nd Edition by Patel... think it's Foundations republished, cause 4O is still the revision for Complex... and all the answers are there on pages 357 - 358...

If the answers to
3. a. 3 - 2i
3. b. locus is either a point (3, -2) or the circle (x - 3) + (y + 2) = 1
4. a. x = 3/2, y = 2 or x = -3/2, y = -2
4. b. i. x + 2y + 2 = 0
4. b. ii. circle x + y + 2x + y = 0, C (-1, -1/2), r = rt5/2
make sense, let me know
 

Azn_Phoenix

Member
Joined
Feb 14, 2004
Messages
50
Gender
Male
HSC
2004
Hey does ne1 have worked solutions 4 Patel?? I really need them...plz
 

redslert

yes, my actual brain
Joined
Nov 25, 2002
Messages
2,373
Location
Behind You!!
Gender
Male
HSC
2003
yes i did the exercies in patel...
on paper, not about to scan it
not about to give it out

would be better if u had problems, post them here and people will get back to you
 

*Pooja*

Member
Joined
Feb 29, 2004
Messages
244
Location
on the other side of the moon
how do u do this one (its question 2 in this book):

the points z1, z2 and z3 are three complex numbers that lie on a circle passing through the origin. Prove that the points which represent 1/z1, 1/z2 and 1/z3 are collinear.
 

CM_Tutor

Moderator
Moderator
Joined
Mar 11, 2004
Messages
2,642
Gender
Male
HSC
N/A
The points are collinear if the line through 1 / z<sub>1</sub> and 1 / z<sub>2</sub> is parallel to the line through 1 / z<sub>1</sub> and 1 / z<sub>3</sub>. Thus, we need to show that arg[(1 / z<sub>1</sub>) - (1 / z<sub>2</sub>)] = arg[(1 / z<sub>1</sub>) - (1 / z<sub>3</sub>)]

Start by taking this statement, and showing it is the same as arg(z<sub>2</sub> - z<sub>1</sub>) - arg z<sub>2</sub> = arg(z<sub>3</sub> - z<sub>1</sub>) - arg z<sub>3</sub>.

Now, draw a diagram of a circle, passing through O and z<sub>1</sub> and z<sub>2</sub> and z<sub>3</sub>, and use it to prove (geometrically) that the above statement is true.

Edit: Note - the above proof requires the points O, z<sub>1</sub>, z<sub>2</sub> and z<sub>3</sub> to be placed in this order around the circle. This may be done without loss of generality. Alternately, if not placed in this order, then the proof needs to be extended to state that either arg[(1 / z<sub>1</sub>) - (1 / z<sub>2</sub>)] = arg[(1 / z<sub>1</sub>) - (1 / z<sub>3</sub>)] or
arg[(1 / z<sub>1</sub>) - (1 / z<sub>2</sub>)] = pi - arg[(1 / z<sub>1</sub>) - (1 / z<sub>3</sub>)]. Depending on the relative positions of O, z<sub>1</sub>, z<sub>2</sub> and z<sub>3</sub>, exactly one of these statements is true (provable geometrically), and either will result in 1 / z<sub>1</sub>, 1 / z<sub>2</sub> and 1 / z<sub>3</sub> being collinear.

Further note - can you construct a geometric proof that the line on which 1 / z<sub>1</sub>, 1 / z<sub>2</sub> and 1 / z<sub>3</sub> lie does not pass through the origin?
 
Last edited:

ngai

Member
Joined
Mar 24, 2004
Messages
223
Gender
Male
HSC
2004
ohh, that question..
consider a moving pt P(z) on the circle, and the fixed pt Q(w), where OQ is the diameter
then u have the points A(1/z) which is variable and B(1/w) which is fixed
using geometry and modulus stuffs, prove OPQ similar to OAB
then using angle in semicircle, angle OPQ = angle OBA = 90 for all z
so for all z on the circle, 1/z lies on the line through B perpendicular to OB
and so its true for z=z1, z2, z3
 

Pianpupodoel

Neurotic Member
Joined
Nov 7, 2004
Messages
73
Location
3. The answer's always 3.
Gender
Male
HSC
2006
Sorry to raise an *extremely* old topic from the dead, i just did a search thingy and found this.

About question two in 4O, what about this example -

Taking the unit circle, z1 = i, z2 = 1, z3 = -i

1/z1 = -i, 1/z2 = 1, 1/z3 = i.

It would seem quite obvious they do not lie on the same line. Hence the question is wrong?
 
I

icycloud

Guest
pianpupodoel said:
It would seem quite obvious they do not lie on the same line. Hence the question is wrong?
Your circle doesn't pass through the origin (as the question stipulates).

In fact it can be proved that if the circle z doesn't pass through the origin, then 1/z is the locus of another circle (there's another thread about that somewhere here).

So basically: if z is a circle which passes through the origin, then 1/z is a line.
if z is a circle which does not pass through the origin, then 1/z is a circle.
 

davidyin92

New Member
Joined
Oct 19, 2008
Messages
12
Gender
Male
HSC
2010
what is the a and b of the hyperbola x^2 and 4y^2 = 1
I can't seem to get the right answer..its in exercise 6B question 3
 

independantz

Member
Joined
Apr 4, 2007
Messages
409
Gender
Male
HSC
2008
davidyin92 said:
what is the a and b of the hyperbola x^2 and 4y^2 = 1
I can't seem to get the right answer..its in exercise 6B question 3
lol, you reopened a 2 year old thread 0_0

i'd think a=1, b=(1/2)

since (y/(1/2))^2=4y^2
 

davidyin92

New Member
Joined
Oct 19, 2008
Messages
12
Gender
Male
HSC
2010
lol, thanks for the help i got it now
I'll make sure not to revive a year old thread again haha :rolleyes:
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top