tough complex locus q's ! >.< (1 Viewer)

[shkspeare]

hi im finding trouble with the following q's

1. Two points - P,Q represent the complex numbers z, 2z+3+i respectively. If P moves on the circle |z|=k, how does Q move?

2. Find locus of z if |z+3i|^2 + |z-3i|^2 = 90 [is there a quick way?]

3. If z(w+1) = w-1, show that as Z describes the y axis, W describes a circle with the origin as centre, and that, as Z describes the x axis, W describes the x axis also

4. Two complex numbers z, Z are related by Z = (2+z)/(2-z). Show that as the point z describes the y axis from the negative end to the positive end, the point Z describes completely the circle x^2+y^2 = 1 in the counter clockwise sense

5. Given t is a real variable, find the locus of the point z on the argand diagram such that z = (2+it)/(2-it)

Finally..

6. If the point z moves in a semicircle, centre origin and radius 2, in an anticlowise direction from the point 2 to point -2, findi the path traced by the point 1/z

thx a lot... i just got these from a really old book i got today..

 

[martin310015]

for question 2. i just let z=x+iy..............than just expanding and the locus was a circle with centre (0,-3) and radius 3sqrt5...hope its rite.

ok this is what i think question 6 is ......
if we let z=x+iy then for the point (2,0)we get z=2
so 1/z would trace out a semi circle with radius 1/2......

 

[ND]

1) For every pt of P, Q is just that times 2 plus a constant, therefore it's a circle with radius twice that of z (i.e. 2k), and just shifted.
2) The expansion shouldn't be more than 5 lines.
3) First get w as the subject, multiply top and bottom by the conjugate of the bottom, then just take cases when z is real and z is imaginary.
4) Same as above.
5) What happens when you divide a complex number by it's conjugate (which is what you'll be doing for every value of t).
6) Consider 1/|z| and arg(1/z).

 

[martin310015]

guess i was wrong for question 6....

 

[Affinity]

q1.) Q-> circle, centered at 3+i, radius 2k.
To help you see this consider the locus of R = w = 2z
then since Q = R + 3+i.. it's just r translated.

q3.) as Z describe the y axis, z=ki for some real k.

ik(w+1) = (w-1)

1+ik = (1-ik)w

w= (1+ik)/(1-ik)

w= (1-k^2 + 2ik) /(1+k^2)

|w| = 1 (check this)

if Z describes the x axis Z=k

kw + k = w-1
(1-k)w = 1+k

w= (1+k)/(1-k)

w can be all reals except -1.


q5.)
it's the unit circle with i, and -i taken away. check the RHS of the expression to mod-arg form


Zzzzz.. I better go, will post the rest tomorrow if not already posted then

 

[ND]

Originally posted by martin310015
guess i was wrong for question 6....

Nope you're right for 6, but you should mention that 1/z goes clockwise as z goes anticlockwise, and so the semicircle is on the bottom.

But i think you're wrong for 2, looking at it (mind you, i haven't written anything on paper for it so i may be wrong), it should have a radius 6.

edit: and be centred at the origin.

 

[ND]

Originally posted by Affinity
Zzzzz.. I better go, will post the rest tomorrow if not already posted then

I think he wanted hints not full solutions. I'm being careful now cos i got a comment from Grey Council cos i was showing full solutions when he wanted just hints.

 

[martin310015]

ok thanx ill go check it tommo

 

[:: ck ::]

3. If z(w+1) = w-1, show that as Z describes the y axis, W describes a circle with the origin as centre, and that, as Z describes the x axis, W describes the x axis also

q3.) as Z describe the y axis, z=ki for some real k.

ik(w+1) = (w-1)

1+ik = (1-ik)w

w= (1+ik)/(1-ik)

w= (1-k^2 + 2ik) /(1+k^2)

|w| = 1 (check this)

we we be allowed to d the q geometrically?
z = (w-1)(w+1)
given that z describes the y axis, arg(z) = +-pi/2
arg(w-1)(w+1) = +-pi/2

therefore the locus of w describes a unit circle, not including the pt (-1,0) (1,0) [im not sure abt 1,0 ...]

back on track : when z describes the x axis, arg(z) = 0, pi
therefore arg(w-1 / w+1 )= 0 is two lines going out on the x axis from 1,0 and -1,0 ... and when arg(w-1 / w+1) = pi... it is the interval from -1,0 [and 1,0 ? - im not sure abt] ... not including these pts as well

OHH and one thing : i found that in most of the newer books, when they give u a q like arg(w-1)(w+1) = +-pi/2... they place an open circle around both (1,0) and (-1,0) .. however my older books (coroneos, terry lee) ONLY place it on (-1,0) ...

what should we put as open and not open? ... hrmmmz >.<"

4. Two complex numbers z, Z are related by Z = (2+z)/(2-z). Show that as the point z describes the y axis from the negative end to the positive end, the point Z describes completely the circle x^2+y^2 = 1 in the counter clockwise sense

similarly for 4, making z the subject, z = (2Z-2)/(1+Z)
if z describes the y axis, arg(z) = +-pi/2

arg(2Z-2)/(1+Z) = 2arg(Z-1 / Z+1) = +-pi/2

can u say thats a unit circle?

 

[Grey Council]

Originally posted by ND
I think he wanted hints not full solutions. I'm being careful now cos i got a comment from Grey Council cos i was showing full solutions when he wanted just hints.

lol, sorry. Its just that, i dunno about others, but if i work through the question myself, I remember how I did it. But if I get the solution, I'm left thinking, hokay, how smart is ND/Affinity. But come exam time, YOUR smartness won't help me.

Hehe, I usually like hints/guidance for a question. Dunno, i s'pose the poster should say if they want solutions or hints.

Sorry, ND, didn't mean to criticise you, was just saying. lol, i get kinda annoyed if i see solutions.

OHH and one thing : i found that in most of the newer books, when they give u a q like arg(w-1)(w+1) = +-pi/2... they place an open circle around both (1,0) and (-1,0) .. however my older books (coroneos, terry lee) ONLY place it on (-1,0) ...

what should we put as open and not open? ... hrmmmz >.<"

also wanting to know the answer to this.

 

[ND]

Originally posted by :: ryan.cck ::
OHH and one thing : i found that in most of the newer books, when they give u a q like arg(w-1)(w+1) = +-pi/2... they place an open circle around both (1,0) and (-1,0) .. however my older books (coroneos, terry lee) ONLY place it on (-1,0) ...

what should we put as open and not open? ... hrmmmz >.<"[/i

Arg(w-1)/(w+1)=+-pi/2
arg(w-1)-arg(w+1)=+-pi/2
it's pretty obvious that when w=1,-1, the args are 0 and pi so you don't include either points.

lol, sorry. Its just that, i dunno about others, but if i work through the question myself, I remember how I did it. But if I get the solution, I'm left thinking, hokay, how smart is ND/Affinity. But come exam time, YOUR smartness won't help me.

Hehe, I usually like hints/guidance for a question. Dunno, i s'pose the poster should say if they want solutions or hints.

Sorry, ND, didn't mean to criticise you, was just saying. lol, i get kinda annoyed if i see solutions.

Hahah yeh it's cool. It's just that i don't think i'm a very good hint giver.. (like most of the time i try and give you a hint you don't get it, then CM or OLDMAN or someone gives you a hint and you can do it =/ )

 

[:: ck ::]

Originally posted by ND
Arg(w-1)/(w+1)=+-pi/2
arg(w-1)-arg(w+1)=+-pi/2
it's pretty obvious that when w=1,-1, the args are 0 and pi so you don't include either points.

blah but when u look at q3

3. If z(w+1) = w-1, show that as Z describes the y axis, W describes a circle with the origin as centre, and that, as Z describes the x axis, W describes the x axis also

doing it geometrically [the way i did it] ... you wouldnt include the points 1,0 and -1,0 [as u argued...]

yet if u did it algebraicly, as affinity did

w= (1+k)/(1-k)

w can be all reals except -1

you would only place it around -1,0 ...

blahhhhhh~!!

 

[ND]

Originally posted by :: ryan.cck ::


"w= (1+k)/(1-k)

w can be all reals except -1"




you would only place it around -1,0 ...

[/B]

w=(1+k)/(1-k) is not a circle, it's a real number...

 

[:: ck ::]

sorry i wasnt being clear.. haha

that was the 2nd half of q3....

. If z(w+1) = w-1, show that as Z describes the y axis, W describes a circle with the origin as centre, and that, as Z describes the x axis, W describes the x axis also

the questoin of where to put the open circles around is still there.. as arg(z) = arg (w+1 / w-1) = 0, pi ...

affinity's method only places -1,0 as an open circle ... [where the locus doesnt lie]

my method [maybe its wrong?] geometrically... would say both 1,0 and -1,0 is an open circle...?

 

[shkspeare]

thx guys!!!

one small q

if arg(z) = @ , |z| = 1
express arg(z^2-z) in terms of @

im lost.. plz help !!

 

[ND]

Originally posted by :: ryan.cck ::
the questoin of where to put the open circles around is still there.. as arg(z) = arg (w+1 / w-1) = 0, pi ...

affinity's method only places -1,0 as an open circle ... [where the locus doesnt lie]

my method [maybe its wrong?] geometrically... would say both 1,0 and -1,0 is an open circle...?

I'm actually doubting whether arg(0) has a value. But that would mean that neither 1, nor -1 exists on w... But 1 can clearly exist when done geometrically.... meh i dunno, maybe CM or OLDMAN or Affinity or someone can explain?

 

[ND]

Originally posted by shkspeare
thx guys!!!

one small q

if arg(z) = @ , |z| = 1
express arg(z^2-z) in terms of @

im lost.. plz help !!

<[ z^2-z=z(z-1)
arg(z(z-1))=arg(z)+arg(z-1)
but arg(z-1)=arg(z^2) (draw it if you wanna see why)
.'. arg(z^2-z)=3@]>

edit: oops i posted the full solution... i'll try to hide it.

edit2: shit that didn't work. hmm oh well... can anyone explain how to hide stuff?

 

[shkspeare]

ahhh i got the answer...

arg(z^2-z) = arg(z) + arg(z-1) = @ + (pi+@)/2 = (3@+pi)/2

i think u mixed up the part where arg(z-1) = arg(z^2) ND.. nevertheless your solution helped me think about it more

arg(z-1) actually equals to pi/2 + @/2 ... coz argz = @, arg (z-1) = arg(z+1) + pi/2 (ext angle fo triangle, diamater cubtends a rightangle with point on circumf)... but arg(z+1) = 1/2 arg(z) [angle at centre is dbl angle at circumf] thus arg(z+1) = @/2

thx a lot guys!

 

[ND]

There's nothing wrong with my solution, so i can only assume that their answer is wrong. (either that or you've made a typo in writing up the question)

 

[shkspeare]

read my edited post

mm welll i checked it with the back of the book.. its (3@+pi)/2

unless... u can point out which part i did wrong >.<