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

[nike33]

argz + arg(z-1)
@ + B

where arg(z-1) = B
draw a diagram, note iso triangle noticeB = (@ + pi) / 2

wich gives the req answre

edit: didnt see u got it

 

[ND]

Hahah sorry it was cos ryan asked me to prove arg(z-1)=arg(z^2) yesturday, but it was only for a certain case, and i forgot that it was only for that certain case:

http://www.boredofstudies.org/community/showthread.php?s=&threadid=30524&perpage=15&pagenumber=3

That was stupid...

 

[shkspeare]

eheh its ok !

mmm one LAST little question (hopefully >.<") on complex locus [i can see a few threads goin on about complex locus here.. haha]

ummm the complex number z is given by z = t + 1/t, where t = rcis@

Find the equation of the locus of the point P which represents z on an argand diagram in the case :

@ = pi/4 and r varies

thanks

 

[ND]

Hmmm, ok:

That comes out to be z=rcis(pi/4) +1/(rcis(pi/4))
It's pretty obvious that this translates to parametric coords of x=p+1/p, y=p-1/p. I think you can take it from here.

 

[:: ck ::]

is it a alobrepyh ralugnatcer ?

[read backwards]

 

[CM_Tutor]

Originally posted by ND
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?

I got PM'd asking me to make a comment on this thread, so here I go...

arg 0 is definitely undefined, as it has no reasonable geometric interpretation - arg z is the direction of the vector from the origin to z, and if z = 0, this vector is a directionless point. Thus, arg 0 has no meaning.

So, if we are looking at a locus like arg(z + 1) - arg(z - 1) = pi / 2, it will be a semicircle with OPEN circles at both the points z = 1 and z = -1.

Does this address the issues that people were wondering about? If not, please post any further questions, and I'll have a look at them sometime over the weekend.

 

[:: ck ::]

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 this geometrically, u'd put open circles around 1,0 and -1.0

but doing this algebraicly, as affinity said

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.

so doing this question geometrically gives u the wrong answer?!

 

[CM_Tutor]

OK, now I understand what the concern is about. Sorry, I've been a bit slow in realising the point at issue.

If z lies on the imag axis, and z(w + 1) = w - 1, then the locus of w is the unit circle with only the intersection with the real axis at -1 missing, whether you do the problem geometricaly or algebraically.

The 'error' arising from the geometric approach is that you have transformed the problem from z(w + 1) = w - 1 to arg (w - 1) - arg(w + 1) = +/- pi / 2, which is a valid transformation for all cases except z = 0, as arg 0 is undefined. The z = 0 case, which leads to w = 1, must be taken separately, and its result added into the locus.

So, the geometric approach should be:

Case 1: z = ki, k <> 0

z = (w - 1) / (w + 1) is purely imag, so arg z = pi / 2 or - pi / 2, so ... working ...

So, the locus of w is the unit circle with both real intercepts missing.

Case 2: z = 0

0 * (w + 1) = w - 1
Hence, w = 1

So, combining the results from the cases, the locus of w is the unit circle with the single point at -1 missing.

When doing a question inolving a transformation, you should always look back to where you started. It is clear that z(w + 1) = w - 1 cannot have w = -1, but any other value of w is possible with an appropriate z. It is like solving an equation like 2log₁₀x = log₁₀(x + 6). You transform it to x² - x - 6 = 0, and solve to get x = -2 or 3, but you still go back to where you started to realise that only x = 3 is valid. Here, the transformation gives two invalid points, but the starting equation shows one of them is, in fact valid, but covered by another case.

 

[:: ck ::]

ohHh thx a lot cm ur a legend

sorry about being unclear....

wow... so all those old dusty old textbooks appear to be "half wrong" in a way

ive tried to approach questions using geometric methods whenever possible... as it is at least 1/3 or even 1/4 of the working in cases where a geometric approach is approachable

now i kno wot i have been missing alll this time... thanks!

 

[study-freak]

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

I thought it was a unit circle without -1??

Proof: let w=2+it
then z=w/conjugate(z)
=w^2/abs(z)^2
then let w=rcis$
=r^2cis2$/r^2
=cis2$
=cis(2(inverse tan t/2)

therefore, abs(z)=1
arg(z)=2(inverse tan t/2)

but this shows that arg(z)cannot=pi
since tan(pi/2) cannot equal to t/2 as it is undefined.

the locus of z is thus a unit circle with centre at the origin but without the point z=-1.(since arg(z) is not equal to pi)