complex No. Q (1 Viewer)

[Sirius Black]

arg(z-2)=-arg(z-3+2i)
Sketch the locus of z


hmm, i did some algebrac calculation on this question, and initially let z=x+iy
and then got sth like: 2xy-2x-y+4=0<---oops,just found out this is wrong...
What does it tell us about the locus of z?
here is my working:

arg(z-2)=-arg(z-3+2i)
-arg(x+iy-2)=arg(x-3+(y+2)i)
-tan^-1(y/(x-2))=tan^-1(y+2)/(x-3)
take tan( ) of both sides
-y/(x-2)=(y+2)/(x-3)
-xy+3y=xy+2x-2y-4
2xy+2x-5y-4=0
and @ Slide Rule: this is not an eqn for a straight line right?

 

[m_isk]

arg(z-2)=-arg(z-3+2i)
Sketch the locus of z

are you sure there is a negative sign on the RHS??

 

[Sirius Black]

are you sure there is a negative sign on the RHS??

yes. that's the problem

 

[LaCe]

Is there an angle in there too?

1.plot the points of the arguments i.e. (2,0) & (3,-2)
2. LHS must = RHS
3. I dunno where to go from there

 

[queenie]

i got a really weird answer; something along the lines of:

x^2 - y^2 - 5x - 2y + 6 + i(2x + 5y + 2xy - 4) = 0

what is the answer - has any1 worked it out?

i dnt know if its right, but:

when multiplying complez no's u multiply the mods and add the arguments so :

arg of :

(z-2)(z-3+2i) = arg (z-2) + arg (z-3+2i)
which is:

arg(z-2) = -arg(z-3+2i)

thats not right is it? crap ive forgotten everything, should it be:
arg(z-2+(z-3+2i)) ??

 

[Slidey]

Ignoring magnitude and position, the vector represented by arg(z-2) is a reflection of the vector represented by arg(z-2+3i), about the x-axis.

The vector represented by PA+PB lies on the locus. The vector PA+PB is one of the diagonals of a paralleogram (parallelogram law), and we know the other diagonal goes from A to B (it coincidentally represents the subtraction of the two vectors).
Let:
P(s,t)
A(2,0)
B(3,-2)

Vector AP has its head on the locus at (s-2,t)
Vector BP has its head on the locus as (s-3,t+2)
.'. (t-t-2)/(s-2-s+3)=-2 is the gradient.
C(5/2,-1) (intersection of diagonals) also lies on the locus, so:
y+1=-2(x-5/2)
y+1=2x+5
y=2x+4

 

[Sirius Black]

The vector represented by PA+PB lies on the locus.

er, would u plz explain this bit?
thanx

 

[Slidey]

er, would u plz explain this bit?
thanx

Er. Yes. Eheh. Well. You'll have to actually put some effort in and draw the thing, but...

You know the parallelogram rule of vector addition and subtraction, right?

Given 3 points, A, B, C, then, all the following vectors: AB + BC = AC

Here, we have A, B, P. Now, PA is a reflection of PB, right? But recall that a vector is a vector regardless of where it is located (generally). So you can translate the vector PB such that the head of PA (head at A) is the tail of PB. The head of the translated PB is the head of your new vector, PA+PB, which is the diagonal of the parallelogram formed with 2 lots of vectors PA and PB.

You can alter the position of P all you like, so long as PA remains a relfection of PB.

If this is not helping, revise vectors.

 

[Sirius Black]

Er. Yes. Eheh. Well. You'll have to actually put some effort in and draw the thing, but...

You know the parallelogram rule of vector addition and subtraction, right?

Given 3 points, A, B, C, then, all the following vectors: AB + BC = AC

Here, we have A, B, P. Now, PA is a reflection of PB, right? But recall that a vector is a vector regardless of where it is located (generally). So you can translate the vector PB such that the head of PA (head at A) is the tail of PB. The head of the translated PB is the head of your new vector, PA+PB, which is the diagonal of the parallelogram formed with 2 lots of vectors PA and PB.

You can alter the position of P all you like, so long as PA remains a relfection of PB.

If this is not helping, revise vectors.

i understand what u just said but the thing is point A, B are fixed two points, as u said, PA has to be the reflection of the PB about x-axis, this doesn't seem to work when A,B are fixed.

 

[Slidey]

Draw A and B accurately. Plot a point P somewhere (I suggest to the left or right of the two points, and half-way between them). Connect the vectors PA and PB. Draw an origin at P. Is PA a reflection of PB (angle formed by positive x-axis and vector PA the same, but negative, as the angle form by the positive x-axis and PB)? If not move P until it is.

 

[ngai]

Ignoring magnitude and position, the vector represented by arg(z-2) is a reflection of the vector represented by arg(z-2+3i), about the x-axis.

The vector represented by PA+PB lies on the locus. The vector PA+PB is one of the diagonals of a paralleogram (parallelogram law), and we know the other diagonal goes from A to B (it coincidentally represents the subtraction of the two vectors).
Let:
P(s,t)
A(2,0)
B(3,-2)

Vector AP has its head on the locus at (s-2,t)
Vector BP has its head on the locus as (s-3,t+2)
.'. (t-t-2)/(s-2-s+3)=-2 is the gradient.
C(5/2,-1) (intersection of diagonals) also lies on the locus, so:
y+1=-2(x-5/2)
y+1=2x+5
y=2x+4

lol i got nfi whats going on...
whats the 'vector' represented by arg(z-2)?...arg(z-2) is an argument, an 'angle'...a real number
do u mean the vector represented by z-2 and z-2+3i?...but those arent reflections about the x-axis?
"PA+PB lies on the locus"....what is P?...any point on the locus?...
finally, you get an answer y=2x+4, which implies x=-2,y=0 is on the locus, ie z=-2 satisfies arg(z-2)=-arg(z-3+2i)...ie. arg(-4) = -arg(-5+2i)?

 

[Sirius Black]

But note that it IS possible for a point P to exist which creates two vectors AP and BP which have the opposite arguments.

yes that is true,but at this time, PA is not a reflection of PB about x-axis but about the horiz. line y=a where a =y-ordinate of P

 

[KFunk]

This may be something of a punt but I expect the range of the locus of P would be -2 < y < 0 .

EDIT: Since outside of this range arg(z-2) and arg(z-3+2i) will be of the same sign hence arg(z-2)=-arg(z-3+2i) wont hold. And when y= -2 or 0 weird stuff happens . (Just to justify my statement)

 

[Slidey]

Let us try a distinctly different approach, eh?
arg(z-2)=-arg(z-3+2i)
arg(x-2+yi)=-arg(x-3+(y+2)i)

Removing the negative sign gives two cases:
arg(x-2+yi)=arg(x-3-(y+2)i), or
arg(-x+2+yi)=arg(-x+3-(y+2)i)

Case 1:
x-2+yi=k(x-3-(y+2)i)
x-2=k(x-3)
k=(x-2)/(x-3)
y=-k(y+2)
y=-(x-2)(y+2)/(x-3)

Case 2:
arg(-x+2+yi)=arg(-x+3-(y+2)i)
-x+2+yi=k(-x+3-(y+2)i)
-x+2=k(-x+3)
k=(x-2)/(x-3)
y=-k(y+2)
y=-(x-2)(y+2)/(x-3)

Both cases produce the same thing:

y=-(x-2)(y+2)/(x-3)

Now you can punch big giant holes in my working.

 

[KFunk]

Both cases produce the same thing:

y=-(x-2)(y+2)/(x-3)

Now you can punch big giant holes in my working.

Where -2 < y < 0 (or rather y < 0 working with the natural tendencies of the graph) ... nice work.

 

[Slidey]

If my solution is correct, then credit should go to Sirius for answering his own question in the very first post.

Could you explain why you placed those restrictions on the range? I didn't really understand you the first time.

Sirius: It is the equation for a... hyperbola, I think.

 

[Will Hunting]

You guys seem to have taken this question to the wildest heights ever! It's just about wrapped up, but I'll throw my two cents in anyhow

here is my working:

arg(z-2)=-arg(z-3+2i)
-arg(x+iy-2)=arg(x-3+(y+2)i)
-tan-1(y/(x-2))=tan-1(y+2)/(x-3)
take tan( ) of both sides
-y/(x-2)=(y+2)/(x-3)
-xy+3y=xy+2x-2y-4
2xy+2x-5y-4=0

Absolutely spot on, man. In fact, this has been the most direct method to see light in the thread thus far. Rearranging, the locus, in terms of x, reads:

y = (-2x + 4)/(2x - 5)

x-intercept: x = 2
y-intercept: y = -4/5
vertical asymptote: x = 5/2
horizontal asymptote: y = -1
For x ---> &infin;, lim y = -1^-
For x ---> -&infin;, lim y = -1⁺
There is no symmetry, since neither f(x) = f(-x), nor -f(x) = f(-x) hold for this locus
There are no st. pts. since f'(x) &ne; 0 for any real x
Similarly for inflexion pts.

The locus obviously resembles a rectangular hyperbola in shape. This should be enough to get you going with a sketch. Have fun!

Is there an angle in there too?

No, an angle is a constant. Both sides of the equation contain a complex variable.

when multiplying complez no's u multiply the mods and add the arguments so :

arg of :

(z-2)(z-3+2i) = arg (z-2) + arg (z-3+2i)
which is:

arg(z-2) = -arg(z-3+2i)

thats not right is it?

You're right on the mark, dude. This is very discerning reasoning and could possibly be the most judicious approach to solving this problem. Seeing your groundwork through, the deduction would unfold thusly,

(z - 2)[z - (3 - 2i)] =
z² -z(5 - 2i) + 6 - 4i =
x² - y² - 5x - 2y + 6 + i(2xy + 2x - 5y - 4)

Now, arg[(z - 2)(z - 3 + 2i)] = 0
&there4; (z - 2)[z - (3 - 2i)] purely real
i.e. Im[(z-2)(z - 3 + 2i)] = 0
2xy + 2x - 5y - 4 = 0
y = (-2x + 4)/(2x - 5)

From this, it is clear that your method is as valid as any. Solid.

 

[KFunk]

If my solution is correct, then credit should go to Sirius for answering his own question in the very first post.

Could you explain why you placed those restrictions on the range? I didn't really understand you the first time.

Sirius: It is the equation for a... hyperbola, I think.

I didn't really think much about the range, it's just a kind of an assumption I came to at first glance given the question asked. In your answer I believe the curve passes through (2,0) at which point arg(z-2) is undefined i.e ---> arg(z-2)=-arg(z-3+2i) falls apart. When the curve goes above y=0 then the arguments both become positive i.e the equality cannot hold for any of the points above y=0.

In any case I believe, as Will has shown, that Sirius' original answer was correct (just needing a bit of rearranging). In this case a hyperbola is what you would expect since by knowing the points and the relationship between the args you will end up with points extending to infinity in either direction.

Once again the curve y = (-2x + 4)/(2x - 5) passes through (2,0) but also (3,-2) and at each of these points arg(z-2) and arg(z-3+2i), respectively, are undefined. Above y=0 arg(z-2) and arg(z-3+2i) will be of the same (+'ve) sign and below y = -2 arg(z-2) and arg(z-3+2i) will be of the same (-'ve) sign. That is my logic for why arg(z-2)=-arg(z-3+2i) cannot hold outside of -2 < y < 0 .

Props to Will for finally getting some clarity into all of this (and to Slide for attacking it in every damn way imaginable ).

 

[m_isk]

Let us try a distinctly different approach, eh?
arg(z-2)=-arg(z-3+2i)
arg(x-2+yi)=-arg(x-3+(y+2)i)

Removing the negative sign gives two cases:
arg(x-2+yi)=arg(x-3-(y+2)i), or
arg(-x+2+yi)=arg(-x+3-(y+2)i)

I don't really understand how you got this. COuld you please explain it to me?

 

[Slidey]

It's just some basic mucking around with quadrants.

Draw up a small Cartesian plane. Draw a ray at 45 degrees, it will be in the 1st quadrant, right?
1st quad, (x,y)=(+,+)
Its reflection will be -45 and in the 4th qudrant
4th quad reflection, (x,y)=(+,-)
Similar for 2nd and 3rd:
2nd quad: (x,y)=(-,+)
3rd quad relfection: (x,y)=(-,-)
So you apply this two the x and y components of each argument and can then remove the negative sign.

You'll also notice I multiplied by k when I removed the arguments. This is because arg(2+2i) is the same as arg(1+1i).