Complex Locus (1 Viewer)

[cyl123]

arg(z-2)=k.arg(z^2-2z)
If mod(z-2)=2, then;
arg(2)=k.arg(2z) [idk if this is correct]
But, mod(z-2)=2
ie. z-2=2 or z-2=-2
Thus, z=4 or 0 but arg(0) is undefined ie. z=4

arg(2)=k.arg(8)
arg(2)=3k.arg(2)
k=1/3

arg(2)=0 btw..... so that doesnt work (division via 0)
|z-2|=2 doesnt imply z-2=2 or -2.... z-2 could be 1+sqrt(3)i

An algebraic solution would be

|z-2|=2 --> z-2=2cisx for some x (ie arg(z-2)=x)
z=2(cosx+1+isinx)

Note cosx+1=2cos^2(x/2) and sinx=2cos(x/2)sin(x/2)
z=2(2cos(x/2))cis(x/2) after simplification
so arg(z)=x/2

so arg(z-2)=k.arg(z^2-2z)=karg(z)+karg(z-2)
so x=kx/2+kx
So 1=3k/2---> k=2/3

This is basically the same as the geometric solution where we get 2arg(z)=arg(z-2)

 

[addikaye03]

if this is the question
'find k if arg(z-2)=k arg(z^2-2z)' and there are no more conditions, then k varies depending on z doesnt it? at least that's what Simon and I think... but our 4U knowledge is getting rusty so yeh...

Nah you are right, i viewed the Question wrong. The extra information that was provided after i posted that attempt would have been useful! lol Besides i'm not going to disagree with you+simon lol

 

[shaon0]

arg(2)=0 btw..... so that doesnt work (division via 0)
|z-2|=2 doesnt imply z-2=2 or -2.... z-2 could be 1+sqrt(3)i

An algebraic solution would be

|z-2|=2 --> z-2=2cisx for some x (ie arg(z-2)=x)
z=2(cosx+1+isinx)

Note cosx+1=2cos^2(x/2) and sinx=2cos(x/2)sin(x/2)
z=2(2cos(x/2))cis(x/2) after simplification
so arg(z)=x/2

so arg(z-2)=k.arg(z^2-2z)=karg(z)+karg(z-2)
so x=kx/2+kx
So 1=3k/2---> k=2/3

This is basically the same as the geometric solution where we get 2arg(z)=arg(z-2)

Oh ok, thanks for the correction.

 

[untouchablecuz]

arg(z-2)=k.arg(z^2-2z)
If mod(z-2)=2, then;
arg(2)=k.arg(2z) [idk if this is correct]
But, mod(z-2)=2
ie. z-2=2 or z-2=-2
Thus, z=4 or 0 but arg(0) is undefined ie. z=4

arg(2)=k.arg(8)
arg(2)=3k.arg(2)
k=1/3

This solution is probably incorrect as I'm rusty with 4u
Gurmies, how did you do it geometrically?

 

[adomad]

another question....


find the locus of z if:

Arg(z-i) - arg(z+1) = pi/2

is the answer a semi circle that has the points i and -1 at the ends?? (2nd quad)

 

[study-freak]

another question....


find the locus of z if:

Arg(z-i) - arg(z+1) = pi/2

is the answer a semi circle that has the points i and -1 at the ends?? (2nd quad)

Yep, but exclude (-1,0), (0,1)

 

[adomad]

Yep, but exclude (-1,0), (0,1)

so just sketching hte locus would be fine?, or do you think i need to find it out using x and y.... its worth 2 marks... what do you think that marks are awarded for??

i got it from some 4U paper 2006

 

[study-freak]

so just sketching hte locus would be fine?, or do you think i need to find it out using x and y.... its worth 2 marks... what do you think that marks are awarded for??

i got it from some 4U paper 2006

Well, use geometric argument and just write the equation of the circle with restrictions.
That's prob too much for a 2 marker but just to be safe.