Stuck on FP2 complex numbers

a) Show that in an Argand diagram the equation arg(z-2)-arg(z-2i)=3pi/4 represents and arc of a circle and that mod((z-4)/(z-1)) is constant on this circle.

b) Find the values of z corresponding to the points in which this circle is cut by the curve given by mod(z-1)+mod(z-4)=5.

I've completed a)

For b)
I divided both sides by mod(z-1) to give 1+mod((z-4)/(z-1))=5/mod(z-1). Then using the identity formed in a), I've managed to get down to mod(z-1)=5/3 .How do I solve this to find the values of z?

I am not quite following the part (b) but can you see the second locus is a ellipse?

I see slthough we haven't covered ellipses (its Fp3 I think) so what do I do with it now/. The answer's 10/9 +- 4root(14)i/9

my advice is to switch to cartesian

the circular arc do you have a Cartesian form? or just a complex locus?


The ellipse I will explain how to change it to Cartesian

For mod(z-1)=5/3 ,

Converting to Cartesian gives: (x-1)^2+y^2=25/9

I'm not sure how to convert the circular arc to cartesian

it is not correct...

the second curve is an ellipse with Cartesian equation x²/25+y²/4=1.


Secondly how have you proven that the first part is an arc of a circle without Cartesian?
(just a geometric argument?)

For the second curve, I did:
mod(z-1)=5/3
So mod((x-1)+iy)=5/3
So root[(x-1)^2 +y^2 )]=5/3
So (x-1)^2 +y^2 =25/9

For proving its an arc of a circle, I just drew an Argand diagram which consisted of a minor arc of a circle with centre(0,0) and radius=1

i have no idea where mod(z-1)=5/3 came from.


(x-1)^2 +y^2 =25/9 might be the correct answer if mod(z-1)=5/3

but is not the correct answer for |z-1|+|z-4|=5.


for the circle give me 10 minutes or so to work out Cartesian

Both parts are linked-
The equation of the full circle is x^2 + y^2 =4 from the diagram, since it has radius=2 and centre (0,0).

To show mod[(z-4)/(z-1] is a constant, I said:
let mod[(z-4)/(z-1)]=k
so mod(z-4)=kmod(z-1)
z=x+iy
so mod[(x-4)+iy]=kmod[(x-1)+iy]
so root[(x-4)^2+y^2]=kroot[(x-1)^2+y^2]
Expanding out,rearranging and subbing in x^2+y^2=4.
I ended up with 4(5-2x)=k^2(5-2x)
so k=2.

Hence,
mod[(z-4)/(z-1)]=2

Now, for part b:
mod(z-1)+mod(z-4)=5.
so 1+mod[(z-4)/(z-1)]=5/mod(z-1)

From part a, mod[(z-4)/(z-1)]=2
so 1+2=5/mod(z-1)
so mod(z-1)=5/3

sorry for the delay I just worked out Cartesian for the arc

x²+y²=4 in the 1st quadrant

(you mentioned earlier x²+y²=1 )

Oh, sorry! I did write out x²+y²=4 on paper

So, how do i solve :
mod(z-1)=5/3

to end up with 10/9 +- 4root(14)i/9

this is strange because I cannot see a mistake in your workings ...

and |z-1|=5/3 is a circle centre (1,0), radius 5/3

there is nothing to solve, A circle has infinite point to satisfy it

Oh, sorry! I did write out x²+y²=4 on paper

So, how do i solve :
mod(z-1)=5/3

to end up with 10/9 +- 4root(14)i/9

I solved it but not your way

i got 10/9+4root14/9i (no plus minus)


EDIT :did it your way too and I got the same answer

Ooh...how did you do it my way?

solved simult

mod(z-1)=5/3 with x²+y²=4, x>0, y>0

mod(z-1)=5/3 is (x-1)²+y²=25/9


did it with my ellipse too ...

You can show me the ellipse way as well if you want

Just finished it using my way, thanks