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# P6 Complex numbers watch

1. Hello, helpful math folk, I am stuck on question 14 p.43 of heinemann book, part b:

Given that arg((z-1)/(z+1)) = pi/4

(a) show that thte point P, which represents z on an Argand diagram, lies on an arc of a circle

(b) Find the centre and radius of the circle and sketch the locus of P

All help appreciated, as time running out, thanks!
2. (Original post by Rixius)
Hello, helpful math folk, I am stuck on question 14 p.43 of heinemann book, part b:

Given that arg((z-1)/(z+1)) = pi/4

(a) show that thte point P, which represents z on an Argand diagram, lies on an arc of a circle

(b) Find the centre and radius of the circle and sketch the locus of P

All help appreciated, as time running out, thanks!
Ther was a question similar to this yesterday.
See here
Look at the link in post #2. I think that may help.
3. It was the same area of p6, but i still dont know how to find the centre of the circle
4. (Original post by Rixius)
It was the same area of p6, but i still dont know how to find the centre of the circle
Well, if you draw the argand diagram, you can work that out from trig, i would think.
5. Ive just seen that im too late, but ive done it now.

This is quite a long question, maybe ive used a stupid method, but anyways
When you multiply compex numbers, you add the arguments, so multiply out the z+1 on the bottom;
arg(z-1) = arg(z+1) + pi/4

z = a + bi, arg(z) = arctan(b/a), so

arctan(b/a-1) = arctan(b/a+1) + arctan(1)
tan each side;
b/(a-1) = [b/(a+1) + 1]/[1 - (b/a+1) *1]
Simplify and multiply out to get

a^2 + (b-1)^2 = 2
6. I did some trig and I got
centre = (0,(1+√2)/3)
7. I did it and got a different answer.

If arg((z-1)/(z+1)) = pi/4

then arg(z-1)-arg(z+1)=pi/4
so the coords of A1,0) and B-1,0)
The sector of the circle where the point P lies is therefore drawn from A to B where AP to BP is clockwise.

Because the angle is pi/4,from circle laws we know the angle at the centre is twice the angle at the circumference, so wouldn't the angle at the centre be pi/2. If this is the case, I used trig to work out the opp (which would be the radius) and got r=2sin(pi/2)=2
Using Pythagoras then to get the y coord of the height (because the x coord is 0): root of (1+2^2)

So I get the centre to be (o,root5) and the radius to be 2.

I don't have great credibility though, and would very much appreciate a genius to swoop in and tell me how to do this question too!
8. (Original post by JamesF)
...

z = a + bi, arg(z) = arctan(b/a), so

arctan(b/a-1) = arctan(b/a+1) + arctan(1)
I didn't think you could do this with the arguments, but I've just done it the long way round and I get the same as you,

x² + (y-1)² = 2

So I must've messed up my trig!
9. (Original post by Hope)
I did it and got a different answer.

If arg((z-1)/(z+1)) = pi/4

then arg(z-1)-arg(z+1)=pi/4
so the coords of A1,0) and B-1,0)
The sector of the circle where the point P lies is therefore drawn from A to B where AP to BP is clockwise.

Because the angle is pi/4,from circle laws we know the angle at the centre is twice the angle at the circumference, so wouldn't the angle at the centre be pi/2. If this is the case, I used trig to work out the opp (which would be the radius) and got r=2sin(pi/2)=2
Using Pythagoras then to get the y coord of the height (because the x coord is 0): root of (1+2^2)

So I get the centre to be (o,root5) and the radius to be 2.

I don't have great credibility though, and would very much appreciate a genius to swoop in and tell me how to do this question too!
Check your trig - you should get the radius as √2 and the centre at y=1, x=0.
10. Oh yeh!! I used 2sin(pi/2) when it should have been 1/tan(pi/4)

(heehee - my coordinates came out as smilies!! That'll teach me to use colons and brackets next to each other without wanting to make faces!)
11. (Original post by Hope)
I don't have great credibility though, and would very much appreciate a genius to swoop in and tell me how to do this question too!
How to do it!!

arg[(z-z1)/(z-z2)] = pi/4

where z1 = 1 and z2 = -1 and z = x + iy

z-z1 = (x-1) + iy
z-z2 = (x+1) + iy

let z-z1 = r1ø1 and z-z2 = r2ø2

ø1 = arg(z-z1) = arctan(y/(x-1))
ø2 = arg(z-z2) = arctan(y/(x+1))

then,

(z-z1)/(z-z2) = (r1/r2)(ø1 - ø2)
arg[(z-z1)/(z-z2)] = (ø1 - ø2) = pi/4

take tan of both sides,

(tanø1 - tanø2)/(1+tanø1tanø2) = tan(pi/4)
(y/(x-1) - y/(x+1))/(1+y²/(x²-1)) = 1
y(x+1) - y(x-1) = (x²-1) + y²
yx + y - yx + y = x² - 1 + y²
x² + y² - 2y - 1 = 0
x² + (y-1)² - 2 = 0
x² + (y-1)² = 2
==========

centre at (0,1) r= √2
=============

Edit: of course, james F's method is far quicker!

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