Fp1 argand diagram

Need help with this question.

Basically how do you sketch |z+1|=|z+i|.
Then it says write down the complex numbers represented by the points of intersection of |z-2i|=2 and the one I need help with.
I know what |z-2i|=2 looks like but not sure about the other one.
Cheers

Reply 1

TeeEm

19

Original post by Super199

Need help with this question.

Basically how do you sketch |z+1|=|z+i|.
Then it says write down the complex numbers represented by the points of intersection of |z-2i|=2 and the one I need help with.
I know what |z-2i|=2 looks like but not sure about the other one.
Cheers

let z = x + iy

Reply 2

Student403

19

This is FP1? :redface:

Reply 3

Zacken

22

Original post by Super199

Need help with this question.

Basically how do you sketch |z+1|=|z+i|.
Then it says write down the complex numbers represented by the points of intersection of |z-2i|=2 and the one I need help with.
I know what |z-2i|=2 looks like but not sure about the other one.
Cheers

That says: the distance between z and 1 is the same as the distance between z and i. Can you think of what that would look like? *coughs* perp. bisector *cough*

Original post by Student403

This is FP1? :redface:

This is C3 in CIE. :lol:

Reply 4

Ayman!

15

Original post by Super199

Need help with this question.

Basically how do you sketch |z+1|=|z+i|.
Then it says write down the complex numbers represented by the points of intersection of |z-2i|=2 and the one I need help with.
I know what |z-2i|=2 looks like but not sure about the other one.
Cheers

This is a standard form for a certain kind of loci. Maybe you could even try seeing what it maps to on a cartesian plane using algebra like TeeEm suggested?

Reply 5

B_9710

18

Transform complex plane to Cartesian an plane. You could then find Cartesian equation of the line and circle and go from there.
There may be a much simpler method but I haven't looked at it at all really.

Reply 6

the bear

20

Original post by Zacken

That says: the distance between z and 1 is the same as the distance between z and i. Can you think of what that would look like? *coughs* perp. bisector *cough*

This is C3 in CIE. :lol:

probably foundation level GCSE in Hong Kong

Reply 7

Student403

19

Original post by Zacken

That says: the distance between z and 1 is the same as the distance between z and i. Can you think of what that would look like? *coughs* perp. bisector *cough*

This is C3 in CIE. :lol:

Original post by the bear

probably foundation level GCSE in Hong Kong

LOL

Reply 8

Super199

OP

20

Original post by Zacken

That says: the distance between z and 1 is the same as the distance between z and i. Can you think of what that would look like? *coughs* perp.

This is C3 in CIE. :lol:

Sound. Now how do i solve this thing

Reply 9

Zacken

22

Original post by Super199

Sound. Now how do i solve this thing

To find the intersections, you'd be best solving it algebraiclly:

You have a circle centre (0, 2) (I think, modify if not) and radius sqrt(2), so:

x^2 + (y-2)^2 = 2

Now find the equation of the perp bisector of 1 and i in cartesian form and plug it into the above equation, find x and y and then your complex numbers will be x + iy

Reply 10

Super199

OP

20

Original post by Zacken

To find the intersections, you'd be best solving it algebraiclly:

You have a circle centre (0, 2) (I think, modify if not) and radius sqrt(2), so:

x^2 + (y-2)^2 = 2

Now find the equation of the perp bisector of 1 and i in cartesian form and plug it into the above equation, find x and y and then your complex numbers will be x + iy