complex question (1 Viewer)

[Gruma]

Q: draw neat labelled sketches to indicate the subsets on an argand diagram.

z: lz-1l < lz+1l

 

[McLake]

A simple way of looking at this is "what points are an equal distance from 1 and -1"? The answer is y = 0 (ie: x-axis).

More formally:

let z = x + iy

so |x + iy - 1| < |x + iy + 1|
|(x - 1) + i(y)| < |(x + 1) + i(y)|
sqrt((x - 1)^2 - y^2) < sqrt((x + 1)^2 - y^2)
(x^2 - 2x + 1 - y^2) < (x^2 + 2x + 1 - y^2)
-2x < 2x
x > - x

which has no sol in x.
so the solution must be y = 0.

 

[BlackJack]

Lol, [doesn't] | z-1 | = | z+1 | equal the y axis? Draw a graph, |z-1| is the distance from 1+0i and the other is from -1+0i

Then | z-1 | < | z+1 | is the right side of the complex field.

At least that's what I think.

 

[SoFTuaRiaL]

Originally posted by BlackJack
Lol, isn't | z-1 | = | z+1 |

no pal, |z-1| = sqrt ((x-1)^2 + y^2) wheras |z+1| = sqrt((x+1)^2 + y^2) .....
so, it basically comes down to (x-1)^2 = (x+1)^2
which results in x=0, ie, the y axis. mclake is rite ...

 

[BlackJack]

Ah. There's an extra paragraph marker there, I meant it's the y axis.

 

[spice girl]

Originally posted by Gruma
z: lz-1l < lz+1l

The question translates to: find the region where z is closer to the point 1 than to the point -1. Thus it's the region {x,y| x > 0}

 

[SoFTuaRiaL]

Re: Re: complex question

Originally posted by spice girl


The question translates to: find the region where z is closer to the point 1 than to the point -1. Thus it's the region {x,y| x > 0}

oh yea, ure rite ........ i read the < as an = ... bah !

 

[Rahul]

for this type of q's i use the method that mclake used....algebraically....i find doing it using vectors hard