Help with complex numbers FP1

On a sketch of the complex plane, shade the region represented by the inequality |z-(2+i)| < |z+1|

how would I go on tackling this?

Reply 1

TVIO

Think about them as seperate regions then worry about the inequality signs. Also might be useful to think about the latter one as |z-(-1)|. Where have you gotten to by now?

Reply 2

unclefred

Original post by AG Singh

On a sketch of the complex plane, shade the region represented by the inequality |z-(2+i)| < |z+1|

how would I go on tackling this?

Mod(z - a) in the complex plane represents the distance of z from the complex number a. So mod(z-a) = mod(z-b) tells you that z is equidistant from the complex numbers a and b. What locus is this? The inequality puts you on one side of this locus.

Reply 3

AG Singh

Original post by TVIO

Think about them as seperate regions then worry about the inequality signs. Also might be useful to think about the latter one as |z-(-1)|. Where have you gotten to by now?

Well I done many different things many times but the most recent attempt i did was : |z^2-[(x+2)+(y+1)]^2| < |(x+1)^2+y^2|
and arrived at : 4 < x^2+4x+y^2+2y
Which then I got : 9 < (x+2)^2 + (y+1)^2
Then i sketch the circle and shade

Reply 4

unclefred

Original post by AG Singh

You are trying to convert this to cartesian coordinate geometry. Also by squaring you introduce unallowabe answers. Don't!!

It is analagous to vectors. AB = b-a so mod(b-a) is the length of the vector AB. In the complex plane mod(z-a) is the length of the vector AZ ie the distance from the complex number z to the complex number a. What is the locus of a point that is equidistant from two fixed points - I assure you it is very very simple

Reply 5

AG Singh

Original post by unclefred

I looked through some old notes and is the locus a Perpendicular bisector?

Reply 6

unclefred

Original post by AG Singh

I looked through some old notes and is the locus a Perpendicular bisector?

You've got it! The locus of a variable point that is equidistant from two fixed points is the perpendicular bisector. Draw it and you'll see how obvious it is. Your final problem now is to decide which side of the line is your required region - just choose a side, pick any point in it and see if it fits the inequality - if it does - bingo! If it doesn't then its the other sode of the line you want.

Reply 7

AG Singh

Original post by unclefred

So I should plot (2,1) & (-1,0) cut it with a Perpendicular bisector? Is that all there is to it? :redface:

Reply 8

unclefred

Original post by AG Singh

I looked through some old notes and is the locus a Perpendicular bisector?

One further thought after my last post - you will find that complex loci, vector geometry and pure euclidean geometry are very closely linked - understand one and the others are that bit easier. In this case mod(z-a) = mod(z-b) is saying z is equidistant from a and b - you are now in the realms of pure geometry - no need for any complex analysis and certainly no need to try to convert it to coordinate geometry.

By the way - if you get mod(z-a) = 2mod(z-b) you are saying that the distance of z from a is twice the distance of z from b. The locus must be symetric about AB - and there is a point between a and b and another on the line AB but the other side of B. In fact it's an ellipse - draw it and see.

Reply 9

unclefred

Original post by AG Singh

So I should plot (2,1) & (-1,0) cut it with a Perpendicular bisector? Is that all there is to it? :redface:

That's it!! Just have to decide which side you want! Remember the original question was an inequality.

Reply 10

AG Singh

Original post by unclefred

That's it!! Just have to decide which side you want! Remember the original question was an inequality.

Thank you so much, I have spent days attempting different things on this one 3 marker question, Can't thank you enough!

Reply 11

unclefred

Original post by AG Singh

Thank you so much, I have spent days attempting different things on this one 3 marker question, Can't thank you enough!

Delighted to help and thank you so much for your response.

Further thoughts - you will come across arg(z-a). Think of it as the angle that the vector AZ makes with the real axis. So - arg(z-a) = pi/4 is a line from a at 45 degrees to the real axis (with a hole at a - so is an open ended half line). arg((z-a)/(z-b)) = pi/4 is arg(z-a)-arg(z-b) = pi/4 ie the angle that the vector AZ makes with BZ is 45 degrees - the major arc of a circle on AN as chord ( a result from pure geometry).

You will find all these complex loci problems easier if you think geometrically

Reply 12

AG Singh

Original post by unclefred

Thank you, I will keep that in mind as the next question is indeed another sketching question with arg involved

Reply 13

unclefred

Original post by AG Singh

Thank you, I will keep that in mind as the next question is indeed another sketching question with arg involved

One final thought before I go to bed! The fact that it was worth three marks only tells you an awful lot. It's not difficult and certainly not worth getting involved with complicated coordinate geometry - the examiners have three marks to play with - one for some idea that you are dealing with distance of z from two points a and b, one for perpendicular bisector and one for the correct side of the line. You will find that the marks allocated can be a helpful clue by themselves. Sherlock Holmes!!!