Hey i got a complex # q. (1 Viewer)

[currysauce]

hey... like the title says...

if u have the book, cambridge, pg 72 qu.6

says

6. z = x + iy is such that z - i / z + 1 is purely imaginary. Find the equation of the locus of point P representing z and show this locus on an Argand Diagram.

thanks all

 

[McLake]

[x + i(y - 1)]/[(x + 1) + iy] = ai
so [(x + i(y - 1))((x + 1) - iy)]/[(x + 1)^2 + y^2] = ai
[(x^2 + x) - ixy + i(xy + y - x - 1) + (y^2 - y)]/[(x + 1)^2 + y^2] = ai

so just imginary bit
[- xy + (xy + y - x - 1)]/[(x + 1)^2 + y^2] = a
(y - x - 1) = a[x^2 + 2x + 1 + y^2]
ax^2 + ay^2 + 2ax + y + 2 = 0

and just the real bit
[(x^2 + x) +(y^2 - y)]/[(x + 1)^2 + y^2] = 0
so x^2 + x + y^2 - y = 0
so (x + 0.5)^2 - 0.25 + (y - 0.5)^2 - 0.25 = 0
so (x + .5)^2 + (y - .5)^2 = .5

Circle!

 

[Estel]

(z - i)/(z + 1) is imaginary
ie arg(z-i)-arg(z+1) = +- pi/2
:. the join of i and -1 is the diameter of the circle of locus P
Midpt: (-1/2, 1/2)
Since the distance between the two points is rt2, radius is 1/rt2
locus is (x+1/2)^2+(y-1/2)^2 = 1/2 , z =/ i or -1.

 

[currysauce]

estel... could u elaborate... move slower

can u rewrite with extended response please!.. i am slow.... lol

also the answer in the back is x(x+1) + y(y-1) = 0

 

[Estel]

Let w = (z-i)/(z+1)
Arguments of imaginary numbers are +-pi/2.
so take the argument of w
you know it will be +-pi/2
arg w = arg(z-1) - arg(z+1) [remember ur arg rules]
Join i and -1. They subtend +-pi/2 at P(w) [exterior angles- use a diagram]

:. the locus of w is a circle with diameter the join of i and -1 [ur circle geo theory abt angle in a semicircle]
as you know, the centre of a circle is it's diameter, and the radius is half the diameter.
The rest follows.

If you are uncomfortable with this sort of argument, use algebra.
Personally, I prefer geometry.

 

[McLake]

Or my way:

[x + i(y - 1)]/[(x + 1) + iy] = ai
so [(x + i(y - 1))((x + 1) - iy)]/[(x + 1)^2 + y^2] = ai
[(x^2 + x) - ixy + i(xy + y - x - 1) + (y^2 - y)]/[(x + 1)^2 + y^2] = ai

and just the real bit
[(x^2 + x) +(y^2 - y)]/[(x + 1)^2 + y^2] = 0
so x^2 + x + y^2 - y = 0
so (x + 0.5)^2 - 0.25 + (y - 0.5)^2 - 0.25 = 0
so (x + .5)^2 + (y - .5)^2 = .5

Circle!

Just go back to the line in blue
x^2 + x + y^2 - y = 0
x(x+1) + y(y-1) = 0

 

[currysauce]

thanks estel... but i prefer algebra... so i'll use mclakes... i sorta understand urs though


and mclake... why would u use the real part... when its a quesiton saying blah blah is purely imaginary?

just a thought

 

[McLake]

and mclake... why would u use the real part... when its a quesiton saying blah blah is purely imaginary?

just a thought

Ok, you may have seen my earlier atempt before I edited it using the imaginary part. The problem with that is that it leaves an unknow variable (that imaginary part), which mean your locus is in three variables, not two, and so not drawable on an Argand plane.

When I use the real part I eliminate the extra variable, since I can set the expression to 0. You will find this comes up a lot in complex numbers, and you should always pick the easier path, the one that isn't there. I forgot that 'cause I am out of practice ...