complex no. question (1 Viewer)

[fashionista]

HII!!!! omg im beating my brain to the point of not being to think straight over this question

k can sum1 please tell me how to figure this one out?
find the locus of z if (z+1)/(z-i) is
(a) purely real
(b) purely imaginary

THanking u muchly

oooo i got another one

if z=x+iy and w=(z+1)/z, prove that if z describes the unit circle then w describes another circle whose radius and centre are to be determined.

 

[fashionista]

i have another question which im half way thru and then i got stuck

it goes find the locus of w if w=(z-2)/z, if w is purely imaginary. yeh so i went like this
z(w-1)=-2 --------> z=2/(1-w)
|z|=2/|1-w| ----------> but |1-w|=|w-1|
and now im stuck
pllleeeeease help!!

 

[Xayma]

Originally posted by fashionista
k can sum1 please tell me how to figure this one out?
find the locus of z if (z+1)/(z-i) is
(a) purely real

Let z=x+iy

(z+1)/(z-i)
=((x+iy)+1)(x+iy-i)
=((x+1)+iy)/(x+(y-1)i)
=((x+1)+iy)(x-(y-1)i)/(x²+(y-1)²)
=[ (x(x+1)+y(y-1))+(-x(y-1)+xy)i]/
(x²+(y-1)²)

For it to be completley real the imaginary component doesnt exist
therefore,
-x(y-1)+xy=0
xy-xy+x=0
Therefore x=0
x was the real component of z

Hence the locus of z is purely imaginary. (I probably made a mistake somewhere havent touched it in a while)

The second one I cant seem to do that way because I end up with x²+x=y-y²

 

[Giant Lobster]

xayma thats a circle

and for the 2nd post, u pretty much answered ur own question.

it goes find the locus of w if w=(z-2)/z, if w is purely imaginary.

if w E M then w is a vertical line. I think u mean if z E M -or- w E M, find locus of z. Which?

 

[turtle_2468]

Part (b): I'm lazy, so assuming xayma is right...
x^2+x=y-y^2
So x^2+x+y^2-y=0
So (x+0.5)^2+(y-0.5)^2=0.5
So locus is circle on argand diagram with centre (-0.5+0.5i) with radius 1/root(2).

 

[Xayma]

Originally posted by Giant Lobster
xayma thats a circle

I never got that far. So I had to just try x+iy.

That circle wouldnt be purely imaginary then, hmm Ill look back over the working soon

 

[Giant Lobster]

aye? i no get.
Man wot r u doin ere anyway?? you dont even do 4u and ur helping those who are? DO 4U! Screw your 'via correspondence' shit, move to sydney - woteva. we need more ppl like u doin this. friggin

 

[Xayma]

Originally posted by Giant Lobster
Screw your 'via correspondence' shit, move to sydney - woteva. we need more ppl like u doin this. friggin

The first one was easy, its just complex conjugates which is covered in the first few exercises.

 

[fashionista]

Originally posted by turtle_2468
Part (b): I'm lazy, so assuming xayma is right...
x^2+x=y-y^2
So x^2+x+y^2-y=0
So (x+0.5)^2+(y-0.5)^2=0.5
So locus is circle on argand diagram with centre (-0.5+0.5i) with radius 1/root(2).

YAY thats wut i got

 

[fashionista]

okie..i have another question...how do u prove something is a circle on the argand diagram? like if theres a question that goes
(z-2)(Conjugate(z)-2)=4...gah i wish we'd been taught this properly

 

[freaking_out]

Originally posted by fashionista
okie..i have another question...how do u prove something is a circle on the argand diagram? like if theres a question that goes
(z-2)(Conjugate(z)-2)=4...gah i wish we'd been taught this properly

well let z=x+iy, and then sub in, and show how the equation turens out to b a circle.

 

[fashionista]

okie dokie...tanking u muchly