Complex locus question (1 Viewer)

[blackops23]

Hi I have a question here which seems pretty simple (only 2 marks) but my brain is a little fuzzled...

Question: If w = z/(z+2), where z=x+iy, x,y real
Find the locus of w if it is PURELY IMAGINARY

Ok so here's what I did:

w= (x+iy)/[(x+2)+iy]
= {(x+iy)*[(x+2) - iy]}/{[(x+2)+ iy]*[(x+2) - iy]} <<----- REALISING IT

=[x(x+2) + y^2 + 2iy]/[(x+2)^2 + y^2]

=X + iY

Equating....

Ok now what do I do? If it asked for the locus of z, it would be easy enough, but it's asking for the locus of w, and I've no idea what to do.

Help greatly appreciated, thanks.

 

[jyu]

the locus of w is the Im axis, perhaps without the origin O

 

[blackops23]

the locus of w is the Im axis, perhaps without the origin O

how did you get this?

 

[HyperComplexxx]

if its purely imaginary, you make the real equal to zero then solve to find locus of w

 

[blackops23]

if its purely imaginary, you make the real equal to zero then solve to find locus of w

real part: x(x+2) + y^2 = 0 (excluding (2,0)

But isn't the above the LOCUS OF z, and not w?

 

[HyperComplexxx]

my bad
when you equate the X + iY, you get X = 0 (because it Re(W) = 0) and Y = 2y/[(x+2)^2 + y^2]
because X = 0, the locus of W will always be on the Im axis and since it is purely Im, Y cannot equal zero
then locus of W is the Im axis excluding (0,0)

 

[blackops23]

my bad
when you equate the X + iY, you get X = 0 (because it Re(W) = 0) and Y = 2y/[(x+2)^2 + y^2]
because X = 0, the locus of W will always be on the Im axis and since it is purely Im, Y cannot equal zero
then locus of W is the Im axis excluding (0,0)

aaaaaaaahhh..... I seee.... thanks mate

 

[rawrence]

 

[blackops23]

That is the locus of z i.e (x+1)^2 + y^2 = 0.

Here's what i got out of the question:
ignore the z/z+2 part, concentrate on w, as we are finding the locus of w.
Now if w is purely imaginary would t not make sense that the locus is the imaginary axis?

 

[rawrence]

Oh. The locus of w.

That's actually a very simple question then, you subbed in x+iy in your first line of working as z but that's finding locus of z, to find w, just set the condition Re(w)=0 therefore x=0 is the locus.

That's only a 1 marker in tests anyway, I got confused in my test with that as well

 

[nohelp]

That is the locus of z i.e (x+1)^2 + y^2 = 0.

Here's what i got out of the question:
ignore the z/z+2 part, concentrate on w, as we are finding the locus of w.
Now if w is purely imaginary would t not make sense that the locus is the imaginary axis?

Um... this may be just me but when your finding the locus of w, z, x etc. the imaginary and real axis don't apply. Which means that rawrence is still correct. What I mean to say is that the locus will never the Re, x or Im, y axis. Am I making any sense?

 

[ZachBC_94]

I wasn't aware that you could have loci on the imaginary axis, just vectors.