is it just me?

[Dahello]

I thought i was able to do this question, but maybe not???

W is the subset of complex plane:
z: I z-2-2i I - I z+2+21 I = 4
a) find the cartesian equation:

so what i got was x+y = -2? Is that what you guys got too?

but the cause in part c) explain why graph of W consists of only one arm of hyperbola??

um...

[drake]

in complex numbers, equations in the form: |z - a| - |z - b| = c will generally give you an hyperbola. try again with finding your cartesian equation, it's lots of algebra!

[Dahello]

i didn't get a hyperbola. This is what i did:

I x+yi-2-2i I - I x+yi+2+2i I = 4
squareroot( (x-2)^2 + (y-2)^2 ) - squareroot( (x+2)^2 + (y+2)^2 ) = 4
(x-2)^2 + (y-2)^2  - (x+2)^2 + (y+2)^2 = 16....

[kinslayer]

i didn't get a hyperbola. This is what i did:

I x+yi-2-2i I - I x+yi+2+2i I = 4
squareroot( (x-2)^2 + (y-2)^2 ) - squareroot( (x+2)^2 + (y+2)^2 ) = 4
(x-2)^2 + (y-2)^2  - (x+2)^2 + (y+2)^2 = 16....

You left out this term when expanding the square root on the left:

Like drake said it is a lot of algebra. Good luck!

[Dahello]

I x+yi-2-2i I - I x+yi+2+2i I = 4
squareroot( (x-2)^2 + (y-2)^2 ) - squareroot( (x+2)^2 + (y+2)^2 ) = 2
(x-2)^2 + (y-2)^2  - (x+2)^2 + (y+2)^2 = 2
-8x - 8y = 2
divide both sides by 2, isn't that still a straight line?

[kinslayer]

If z = x + yi then:



implies:



Now you can square both sides and continue. Remember the cross product term I mentioned in my post above.

[keltingmeith]

What you're saying is:



This is equivalent to saying that:



Which isn't true, as we know through binomial expansion:



Everyone's said it, I'll just re-iterate - expect A LOT of algebra. At least half a page. Any less, and you've probably made a mistake somewhere.

[Dahello]

oh then how would i do it then.. can you start it off? just after the a^2 - b^2 part you were expalining and then i'll do the rest... not sure how to start ... sorry