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???|z-3| + |z+3| = 5
|x-3 + iy| + |x+3 + iy| = 5
(x-3)^2 + y^2 + (x+3)^2 + y^2 = 25
x^2 + 18 + 2y^2 = 25
x^2+ 2y^2 = 7 (ellipse)
There's no way that could be an ellipse.Need help with a locus question, question below
Find the locus of "modulus(z-3) + modulus(z+3) =5"
Looks like an ellipse but when i let z=x+iy it comes out as a hyperbola........
Any ideas people?
arg(z-2)=k arg(z^2-2z)for this question, find the locus of z if Re(z) =|z|. can't you just say that it is only logical if the locus is the Re(z) axis where x>=0??
and how the fiffie cakes do you do this one
find the value of k if arg(z-2)=k arg(z^2-2z)?
Just because their arguments are equal, it doesn't mean that modulus of the expressions inside the brackets are also equal.arg(z-2)=k arg(z^2-2z)
z-2=(z^2-2z)^k
hahahahaha well saying "that helps" wud be a huuuuuuuuuuuuuuuge UNDERSTATEMENTLOL. i forgot the read the start of the question. just before i was going to rage quit, i scrolled up
it states that |z-2| = 2 and 0< arg(z) <90 degrees
i got it now. thanks guys.
Damn, i got k=1/3.k = 2/3
if this is the questionarg(z-2)=k arg(z^2-2z)
z-2=(z^2-2z)^k
z-2=z^k(z-2)^k
(z^k) (z-2)^(k-1)=1
Got it from there?
I think geometric way is the best way.Could somebody post an algebraic solution, mine was geometric?