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complex q. (1 Viewer)
- Thread starter spikestar
- Start date
Mountain.Dew
Magician, and Lawyer.
i wont tell u the answer, BUT i will tell you how u should approach it...
think about it this way: |z-@| / |z-$| = |(z-@)/(z-$)| <== then u 'realise' the denominator once u substitute z=x+iy, then square both sides of equation, and u have ur locus. fiddle around with algebra a bit, and it should turn out nice.
think about it this way: |z-@| / |z-$| = |(z-@)/(z-$)| <== then u 'realise' the denominator once u substitute z=x+iy, then square both sides of equation, and u have ur locus. fiddle around with algebra a bit, and it should turn out nice.
Most straight forward way of doing this is let z=x+iy, then use the fact that |a+ib|=sqrt(a2+b2), square both sides, then multiply the denominator across to the 2, expand everything, simplify a bit, then complete the square and you end up with the locus of a circle with centre (2,1) and radius 2 units.