Find the plot of the points on the complex plane such that
|4z-2i|< Im(4z-i) and |z-1|<4
I have given this as a follow up to yesterdays' question! (It is simpler than that one though!)
1ans is area intersected b/w the circle with centre (0,1) and r=4 and the parabola x2=1/2(y-3/8)
62are u trying to upload BMP's? tyr oto save the image as .jpg in paint!
1tried all formats maybe its taking longer than i can bear
but anyway is the ans correct
33Distance of z from a point(0,1/2) is less than than that from a line y=1/4 so its the interior of the parabola...
|z-i/2|<Im(z-i/4)
So parabola is one having directrix as y=1/4 and focus at 1/2
So parabola is x1/2=2(y-3/8)..... (edited)
But why Philip is getting x2=1/2(y-3/8).... :(
may be i made a mistake...
*edited
And second is clearly the interior of circle....
33By X+iY answer is quite obvious but i'm not going to solve that... i am a bit lazy :P
33Opsie got my mistake...
Found the focus...
Vertex but wrong equation...
:P
Philip eqn is right..
