Basically how do you sketch |z+1|=|z+i|. Then it says write down the complex numbers represented by the points of intersection of |z-2i|=2 and the one I need help with. I know what |z-2i|=2 looks like but not sure about the other one. Cheers
Basically how do you sketch |z+1|=|z+i|. Then it says write down the complex numbers represented by the points of intersection of |z-2i|=2 and the one I need help with. I know what |z-2i|=2 looks like but not sure about the other one. Cheers
Basically how do you sketch |z+1|=|z+i|. Then it says write down the complex numbers represented by the points of intersection of |z-2i|=2 and the one I need help with. I know what |z-2i|=2 looks like but not sure about the other one. Cheers
That says: the distance between z and 1 is the same as the distance between z and i. Can you think of what that would look like? *coughs* perp. bisector *cough*
Basically how do you sketch |z+1|=|z+i|. Then it says write down the complex numbers represented by the points of intersection of |z-2i|=2 and the one I need help with. I know what |z-2i|=2 looks like but not sure about the other one. Cheers
This is a standard form for a certain kind of loci. Maybe you could even try seeing what it maps to on a cartesian plane using algebra like TeeEm suggested?
Transform complex plane to Cartesian an plane. You could then find Cartesian equation of the line and circle and go from there. There may be a much simpler method but I haven't looked at it at all really.
That says: the distance between z and 1 is the same as the distance between z and i. Can you think of what that would look like? *coughs* perp. bisector *cough*
That says: the distance between z and 1 is the same as the distance between z and i. Can you think of what that would look like? *coughs* perp. bisector *cough*
To find the intersections, you'd be best solving it algebraiclly:
You have a circle centre (0, 2) (I think, modify if not) and radius sqrt(2), so:
x^2 + (y-2)^2 = 2
Now find the equation of the perp bisector of 1 and i in cartesian form and plug it into the above equation, find x and y and then your complex numbers will be x + iy
To find the intersections, you'd be best solving it algebraiclly:
You have a circle centre (0, 2) (I think, modify if not) and radius sqrt(2), so:
x^2 + (y-2)^2 = 2
Now find the equation of the perp bisector of 1 and i in cartesian form and plug it into the above equation, find x and y and then your complex numbers will be x + iy
Now you have In=32n∫01xn−1(1−x)3/2=32n∫01xn−1(1−x)1/2(1−x)2 see if you can work with that? I'm going for dinner, horrid timing, so I might not be able to reply right away.