thomas49th
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What loci is represented by the following equation:
|z+1| = |z-1|
I believe I can get the answer, I am just slightly confused by what is going on?
I turned this to
|z+1| / |z-1| = 1, which means |z+1| = 1 and |z-1| = 1. Doe this mean we can draw 2 circles of radius 1 at the points (1,0) and (-1,0)? The only place they touch is the origin? Is this correct?
OR
Letting z = x + iy
\frac{\sqrt{(x+1)^{2} + y^{2}}}{\sqrt{(x-1)^{2}+y^{2}}} = 1
\frac{(x+1)^{2} + y^{2}}{(x-1)^{2}+y^{2}} = 1
4x = 0
x = 0
That implies y can be anypoint along the y-axis (at x= 0). Is this correct?
Finally are the actual locations of the loci origins actually -1 and 1, because surely z doesn't have to be x + iy, I could be 5x + 6iy
Thanks
Thomas
|z+1| = |z-1|
I believe I can get the answer, I am just slightly confused by what is going on?
I turned this to
|z+1| / |z-1| = 1, which means |z+1| = 1 and |z-1| = 1. Doe this mean we can draw 2 circles of radius 1 at the points (1,0) and (-1,0)? The only place they touch is the origin? Is this correct?
OR
Letting z = x + iy
\frac{\sqrt{(x+1)^{2} + y^{2}}}{\sqrt{(x-1)^{2}+y^{2}}} = 1
\frac{(x+1)^{2} + y^{2}}{(x-1)^{2}+y^{2}} = 1
4x = 0
x = 0
That implies y can be anypoint along the y-axis (at x= 0). Is this correct?
Finally are the actual locations of the loci origins actually -1 and 1, because surely z doesn't have to be x + iy, I could be 5x + 6iy
Thanks
Thomas