MAT334--2020F > Chapter 1

how to solve this problem?

(1/1)

ziyizhou:
Sketch the locus of those points 𝑤 with
|w+2|=|w-2|

ziyizhou:
could you please show the step?

Jessica Long:
This is my solution to the problem. Note that the result can also be derived from the Apollonius circle formula with 𝜌=0, since the y-axis bisects the line segment between (2,0) and (-2,0).

RunboZhang:
I think there are two ways to solve this problem.

Firstly, you can substitute w by x+iy into the equation. Then organize the equation and put real parts together and imaginary parts together. Now you will have the equation |(x+2)+iy|=|(x-2)+iy|. Then take the square of both sides and organize the equation. Finally you will get x=0, which is the y-axis.

The second way to do this problem is to illustrate it geometrically. |w+2| = |w-2| is same as |w-(-2)|=|w-2|, and it asks you to find all w in complex plane that has equal distances to (2,0) and (-2,0). Thus the answer will be the perpendicular bisector of (2,0) and (-2,0), which is still the y-axis.