Complex Locus (1 Viewer)

[shaon0]

1) Sketch: arg(z-(1+i)) = arg(z-(-2+5i))
Is it just the line via the points excluding the points.

2) Sketch: arg(z-(1+i)) = arg(1-(-2+5i))+pi

3) Sketch: arg(z+3-i) = arg(z-(2-i))

Just need some tips at drawing complex locus.

 

[kaz1]

I think 1 and and 3 are 2 lines moving away from each other and 2 is a most of a circle with the complex numbers as a chord which is the boundary of the circle.

 

[Harkaraj]

1 and 3 is perpendicular bisectors of two points and 2 is the locus of an arc on a circle (explained well in cambridge/patel).

 

[kaz1]

1 and 3 is perpendicular bisectors of two points and 2 is the locus of an arc on a circle (explained well in cambridge/patel).

Don't you get perpendicular bisectors when the modulus is equal?

 

[cyl123]

1) line through both 1+i and -2+5i EXCLUDING the points AND the section in between the points.

3) is similar to 1)

2) Note: if the locus is an arc of the circle, that means the angle of the circumference will be pi LOL which is impossible. Rather, the locus is just the interval joining the 2 points 1+i and -2+5i EXCLUDING the 2 points (which is section excluded in part 1)

To get each of these locuses, its easier to visualise them geometrically. You just plot the 2 points (eg in 1) 1+i and -2+5i) firstly and just think where z must lie in order for vector formed with each point and z to be pointing in same direction (ie same argument). BE CAREFUL cos direction of vectors do matter and for points lying between 1+i and -2+5i, the directions of the 2 vectors formed with z and the 2 points respectively with be in opposite direction (satisfying the second question's condition)