Lecture 6 Complex Analysis

The distance between two points Z_1=x_1+iy_1 and Z_2=x_2+iy_2 in complex plan is
given by
|Z_1-Z_2 |=|(x_1+iy_1 )-(x_2+iy_2 ) |
=|(x_1-x_2 )+i(y_1-y_2 ) |
=?((x_1-x_2 )^2+(y_1-y_2 )^2 )

The complex numbers that satisfy the equation |Z|=r where r?R consist of all the points fall on the circle with center at the origin and radius r .
The complex numbers that satisfy the equation |Z-Z_0 |=r where r?R consist the points fall on the circle with center at the points Z_0 and radius r.
The complex numbers that satisfy the inequality |Z-Z_0 |?r where r?R consist the points fall on and interior to the circle with center at the point Z_0 and radius r .
Ex:- Graph the set of points that satisfy the equation
|Z-(1+i) |=1
I(Z ?-i)=2 ,R(Z ?-i)=2
|Z-i|=|Z+i|
|Z-4i|+|Z+4i|=10 H.W
SOL:-
From |Z-Z_0 |=r we find that the set of point is circle with center 1+i and radius 1 .
Since Z ?-i=¯(x+iy)-i=x-iy-i=x-i(y+1)
Thus I(Z ?-i)=I(x-i (y+1) )=-(y+1)
So that -(y+1)=2?y=-3
Thus the set of points is y=-3
Since R(z ?-i)=R(x-i (y+1) )=x
Thus x=2
So that the set of points is x=2
|Z-i|=|Z+i|?|x+iy-i|=|x+iy+i|
?|x+i(y-1) |=|x+i(y+1) |
??(x^2+(y-1)^2 )=?(x^2+(y+1)^2 ) by square two sides .
?x^2+(y-1)^2=x^2+(y+1)^2 ?y^2-2y+1=y_2+2y+1?4y=0?y=0
So that the set of points is y=0 that is real axis .