Q307 Complex Numbers?

Question

If z = x + iy find the cartesian equation of the given line by:
(1 + i)z + (1 - i) ~z = -2

(the '~z' is meant to be a line over the top of z )

Graph this Line

Answer 1

z=x+iy =>~z=x-iy
(1-i)~z
=(1-i)(x-iy)
=x-iy-ix+i^2y
=(x-y)-i(x+y)---(1)
(1+i)z
=(1+i)(x+iy)
=x+iy+ix+i^2y
=(x-y)+i(x+y)----(2)
(1)+(2)=-2
=>(x-y)+(-ix)+(-iy)+x+iy+ix-y=-2
=>2x+(-2y)=-2
=>x-y+2=0---(3)

To graph this line:
Take arbitrary value of x, say x=1
(3)=> 1-y+2=0
=>y=3
Hence point (1,3) lies on line given by eq.(3)
Similarly, let x=2
(3)=>2-y+2=0
=>y=4
Other point that lies on this line given by eq.(3) is (2,4)
Now locate points (1,3) and (2,4) on the graph and draw the line passing through these points.
This is the line given by eq (3)

Answer 2

(1 + i).(x + iy) = x + i (x + y) - y
(1 - i).(x - iy) = x - i (x + y) + y
Sum = 2x

Equation is then:-
2x = - 2
x = - 1

This is a vertical line that passes thro` (-1,0)