What would the graph of this w-plane look like?

[bean9321]

so I have f(z)= z^2 and we have to let the x-value be a constant

so z=x(0) + iy(t) were x(0) is equal to a constant

so f(x(0)+iy(t))= (x(0)+iy(t))^2= x^2(0) + 2x(0)iy(t)-y^2(t)

and letting x(0)=c with c as a constant and A=c^2 we will have

F(x(0)+iy(t))= A+2ciy(t)-y^2(t)


what would this look like on the w-plane?



Thank You in advance.

 

[Ishuda]

so I have f(z)= z^2 and we have to let the x-value be a constant

so z=x(0) + iy(t) were x(0) is equal to a constant

so f(x(0)+iy(t))= (x(0)+iy(t))^2= x^2(0) + 2x(0)iy(t)-y^2(t)

and letting x(0)=c with c as a constant and A=c^2 we will have

F(x(0)+iy(t))= A+2ciy(t)-y^2(t)


what would this look like on the w-plane?

Thank You in advance.

Let
u = Rl(F(z) = A - y²
v = Im(F(z) = 2 c y
and see if you can find a relationship between u and v. As a hint, you will need to consider two cases; c equal to zero and c not equal to zero.
Highlight the area between the >>>>>>>>>>>>> and <<<<<<<<<<<<<< if you really get stuck but try to work it out for yourself.

>>>>>>>>>>>>>
(1)c{tex]\ne[/tex]0
In this case
u = A - (v/2c)²
which is a parabola with vertex of u=A at v=0 opening downward.

(2) c=0
u = A - y²
v = 0
A straight line where both the right and left hand sides (or top and bottom if you prefer) of the parabola coalesce into a straight line.
<<<<<<<<<<<<<<