Need help with circle diagram in Re and Im numbers?

Question

I have been helped with a question and now have the answer, but I'm not sure of the method to draw a circle (family of circles) from co-ordiantes given in terms of u and v

I've got u(x,y) = (x^2+y^2-1) / ((x+1)^2 + y^2)
v(x,y) = 2y / ((x+1)^2 + y^2)

Now I would like to think that I know something about this so I figure the 1st equation u the top is a circle with center (0,0) radius 1 - but because it's part of a fraction what does this mean? Because on it's own I believe it is a circle with center (-1,0) and radius, I don't know?

Can someone please help me out with the method of working out because I'd like to know how to do it, thanks

Answer 1

I gave you those results
If you put u(x,y)= k k any value .For each value you get a circle Lets put k= 2 You get

x^2+y^-1 = 2x^2 +4x +2+2y^2
so x^2 +y^2 +4x+3 =0 which represents a cf. with center (-2,0) and radius r^2=4-3 = 1

For each value of k you'll get a cf.

For some values of k the cf could be imaginary (that means that r^2 is negative)
For k=0 you get x^2+y^2-1=0
Do the same for v(x,y)

Answer 2

To solve these, we're going to create a series of level curves.

Let's start with u(x,y). To create a level curve, we set the function equal to a constant and determine the function. As we change the constant, what do the variety of curves look like.

(x^2+y^2-1)/((x+1)^2+y^2)=C

multiply, simplify and put into canonical form (if it exists)

x^2+y^2=c(x+1)^2+cy^2
x^2+y^2=c(x^2+2x+1)+cY^2
(1-c)x^2-2cx+(1-c)y^2=c

complete the square

(1-c)(x^2-2c/(1-c)x+c^2/(1-c)^2)+(1-c)y^2=c+c^2/(1-c)

(1-c)(x-c/(1-c))^2+(1-c)y^2=(c(1-c)+c^2)/(1-c)

(x+c/(c-1))^2+y^2=c/(c-1)^2

so as c changes, the center moves along the x axis and the radius changes. You now can take a few sample values of c and see what these level curves look like. You should now be able to repeat this for v(x,y)