find the sketch of y=1 under the mapping $f(z)=z^2$

Question

I am being asked to find and sketch the image of the horizontal line y=1 under the mapping $f(z)=z^2$ This is what I have so far,

$u(x,y)+iv(x,y)=f(z)=z^2=(x+iy)^2=x^2 - y^2 + 2ixy$

so we then have $u(x,y)=x^2 - y^2$ and $v(x,y)=2xy$ which gives me the line $y=1$ which corresponds to $u(x,1)=x^2 - 1$ and to $v(x,1)=2x$ This is where im not sure on what else to do.

Answer 1

Hint:

Consider the point on $y=1$ in the form $re^{i\theta}$. Under $ f$ the arguement will be doubled and the modulus is incresed from $r$ to $r^2$. Now stech the graph separately in modulus form & in the arguement form. Observe there may be some repeatation in the arguement graph. If you can see them separately then join them, the picture will be clear to you.

Answer 2

You have nearly gone all the steps. As you have obtained, $$\begin{array}{l}u = {x^2} - 1\\v = 2x\end{array}$$Now, $x$ is a dummy variable which we could eliminate in these two relations to obtain $u = \frac{{{v^2}}}{4} - 1$. Note that $u$ plays the same role as $x$ in the mapped plane and $v$ has the role of $y$, so that it might be more clear to write $$v = \pm 2\sqrt {u + 1} $$