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

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.

• Write $u$ as a function of $v$ and sketch that in the $uv$-plane. – Daniel Fischer Oct 5 '14 at 23:09
• @DanielFischer do you mean to substitute u into v and solve that? – cele Oct 5 '14 at 23:20
• The other way round, rather. There isn't much to solve then. You have $v = 2x,\; u = x^2-1$, so as a function of $v$, we have $u(v) = \,?$ – Daniel Fischer Oct 5 '14 at 23:23
• @DanielFischer 4x^2 -1 – cele Oct 5 '14 at 23:26
• On the one hand, we want $u$ as a function of $v$, on the other, you have the wrong constant. With $v = 2x$, we have $x = \frac{v}{2}$, and hence $x^2-1 = \left(\frac{v}{2}\right)^2-1 = \frac{1}{4}v^2-1$, so $u(v) = \frac{1}{4}v^2-1$. Sketching that should be familiar from school (but be aware of the orientation, the $v$-axis is the vertical one). – Daniel Fischer Oct 5 '14 at 23:29

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 $$v = \pm 2\sqrt {u + 1}$$
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.