Find a value $c$ such that $\left\|\begin{pmatrix} x^2 - y^2\\2xy \end{pmatrix}\right\| \leq |c|\left\|\begin{pmatrix} x\\y \end{pmatrix}\right\|$ This is part of a larger problem where I am trying to find the derivative of a vector valued function. I feel like I'm missing something simple. NOTE: $c$ can be a function of $x$ and $y$.
 A: If you think about where your quadratic function comes from, it is the representation of $z=(x+iy)^2 = (x^2-y^2)+i(2xy)$ where $z=x+iy$. So you want to be considering
$$|z^2| = |z|^2.$$
That is, your "constant" needs to be $|z|$.
A: If $c$ is allowed to be a function of $x,y$ as is stated in the comments then the following would work.
In polar coordiantes the following:
$$\left\|\begin{pmatrix} x^2 - y^2\\2xy \end{pmatrix}\right\| \leq |c|\left\|\begin{pmatrix} x\\y \end{pmatrix}\right\|$$
is:
$$x=r\cos(\theta),\quad y=r\sin(\theta)$$
$$\left\|\begin{pmatrix} r^2\cos(2\theta)\\r^2\sin(2\theta)\end{pmatrix}\right\| \leq |c|\left\|\begin{pmatrix} r\cos(\theta)\\r\sin(\theta) \end{pmatrix}\right\|$$
$$r^2 \leq |c|r$$
so I guess choose $$c=kr=k\sqrt{x^2+y^2}$$
for any constant $k\ge 1$ and with equality holding only for $k=1$. 
Perhaps the condition is expected to hold over some finite circle of radius $r$  (open set) rather than all of space, in which case you can chose any constant $c\ge r$, and it need not be a function of $x,y$.
A: We compute
$$(x^2 - y^2)^2 + (2xy)^2 = x^{4} - 2x^{2}y^{2}+ y^{4} + 4x^{2}y^{2} = x^{4} + 2x^2 y^2 + y^4 = (x^2 + y^2)^2$$
Hence
$$\sqrt{(x^2 - y^2)^2 + (2xy)^2} = \sqrt{(x^2 + y^2)^2} = x^2 + y^2$$
We are trying to find $c$ for which
$$ (x^2 + y^2) \leq c \sqrt{x^2 + y^2}$$
If we let $z = x^2 + y^2$ we obtain
$$ z \leq c \sqrt{z}$$
Does this seem possible?
