I am trying to find a matrix $G$ that satisfies the following relation:
$$\left(\begin{array}{ll}{g_{11}} & {g_{12}} \\ {g_{21}} & {g_{22}}\end{array}\right)=\left(\begin{array}{cc}{-g_{22}} & {-z} \\ {-z^{*}} & {-g_{11}}\end{array}\right)^{-1}$$ Where $z=x+\mathrm{i}y$ with $x,y\in\mathbb{R}$ and $z^*=x-\mathrm{i}y$.
Using the fact that: $$ \left(\begin{array}{cc}{g_{11}} & {g_{12}} \\ {g_{21}} & {g_{22}}\end{array}\right)\left(\begin{array}{cc}{-g_{22}} & {-z} \\ {-z^{*}} & {-g_{11}}\end{array}\right)=\left(\begin{array}{cc}{1} & {0} \\ {0} & {1}\end{array}\right)$$
This is a simple system of 4 equations. We can easily find that two kind of solutions are possible: $$ \begin{array}{l}{g_{21}=-\frac{1}{z} \text { and } g_{12}=\frac{1}{z} \text { and } g_{22}=g_{11}=0} \\ {g_{21}=-z^{*} \text { and } g_{12}=-z \text { and } g_{11} \neq 0 \text { and } g_{22}=\frac{|z|^{2}-1}{g_{11}}}\end{array}$$
However my teacher told us that the second solution appears only when $|z|<1$. When $|z|>1$ only the first solution (with $g_{22}=g_{11}=0$) is possible.
I don't understand why. He told us that we could show it by analysing the stability of the solutions. However since this is not a differential equation I feel a bit confused on how to analyse the stability.
Here is what I tried:
Expanding in powers of $g_{22}$ and $g_{11}$ for small $g_{22}$ and $g_{11}$:
$$\left(\begin{array}{ll}{g_{11}} & {g_{12}} \\ {g_{21}} & {g_{22}}\end{array}\right)\left(\begin{array}{cc}{-g_{22}} & {-z} \\ {-z^{*}} & {-g_{11}}\end{array}\right)^{-1}\approx\left(\begin{array}{cc}{\frac{g_{11}}{|z|^2}} & {\frac{1}{-z^*}-g_{11}g_{22}}\frac{z}{|z|^4} \\ {\frac{1}{-z}-g_{11}g_{22}}\frac{z^*}{|z|^4} & {-\frac{g_{22}}{|z|^2}}\end{array}\right)$$ Therefore $$ \begin{cases} g_{11}=\frac{g_{11}}{|z|^2} \\ g_{22}=-\frac{g_{22}}{|z|^2} \end{cases}$$ It seems to me that the second case of solutions ($g_{21}=-z^*$ with $g_{11}\neq 0$) is valid even when $|z|>1$... Any ideas?