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\begin{document}

$$
\left \lbrack \begin{array}{cc}
x & y \\
\overline{y} & z
\end{array} \right \rbrack
\left \lbrack \begin{array}{c}
u_{1,j} \\
u_{2,j}
\end{array} \right \rbrack
=
\left \lbrack \begin{array}{c}
a_{1,j} \\
a_{2,j}
\end{array} \right \rbrack
$$
Multiply on the left by the inverse of the pivot block.
$$
\left \lbrack \begin{array}{c}
u_{1,j} \\
u_{2,j}
\end{array} \right \rbrack
=
\frac{1}{xz - y\overline{y}}
\left \lbrack \begin{array}{cc}
z & -y \\
-\overline{y} & x
\end{array} \right \rbrack
\left \lbrack \begin{array}{c}
a_{1,j} \\
a_{2,j}
\end{array} \right \rbrack
=
\frac{1}{xz - y\overline{y}}
\left \lbrack \begin{array}{c}
z a_{1,j} - y a_{2,j} \\
- \overline{y} a_{1,j} + x a_{2,j}
\end{array} \right \rbrack
$$
Look at each of the two entries in the rightmost vector.
\begin{eqnarray*}
z a_{1,j} - y a_{2,j} 
& = & 
\Re(z) ( \Re(a_{1,j}) + i\  \Im(a_{1,j}) )
- ( \Re(y) + i\  \Im(y) )( \Re(a_{2,j}) + i\  \Im(a_{2,I}) )
\\
& = & 
( 
\Re(z) \Re(a_{1,j}) - \Re(y) \Re(a_{2,j}) + \Im(y) \Im(a_{2,I}) 
) \\
& & 
\qquad +\  i( 
\Re(z) \Im(a_{1,j}) - \Re(y) \Im(a_{2,I}) - \Im(y) \Re(a_{2,j})
)
\\
- \overline{y} a_{1,j} + x a_{2,j} 
& = &
-( \Re(y) - i \Im(y) ) ( \Re(a_{1,j}) + i \Im(a_{1,j}) )
+ \Re(x) ( \Re(a_{2,j}) + i \Im(a_{2,I}) )
\\
& = & 
(
-\Re(y) \Re(a_{1,j}) - \Im(y) \Im(a_{1,j}) + x \Re(a_{2,j})
) \\
& & 
\qquad + i\  (
-\Re(y) \Im(a_{1,j}) + \Im(y) \Re(a_{1,j}) + x \Im(a_{2,I})
)
\end{eqnarray*}
\begin{eqnarray*}
\Re(u_{1,j}) 
& = &
\frac{ \Re(z) \Re(a_{1,j}) - \Re(y) \Re(a_{2,j}) 
+ \Im(y) \Im(a_{2,j}) }
{ xz - y\overline{y} }
\\
\Im(u_{1,j}) 
& = &
\frac{ \Re(z) \Im(a_{1,j}) - \Re(y) \Im(a_{2,I}) - \Im(y) \Re(a_{2,j}) }
{ xz - y\overline{y} }
\end{eqnarray*}
Taking absolute values and use the triangle inequality.
\begin{eqnarray*}
\Re(u_{1,j}) 
& = &
\left |
\frac{ \Re(z) \Re(a_{1,j}) - \Re(y) \Re(a_{2,j}) 
+ \Im(y) \Im(a_{2,j}) }
{ xz - y\overline{y} }
\right |
\\
& \le &
\frac{
|\Re(z)|\ |\Re(a_{1,j})| + |\Re(y)|\ |\Re(a_{2,j})|
+ |\Im(y)|\ |\Im(a_{2,j})|
} {
\left | xz - y\overline{y} \right |
}
\\
\max_{j} |\Re(u_{1,j})|
& \le &
\frac{
|\Re(z)|\ \max_j|\Re(a_{1,j})| + |\Re(y)|\ \max_j|\Re(a_{2,j})|
+ |\Im(y)|\ \max_j|\Im(a_{2,j})|
} {
\left | xz - y\overline{y} \right |
}
\\
\end{eqnarray*}

\end{document}
