ok... so, we know that both pairs are conjugate of each other,
so.. both are complex solutions, namely, both are a+bi type
what's "a", we dunno, what is "b", we dunno either
but for z1, we know that a = m-2n-1 and b = 5m-4n-6
for z2, a = 3m-n-6 and b=m-5n-3
for those two folks to be conjugate of each other, they'd be
(a+bi)(a-bi)
notice, the "a"s are the same on z1 as well as z2
the "b" are the same also BUT the z2 "b" is negative,
in order to be a conjugate, regardless of that, the "b"s are the same
so, whatever (m-2n-1) is, is the same as (3m-n-6), since a = a
and whatever (5m-4n-6) is, will be the same value, since b =b,
the "a" and "b" in each conjugate, is the same value, thus one can say
\(\begin{array}{cccll}
z1=&(m-2n-1)+i&(5m-4n-6)\\
&\uparrow &\uparrow \\
&a&b\\
&\downarrow &\downarrow \\
z2=&(3m-n-6)+i&(m-5n-3)
\end{array}
\\ \quad \\
\begin{cases}
m-2n-1=3m-n-6\implies 0=2m+n-5\\
5m-4n-6=m-5n-3\implies 4m+n-3=0
\end{cases}
\\ \quad \\
\textit{now, solving that by using elimination}\)
\(\begin{array}{llcll}
2m+n-5=0&\leftarrow \times -2\implies &-4m-2n+10=0\\
4m+n-3=0&&4m+n-3=0\\
&&\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\\
&&-n+7=0
\end{array}
\\ \quad \\
or\qquad 7=n\)
and pretty sure you can find "m" from there
once you have both, substitute in z1 or z2, to get the values for "a" and "b"
