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 <span>(3m-n-6), since a = a
and whatever </span><span>(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
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![\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}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bcccll%7D%0Az1%3D%26%28m-2n-1%29%2Bi%26%285m-4n-6%29%5C%5C%0A%26%5Cuparrow%20%26%5Cuparrow%20%5C%5C%0A%26a%26b%5C%5C%0A%26%5Cdownarrow%20%26%5Cdownarrow%20%5C%5C%0Az2%3D%26%283m-n-6%29%2Bi%26%28m-5n-3%29%0A%5Cend%7Barray%7D%0A%5C%5C%20%5Cquad%20%5C%5C%0A%0A%5Cbegin%7Bcases%7D%0Am-2n-1%3D3m-n-6%5Cimplies%200%3D2m%2Bn-5%5C%5C%0A5m-4n-6%3Dm-5n-3%5Cimplies%204m%2Bn-3%3D0%0A%5Cend%7Bcases%7D%0A%5C%5C%20%5Cquad%20%5C%5C%0A%5Ctextit%7Bnow%2C%20solving%20that%20by%20using%20elimination%7D)
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![\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](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bllcll%7D%0A2m%2Bn-5%3D0%26%5Cleftarrow%20%5Ctimes%20-2%5Cimplies%20%26-4m-2n%2B10%3D0%5C%5C%0A4m%2Bn-3%3D0%26%264m%2Bn-3%3D0%5C%5C%0A%26%26%5Ctextendash%5Ctextendash%5Ctextendash%5Ctextendash%5Ctextendash%5Ctextendash%5Ctextendash%5Ctextendash%5Ctextendash%5Ctextendash%5Ctextendash%5Ctextendash%5Ctextendash%5Ctextendash%5C%5C%0A%26%26-n%2B7%3D0%0A%5Cend%7Barray%7D%0A%5C%5C%20%5Cquad%20%5C%5C%0Aor%5Cqquad%207%3Dn)
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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"</span>