Distinguish a modulus square from a complex square

From the first equation, $z=10-w^*-5i$. Substitution into the second gives $|w|^2+w^*=28-4i$. Put $w=a+bi$, where $a,b$ are real. Then $|w|^2=a^2+b^2$, $w^*=a-bi$, so $a^2+b^2+a-bi=28-4i$. Comparing imaginary parts gives $b=4$; real parts give $a^2+16+a=28$, hence $(a+4)(a-3)=0$.

If $a=3$, $w=3+4i$ and $z=10-(3-4i)-5i=7-i$. If $a=-4$, $w=-4+4i$ and $z=10-(-4-4i)-5i=14-i$. These are paired solutions, not four independent combinations. Their squared moduli are 25 and 32, matching $z+18+i$ respectively.

Factor $z^2-5z+7+i=[z-(2+i)](z-k)$. Comparing the coefficient of $z$ gives $-5=-k-(2+i)$, so $k=3-i$. Its product with $2+i$ is $7+i$, confirming the constant term.

Multiply the equation by $-i$: $-w^2-5iw+7+i=0$, which is $(iw)^2-5(iw)+7+i=0$. Hence $iw=2+i$ or $iw=3-i$. Divide by $i$ to obtain $w=1-2i$ or $w=-1-3i$.