jmc

Let $z=a+bi.$ Then $$z^2=(a+bi{)}^2=a^2+2abi+b^2{i}^2=a^2+2ab-b^2.$$We want this to equal $-77-36i.$ Setting the real and imaginary parts equal, we get $$\begin{array}{rl}{a}^2-b^2& =-77,\\ 2ab& =-36,\end{array}$$so $ab=-18.$ Then $b=-\frac{18}{a}.$ Substituting, we get $$a^2-\frac{324}{a^2}=-77,$$so $a^4+77a^2-324=0.$ This factors as $(a^2-4)(a^2+81)=0,$ so $a^2=4.$

If $a=2,$ then $b=-\frac{18}{a}=-9.$ If $a=-2,$ then $b=-\frac{18}{a}=9.$ Therefore, the solutions are $\boxed{2-9i,-2+9i}.$