### Instruction: Find all complex numbers $z$ such that
\[z^2 = -77 - 36i.\]Enter all complex numbers, separated by commas.### Response: I need to find two complex numbers that, when squared, give me $-77 - 36i$.
I can write any complex number $z$ in the form $z = a + bi$, where $a$ and $b$ are real numbers and $i$ is the imaginary unit that satisfies $i^2 = -1$.
Squaring both sides, I get $(a + bi)^2 = a^2 - b^2 + 2abi$.
Now I can compare this expression with the given one and try to match the real and imaginary parts.
That means I need to solve the system of equations $a^2 - b^2 = -77$ and $2ab = -36$.
I can start by solving for $a$ in terms of $b$ from the second equation: $a = -18/b$.
Then I can substitute this into the first equation and get $(-18/b)^2 - b^2 = -77$.
This simplifies to $324/b^2 - b^2 = -77$.
Multiplying by $b^2$, I get $324 - b^4 = -77b^2$.
Rearranging, I get $b^4 - 77b^2 - 324 = 0$.
This is a quartic equation in $b^2$, but I can factor it as $(b^2 - 81)(b^2 + 4) = 0$.
So $b^2 = 81$ or $b^2 = -4$.
Taking square roots, I get $b = \pm 9$ or $b = \pm 2i$.
But $b$ has to be a real number, so I can ignore the imaginary values and only keep $b = \pm 9$.
Then, using $a = -18/b$, I get $a = \mp 2$.
So the two complex numbers that satisfy the equation are $z = -2 + 9i$ and $z = 2 - 9i$.