1. **State the problem:** We need to find the complex number $z = a + bi$ such that $$z^2 = -77 + 36i,$$ where $i = \sqrt{-1}$ and $a,b$ are real numbers. 2. **Use the formula:** If $z = a + bi$, then $$z^2 = (a + bi)^2 = a^2 + 2abi + (bi)^2 = a^2 - b^2 + 2abi.$$ Here, we used $i^2 = -1$. 3. **Equate real and imaginary parts:** From $$z^2 = a^2 - b^2 + 2abi = -77 + 36i,$$ we get two equations: $$a^2 - b^2 = -77$$ $$2ab = 36$$ 4. **Solve for $b$ in terms of $a$:** From $2ab = 36$, $$b = \frac{36}{2a} = \frac{18}{a}.$$ 5. **Substitute $b$ into the first equation:** $$a^2 - \left(\frac{18}{a}\right)^2 = -77$$ $$a^2 - \frac{324}{a^2} = -77$$ 6. **Multiply both sides by $a^2$ to clear denominator:** $$a^2 \times a^2 - a^2 \times \frac{324}{a^2} = -77 a^2$$ $$a^4 - 324 = -77 a^2$$ 7. **Rearrange to form a quadratic in $a^2$:** $$a^4 + 77 a^2 - 324 = 0$$ Let $x = a^2$, then $$x^2 + 77x - 324 = 0$$ 8. **Solve quadratic equation:** $$x = \frac{-77 \pm \sqrt{77^2 - 4 \times 1 \times (-324)}}{2} = \frac{-77 \pm \sqrt{5929 + 1296}}{2} = \frac{-77 \pm \sqrt{7225}}{2}$$ $$\sqrt{7225} = 85,$$ so $$x = \frac{-77 \pm 85}{2}.$$ Two solutions: $$x_1 = \frac{-77 + 85}{2} = \frac{8}{2} = 4,$$ $$x_2 = \frac{-77 - 85}{2} = \frac{-162}{2} = -81$$ (discard since $a^2$ cannot be negative). 9. **Find $a$:** $$a = \pm \sqrt{4} = \pm 2.$$ 10. **Find $b$ using $b = \frac{18}{a}$:** If $a = 2$, then $$b = \frac{18}{2} = 9.$$ If $a = -2$, then $$b = \frac{18}{-2} = -9.$$ 11. **Write solutions:** $$z = 2 + 9i \quad \text{or} \quad z = -2 - 9i.$$ **Final answer:** $z = 2 + 9i$ or $z = -2 - 9i$. This corresponds to option C.