1. The problem asks for the value of the 4th root of -81, which means finding a number $x$ such that $$x^4 = -81.$$ 2. Recall that raising any real number to the 4th power results in a non-negative number because $$x^4 = (x^2)^2 \geq 0$$ for all real $x$. 3. Since $-81$ is a negative number, there is no real number $x$ such that $$x^4 = -81.$$ 4. However, if we consider complex numbers, we solve $$x^4 = -81.$$ Write $-81$ as $$81 \times (-1) = 81 \times e^{i\pi}.$$ 5. Taking the 4th root means taking the 4th root of magnitude and dividing the argument by 4: $$|x| = \sqrt[4]{81} = 3,$$ and the argument $$\theta = \frac{\pi + 2k\pi}{4}, \quad k=0,1,2,3.$$ 6. The 4 distinct complex roots are then: $$3 e^{i\frac{\pi}{4}}, 3 e^{i\frac{3\pi}{4}}, 3 e^{i\frac{5\pi}{4}}, 3 e^{i\frac{7\pi}{4}}.$$ 7. Writing in rectangular form: - For $k=0$: $$3 \left(\cos\frac{\pi}{4} + i \sin\frac{\pi}{4}\right) = 3 \left(\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}\right) = \frac{3\sqrt{2}}{2} + i \frac{3\sqrt{2}}{2}.$$ - For $k=1$: $$3 \left(\cos\frac{3\pi}{4} + i \sin\frac{3\pi}{4}\right) = 3 \left(-\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}\right) = -\frac{3\sqrt{2}}{2} + i \frac{3\sqrt{2}}{2}.$$ - For $k=2$: $$3 \left(\cos\frac{5\pi}{4} + i \sin\frac{5\pi}{4}\right) = 3 \left(-\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}\right) = -\frac{3\sqrt{2}}{2} - i \frac{3\sqrt{2}}{2}.$$ - For $k=3$: $$3 \left(\cos\frac{7\pi}{4} + i \sin\frac{7\pi}{4}\right) = 3 \left(\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}\right) = \frac{3\sqrt{2}}{2} - i \frac{3\sqrt{2}}{2}.$$ Final answer: The 4th roots of $-81$ are the four complex numbers $$\frac{3\sqrt{2}}{2} \pm i \frac{3\sqrt{2}}{2}$$ and $$-\frac{3\sqrt{2}}{2} \pm i \frac{3\sqrt{2}}{2}.$$