Find all real and imaginary solutions to the equation. \( a^{4}-1=0 \) a

To find the solutions for the equation \( a^{4}-1=0 \), we can first rewrite it as \( a^{4} = 1 \). The number 1 can be expressed in exponential form as \( e^{2\pi i k} \), where \( k \) is any integer. For the fourth roots, we can set \( k = 0, 1, 2, 3 \). This gives us the following solutions: 1. \( a = e^{\frac{2\pi i (0)}{4}} = e^{0} = 1 \) 2. \( a = e^{\frac{2\pi i (1)}{4}} = e^{\frac{\pi i}{2}} = i \) 3. \( a = e^{\frac{2\pi i (2)}{4}} = e^{\pi i} = -1 \) 4. \( a = e^{\frac{2\pi i (3)}{4}} = e^{\frac{3\pi i}{2}} = -i \) Thus, the complete set of solutions is \( a = 1, i, -1, -i \). For an extra sprinkle of magic, if you think about roots of unity, these solutions are indeed the 4th roots of 1! They form a neat little square on the complex plane, marking each vertex at equal angles. So, not only do we have real numbers (1 and -1), but also those quirky imaginary buddies (i and -i) hanging out in a perfect symmetrical structure!