Opere: a. 1+i 1−i − 3+3 i 1+i b. −1− i√3 √3−1 c. 1+3 i1− i −2. 2+3 i 1−2i

Para resolver as operações fornecidas, primeiro precisamos simplificar cada expressão: a. \[ \frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1 + 2i + i^2}{1 - i^2} = \frac{1 + 2i - 1}{1 + 1} = \frac{2i}{2} = i \] \[ \frac{-3+3i}{1+i} = \frac{3(-1+i)}{1+i} = \frac{3(-1+i)}{1+i} = \frac{-3+3i}{1+i} = 3 \] b. \[ \frac{-1-i\sqrt{3}}{\sqrt{3}-1} = \frac{(-1-i\sqrt{3})(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{-\sqrt{3}-1-\sqrt{3}i-i}{3-1} = \frac{-2\sqrt{3}-2-\sqrt{3}i-i}{2} = -\sqrt{3}-1-\frac{1+\sqrt{3}}{2}i \] c. \[ \frac{1+3i}{1-i} - 2 = \frac{(1+3i)(1+i)}{(1-i)(1+i)} - 2 = \frac{1 + 4i + 3i^2}{1 - i^2} - 2 = \frac{1 + 4i - 3}{1 + 1} - 2 = \frac{-2 + 4i}{2} - 2 = 2i - 3 \] Por lo tanto, las respuestas son: a. i b. \(-\sqrt{3}-1-\frac{1+\sqrt{3}}{2}i\) c. \(2i - 3\)