28. Вычислить:

Пошаговое решение:

1. 1) (3-2i)²
   \[ (3-2i)^2 = 3^2 - 2(3)(2i) + (2i)^2 = 9 - 12i + 4i^2 = 9 - 12i - 4 = 5 - 12i \]
2. 2) (1+2i)³
   \[ (1+2i)^3 = 1^3 + 3(1^2)(2i) + 3(1)(2i)^2 + (2i)^3 \]
   \[ = 1 + 6i + 3(4i^2) + 8i^3 \]
   \[ = 1 + 6i + 12(-1) + 8(-i) \]
   \[ = 1 + 6i - 12 - 8i \]
   \[ = -11 - 2i \]
3. 3) (1+i)⁴
   \[ (1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i \]
   \[ (1+i)^4 = ((1+i)^2)^2 = (2i)^2 = 4i^2 = -4 \]
4. 4) (1-i)⁶
   \[ (1-i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i \]
   \[ (1-i)^6 = ((1-i)^2)^3 = (-2i)^3 = -8i^3 = -8(-i) = 8i \]
5. 5) (1+i)³ - (1-i)³
   Из предыдущих пунктов:
   \[ (1+i)^3 = 1 + 3i + 3i^2 + i^3 = 1 + 3i - 3 - i = -2 + 2i \]
   \[ (1-i)^3 = 1 - 3i + 3i^2 - i^3 = 1 - 3i - 3 - (-i) = -2 - 2i \]
   \[ (-2 + 2i) - (-2 - 2i) = -2 + 2i + 2 + 2i = 4i \]
6. 6) (1/2 + i√3/2)² + (1/2 - i√3/2)²
   Заметим, что \(rac{1}{2} + rac{i ext{√}3}{2} = e^{irac{ ext{π}}{3}}\) и \(rac{1}{2} - rac{i ext{√}3}{2} = e^{-irac{ ext{π}}{3}}\).
   \((e^{irac{ ext{π}}{3}})^2 + (e^{-irac{ ext{π}}{3}})^2 = e^{irac{2 ext{π}}{3}} + e^{-irac{2 ext{π}}{3}} = 2 ext{cos}(rac{2 ext{π}}{3}) = 2(-rac{1}{2}) = -1\)
   Или, без использования экспоненциальной формы:
   \[ (\frac{1}{2} + \frac{i\sqrt{3}}{2})^2 = \frac{1}{4} + 2(\frac{1}{2})(\frac{i\sqrt{3}}{2}) + (\frac{i\sqrt{3}}{2})^2 = \frac{1}{4} + \frac{i\sqrt{3}}{2} + \frac{-3}{4} = -\frac{2}{4} + \frac{i\sqrt{3}}{2} = -\frac{1}{2} + \frac{i\sqrt{3}}{2} \]
   \[ (\frac{1}{2} - \frac{i\sqrt{3}}{2})^2 = \frac{1}{4} - 2(\frac{1}{2})(\frac{i\sqrt{3}}{2}) + (\frac{i\sqrt{3}}{2})^2 = \frac{1}{4} - \frac{i\sqrt{3}}{2} + \frac{-3}{4} = -\frac{2}{4} - \frac{i\sqrt{3}}{2} = -\frac{1}{2} - \frac{i\sqrt{3}}{2} \]
   \[ (- \frac{1}{2} + \frac{i\sqrt{3}}{2}) + (- \frac{1}{2} - \frac{i\sqrt{3}}{2}) = -\frac{1}{2} - \frac{1}{2} = -1 \]