a1 23 41 5 2i6 7cdot 8-2i9 10 1112

Apuntes

Let's solve each of the expressions step by step. **a)** \( (1 + 2i) \cdot (-2i) \) To multiply, we distribute: \[ 1 \cdot (-2i) + 2i \cdot (-2i) = -2i - 4i^2 \] Since \( i^2 = -1 \), we have: \[ -4i^2 = -4(-1) = 4 \] Thus, combining the terms: \[ -2i + 4 = 4 - 2i \] So, \[ (1 + 2i) \cdot (-2i) = 4 - 2i \] --- **b)** \( \frac{(1 + 2i)}{3i} \) To simplify, we multiply the numerator and denominator by the conjugate of the denominator, which is \(-3i\): \[ \frac{(1 + 2i)(-3i)}{(3i)(-3i)} = \frac{-3i - 6i^2}{-9} \] Again, since \( i^2 = -1 \): \[ -6i^2 = -6(-1) = 6 \] Thus, the numerator becomes: \[ -3i + 6 = 6 - 3i \] Now, we simplify: \[ \frac{6 - 3i}{-9} = -\frac{2}{3} + \frac{1}{3}i \] So, \[ \frac{(1 + 2i)}{3i} = -\frac{2}{3} + \frac{1}{3}i \] --- **c)** \( (2 + 2i)^{2} \) We can expand this using the formula \( (a + b)^2 = a^2 + 2ab + b^2 \): \[ (2 + 2i)^{2} = 2^2 + 2 \cdot 2 \cdot 2i + (2i)^2 \] Calculating each term: \[ = 4 + 8i + 4i^2 \] Since \( i^2 = -1 \): \[ 4i^2 = 4(-1) = -4 \] Thus, we have: \[ 4 + 8i - 4 = 8i \] So, \[ (2 + 2i)^{2} = 8i \] --- **d)** \( (2 + 2i) - (1 - 3i) \) Distributing the negative sign: \[ (2 + 2i) - 1 + 3i = (2...