Proposition

Find the values of

1. $(1 + 2i)^3$.
2. $\frac{5}{-3 + 4i}$.
3. $\big(\frac{2 + i}{3 - 2i}\big)^2$.
4. $(1 + i)^n + (1 - i)^n$.

Solution

1. $(1 + 2i)^3 = 1 + 3 \cdot 2i + 3 \cdot (2i)^2 + (2i)^3 = -11 - 2i$.
2. $5 / (-3 + 4i) = 5(3 + 4i) / (-9 - 16) = (3 + 4i) / -5$.
3. \[\begin{align*} \Big(\frac{2 + i}{3 - 2i}\Big)^2 &= \Big(\frac{(2 + i)(3 + 2i)}{9 + 4}\Big)^2 \\ &= \Big(\frac{4 + 7i}{13}\Big)^2 \\ &= \frac{-33 + 56i}{169}. \end{align*}\]
4. \[\begin{align*} (1 + i)^n + (1 - i)^n &= \sum_{k=0}^{n} \binom{n}{k} (i^k + (-1)^k) \\ &= 2\sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{2k} i^{2k} \\ &= 2\sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{2k} (-1)^k. \end{align*}\]