150. Find: $i^{28} ; i^{33} ; i^{135}$.
Solution. We have $28=4 \cdot 7$ (no remainder); $33=4 \cdot 8+1$; $135=$ $=4 \cdot 33+3$. Accordingly, we get $i^{28}=1$; $i^{33}=i$; $i^{135}=-i$.

151-157. Calculate: