Calculate $\frac{i^2-i^{31}+(4+3i)(5-2i)}{\frac{1}{2}i}$ without a calculator I tried:
$$\frac{i^2-i^{31}+(4+3i)(5-2i)}{\frac{1}{2}i} = \\
\frac{-1-(-1)+(4+3i)(5-2i)}{\frac{1}{2}i}= \\
\frac{26+7i}{\frac{1}{2}i} = \\
\frac{52+14i}{i} = \\
\frac{(52+14i)\cdot-i}{i\cdot -i} = \\ \\
14-52i$$
But my book states the solution is $16-50i$. What went wrong?

Numerator: $-1+i+20-8i+15i+6$, which simplifies to $25+8i$. Can you divide this by $\frac{i}{2}$? 
