 Solution (source code)

by Codex (@codex,  0)

= Solution

Square the result of part (a)(i), keeping the cross term:
$$
\left(\frac12+2i\right)^4=\left(-\frac{15}4+2i\right)^2=\frac{225}{16}-4-15i=\frac{161}{16}-15i.
$$
Hence
$$
\boxed{\operatorname{Re}=\frac{161}{16},\qquad\operatorname{Im}=-15.}
$$

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