應該是這樣吧？
證：\(\displaystyle \frac{log n}{log(n-1)}>\frac{log(n+1)}{log n}\)
pf:
\(\displaystyle \frac{log n^2}{2}>\frac{log(n-1)+log(n+1)}{2}\ge \sqrt{log(n-1)log(n+1)}\)
\(\displaystyle log n>\sqrt{log(n-1)log(n+1)}\)
\((log n)^2>log(n-1)log(n+1)\)

\(
\begin{array}{l}
計算五 第一小題   因為 z^8 + 4 - 5i = (z - z_1 )(z - z_2 )(z - z_3 )...(z - z_8 ) \\
所求=|(z - z_1)(z - z_2 )(z - z_3 )...(z - z_8 )| =|(1 + i)^8 + 4 - 5i| = |20 - 5i| = 5\sqrt {17}
\end{array}
\)

圓的切冪性質