回覆 10# johncai 的帖子
（2020 AMC 12A 第22題）
\(\displaystyle \frac{a_0b_0}{7^0} =\)「\(\displaystyle \frac{1}{2}\left(\frac{3+4i}{7}\right)^0\) 」的虛部 =「\(\displaystyle \frac{1}{2}\times 1\) 」的虛部 = 0，
\(\displaystyle \frac{a_1b_1}{7^1} =\)「\(\displaystyle \frac{1}{2}\left(\frac{3+4i}{7}\right)^1\) 」的虛部，
\(\displaystyle \frac{a_2b_2}{7^2} =\)「\(\displaystyle \frac{1}{2}\left(\frac{3+4i}{7}\right)^2\) 」的虛部，
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又 \(\displaystyle \left|\frac{3+4i}{7}\right|<1\)
因此
\(\displaystyle \frac{a_0b_0}{7^0}+\frac{a_1b_1}{7^1}+\frac{a_2b_2}{7^2}+\cdots\)
\(\displaystyle =\)「\(\displaystyle \frac{1}{2}\left(\frac{3+4i}{7}\right)^0+\frac{1}{2}\left(\frac{3+4i}{7}\right)^1+\frac{1}{2}\left(\frac{3+4i}{7}\right)^2+\cdots\) 」的虛部
\(\displaystyle =\)「\(\displaystyle \frac{1}{2}\left(\frac{1}{1-\frac{3+4i}{7}}\right)\) 」的虛部
