上次问了一个题目，非常感谢 蒟蒻365 （其实是神犇）为我解答，这次弱弱的我又来提问了：

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题目：

已知$(\sqrt{3}+i)^m=(1+i)^n,m,n\in Z^+$，则 $mn$ 的最小值为_____

我的解答：

$(\sqrt{3}+i)^m=[2(\frac{\sqrt{3}}{2}+\frac{i}{2})]^m=[2(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6})]^m=2^m(\cos(\frac{\pi}{6}m)+i\sin(\frac{\pi}{6}m))$

$(1+i)^n=[\sqrt{2}(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)]^n=[2(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})]^n=\sqrt{2}^n(\cos(\frac{\pi}{4}n)+i\sin(\frac{\pi}{4}n))$

$\therefore2^m(\cos(\frac{\pi}{6}m)+i\sin(\frac{\pi}{6}m))=\sqrt{2}^n(\cos(\frac{\pi}{4}n)+i\sin(\frac{\pi}{4}n))$

$\therefore\begin{cases}2^m(\cos(\frac{\pi}{6}m))=\sqrt{2}^n(\cos(\frac{\pi}{4}n))\cdots1\\2^m(\sin(\frac{\pi}{6}m))=\sqrt{2}^n(\sin(\frac{\pi}{4}n))\cdots2\end{cases}$

2式除以1式：$\tan(\frac{\pi}{6}m)=\tan(\frac{\pi}{4}n)$

$\therefore\frac{\pi}{6}m=\frac{\pi}{4}n+k\pi,k\in Z\rightarrow\frac{m}{6}=\frac{n}{4}+k,k\in Z\rightarrow12|2m-3n$

1式 * 1式 + 2式 * 2式：$n=2m$

$\therefore12|-4n\rightarrow3|n\rightarrow n_{min}=6$，此时$m=3,nm_{min}=18$

但正确答案是72，不知道我哪里错了$QwQ$

求查错