ここで勉強すれば数学検定１級の壁は超えられるか。

$\left(2+2\sqrt {3}i\right) z^{2}=\left| z\right| ^{2}+1$

両辺の絶対値を求めると

$\left| 2+2\sqrt {3}i\right| \left| z^{2}\right| =\left| z^{2}\right| +1$

左辺の$\begin{aligned}\left| 2+2\sqrt {3}i\right| =\sqrt {2^{2}+12}=4\end{aligned}$

したっがって、

$4\left| z\right| ^{2}=\left| z\right| ^{2}+1$

$\begin{aligned}3\left| z\right| ^{2}=1\\ \left| z\right| ^{2}=\dfrac {1}{3}\end{aligned}$

この値を最初の式に代入すると

$\begin{aligned}4\left( \dfrac {1+\sqrt {3}i}{2}\right) z^{2}=\dfrac {4}{3}\\ z^{2}=\dfrac {1}{3}\cdot \dfrac {2}{1+\sqrt {3}i}\end{aligned}$

（この上の計算の詳細は下の参照事項に）
$\begin{aligned}z=\pm \sqrt {\dfrac {1}{3}}\cdot \sqrt {\dfrac {1-\sqrt {3}i}{2}}\\ =\pm \dfrac {1}{\sqrt {3}}\cdot \dfrac {\sqrt {3}-i}{2}=\pm \left( \dfrac {1}{2}-\dfrac {i}{2\sqrt {3}}\right) \end{aligned}$

参照事項

$\begin{aligned}\cdot \\ \dfrac {2}{1+\sqrt {3}i}=\dfrac {2}{1+\sqrt {3}i}\cdot \dfrac {1-\sqrt {3}i}{1-\sqrt {3}i}\\ =\dfrac {1-\sqrt {3}i}{2}\end{aligned}$