ここで勉強すれば数学検定１級の壁は超えられるか。

 

 

 

（１）

 

ＡＢ＝ＡＣより

 

$\left| z^{2}-z\right| =\left| z^{2}-2z\right| \Rightarrow \left| z\right| \left| z-1\right| =\left| z\right| \left| z-2\right|$

 

 

$\left| z-1\right| =\left| z-2\right|$

 

 

 

また、∠Ｃ＝９０°より

 

$\dfrac {z^{2}-z}{z^{2}-2z}=\pm i\Rightarrow \dfrac {z-1}{z-2}=\pm i$

 

 

 

$z-1=\pm i\left( z-2\right) \Rightarrow z \mp zi=1\pm 2i$

 

 

 

$z=\dfrac {1\pm 2i}{1\mp i}\times \dfrac {1\pm i}{1\pm i}=\dfrac {3\pm i}{2}$

 

 

$\left| z\right| ^{2}=\left| \dfrac {3\pm i}{2}\right| ^{2}=\left( \dfrac {3}{2}\right) ^{2}+\dfrac {\left( 1\right) ^{2}}{2}=\dfrac {10}{4}$

 

 

$\left| z-1\right| ^{2}=\left| \dfrac {1\pm i}{2}\right| ^{2}=\dfrac {1}{4}+\dfrac {1}{4}=\dfrac {1}{2}$

 

 

△ＡＢＣ＝$\dfrac {1}{2}\times AB\times AC=\dfrac {AC^{2}}{2}$

 

 

$=\dfrac {\left| z\right| ^{2}\cdot \left| z-1\right| ^{2}}{2}=\dfrac {\dfrac {10}{4}\times \dfrac {1}{2}}{2}=\dfrac {5}{8}$・・・（１)の答え

 

 

 

(2)

 

 

ＡＣ＝２ＢＣより

 

$\left| z^{2}-z\right| =2\left| z^{2}-2z\right|$　

 

 

$\left| z-1\right| =2\left| z-2\right|$

 

 

 

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

 

 

 

$\left| z-1\right| \left| \overline {z-1}\right| =4\left| z-2\right| \left| \overline {z-2}\right|$　

 

 

$\left( z-1\right) \left( \overline {z}-1\right) =4\left( z-2\right) \left( \overline {z}-2\right)$

 

 

 

$z\overline {z}-z-\overline {z}+1=4\left( z\overline {z}-2z-2\overline {z}+4\right)$

 

 

$3z\overline {z}-7z-7\overline {z}+15=0$

 

 

両辺を３で割ると

 

 

$z\overline {z}-\dfrac {1}{3}z-\dfrac {1}{3}\overline {z}+5=0$

 

 

$\left( z-\dfrac {2}{3}\right) \left( \overline {z}-\dfrac {1}{3}\right) =-15+\dfrac {49}{9}=\dfrac {4}{9}$　

 

 

 

$\left| z-\dfrac {7}{3}\right| ^{2}=\left( \dfrac {2}{3}\right) ^{2}$

 

 

$\left| z-\dfrac {1}{3}\right| =\dfrac {2}{3}$

 

 

$\dfrac {2}{3}$・・・（２）の答え