z=r(cosθ+isinθ)　とおきます。
\(\frac{1}{z}\)=(\(\frac{1}{r}\)){cos(-θ)+isin(-θ)}
=(\(\frac{1}{r}\))(cosθ-isinθ)
\(z^{2}\)=\(r^{2}\)(cos2θ+isin2θ)
1/(\(z^{2}\))={1/(\(r^{2}\))}(cos2θ-isin2θ)
よって
\(z^{2}\)+(1/\(z^{2}\))
={\(r^{2}\)+(1/\(r^{2}\))}cos2θ+i{\(r^{2}\)-(1/\(r^{2}\))}sin2θ
\(z^{2}\)+(1/\(z^{2}\))は実数だから
\(r^{2}\)-(1/\(r^{2}\))=0
よって　\(r^{2}\)=1
z={x+y-(\(\frac{1}{2}\))}+(x-y)i で　|z|=r なのだから
\(r^{2}\)={x+y-(\(\frac{1}{2}\))}^2+(x-y\()^{2}\)=1
2\(x^{2}\)+2\(y^{2}\)-x-y-(\(\frac{3}{4}\))=0
\(x^{2}\)+\(y^{2}\)-(\(\frac{x}{2}\))-(\(\frac{y}{2}\))-(\(\frac{3}{8}\))=0
{x-(\(\frac{1}{4}\))}^2+{y-(\(\frac{1}{4}\))}^2=\(\frac{1}{2}\)
よって
中心(\(\frac{1}{4}\),\(\frac{1}{4}\))　半径1/\(\sqrt{\quad}\)2　の円周上にある。

from UnderBird

\(z^{2}\)+(1/\(z^{2}\))が実数⇔\(z^{2}\)+(1/\(z^{2}\))=conj{\(z^{2}\)+(1/\(z^{2}\))}
conjは共役複素数の意味
また、z*conj(z)=|z|^2を用いると
{\(z^{2}\)-conj(z\()^{2}\)}(|z|^2-1)=0より
|z|=1,z=conj(z),z=-conj(z)
zは順に中心１半径１の円、実数、純虚数であればよい。
z=(x+y-\(\frac{1}{2}\))+(x-y)iから
順に(x+y-\(\frac{1}{2}\)\()^{2}\)+(x-y\()^{2}\)=1,x-y=0,x+y-\(\frac{1}{2}\)=0
よって
(x-\(\frac{1}{4}\)\()^{2}\)+(y-\(\frac{1}{4}\)\()^{2}\)=(1/\(\sqrt{\quad}\)2\()^{2}\),y=x,y=-x+\(\frac{1}{2}\)