1√2
12
1√7
13
|α−z₀|=r. (1)
|1/α−z₀|=2r (2)
2|z₀|²=r²+2. (3)
α.1/α=|α|² ⇒1/α=α/|α|²
Putting 1/α=α/|α|² in (2), we get
|α/|α|²−z₀|=2r (4)
Squaring (1), we get, |α|²+|z₀|²−(αz₀*+αz₀)=r² ⇒|α|²+(r²/2+1)−(αz₀+αz₀)=r² (5)
Squaring (4), we get, |α|⁴+|z₀|²−(αz₀/|α|²+αz₀/|α|²)=4r² ⇒1/|α|²(1−(αz₀+αz₀))=4r²−|z₀|²=4r²−(r²/2+1)=7r²/2
⇒1−(αz₀+αz₀)=|α|²(7r²/2)
⇒αz₀+α*z₀=1−|α|²(7r²/2)
Substituting in (5), we get
|α|²+(r²/2+1)−(1−|α|²(7r²/2))=r²
⇒r²/2+|α|²(7r²/2)=r²
⇒|α|²(7r²/2)=r²/2
⇒|α|²=1/7
∴ |α|=1/√7