the word is modulus
|Z|=|1Z|=|1−Z|
first off you should know that
|1Z|=1|Z|so |Z|=|1Z|⇒|Z|=1
|1−Z|=1√(1−Z)(1−Z∗)=1(1−Z)(1−Z∗)=11−Z−Z∗+ZZ∗=1
but ZZ∗=|Z|2=1 so1−Z−Z∗+1=1Z+Z∗=12Re(Z)=1Re(Z)=12
1=12=|Z|2=Re(Z)2+Im(Z)2=(12)2+Im(Z)2Im(Z)2=34Im(Z)=±√32Z=12(1±i√3)