If `z=(a+ib)^5+(b+ia)^5` then prove that `Re(z)=Im(z),` where `a,b in R.`

If z!=0 is a complex njmber,then prove that Re(z)=0rArr Im(z^(2))=0

If z is a complex number such that Re(z)=Im(2), then

If z=a+ib and |z-2|=|2z-1|prove that a^2-b^2=1

If z=a+ib and Z=A+iB then show thatif z=(i(Z+1))/(z-1) then a^(2)+b^(2)-a=((A^(2)+B^(2)+2A-2B+1)/((A-1)^(2)+B^(2)))