For any complex number $z$,$\bar{z} = \frac{1}{z}$ if and only if

The complex number $\frac{1+2i}{1-i}$ lies in

If $(a+ib)(c+id)(e+if)(g+ih)=A+iB,$ then show that: $(a^{2}+b^{2})(c^{2}+d^{2})(e^{2}+f^{2})(g^{2}+h^{2})=A^{2}+B^{2}$

If $m_1, m_2, m_3$ and $m_4$ respectively denote the moduli of the complex numbers $1+4i, 3+i, 1-i$ and $2-3i$,then the correct relation among the following is:

If $(a + ib)(c + id)(e + if)(g + ih) = A + iB$,then $({a^2} + {b^2})({c^2} + {d^2})({e^2} + {f^2})({g^2} + {h^2}) =$

Let $z = a - \frac{i}{2}$,where $a \in R$. Then $|i + z|^2 - |i - z|^2$ is equal to