Question 1103991
{{{z=a+bi}}}
{{{abs(z)=sqrt(a^2+b^2)}}}
{{{(abs(z))^2=a^2+b^2}}}
and
{{{z^2=(a+bi)(a+bi)=a^2+abi+abi+b^2i^2=(a^2-b^2)+(2ab)I}}}
{{{abs(z^2)=sqrt((a^2-b^2)^2+(2ab)^2)=sqrt(a^4-2a^b^2+b^4+4a^b^2)=sqrt(a^4+2a^b^2+b^4)=sqrt((a^2+b^2)^2)=a^2+b^2}}}
So,
{{{(abs(z))^2=abs(z^2)}}}