Recall that, #e^(itheta)=costheta+isintheta..............(square)#.
#:. z=((e^(2+ipi/2))/(1+3i))^2#,
#=(e^(2+ipi/2))^2/(1+3i)^2#,
#=e^(2*(2+ipi/2))/(1+3i)^2#,
#=e^(4+ipi)/(1+3i)^2#,
#=(e^4*e^(ipi))/(1+3i)^2#,
#={e^4*(cospi+isinpi)}/(1+3i)^2#,
#={e^4(-1+i*0)}/(1+3i)^2#,
#=-e^4*1/(1+3i)^2*(1-3i)^2/(1-3i)^2#,
#=-{e^4(1-3i)^2}/{(1+3i)(1-3i)}^2#,
#=-{e^4(1-3i)^2}/(1-9i^2)^2#,
#=-(e^4(1-6i+9i^2))/{1-9(-1)}^2#,
#=-(e^4(1-6i-9))/(10)^2#,
#=-(e^4(-8-6i))/100#,
#=(e^4(4+3i))/50#.
#rArr Rl(z)=2/25e^4, and Im(z)=3/50e^4#.