Fast way to solve $(s-2i)^2 (s+2i)^2$ $(s-2i)^2 (s+2i)^2=$
$(s^2-4si-4) (s^2+4si-4)=$
$s^4+4s^3i-4s^2-4s^3i+16s^2+16si-4s^2-16si+16 =$
$(s^4-8s^2+16) =$
$(s^2+4)^2$ 
Is there a quicker way to see that $(s-2i)^2 (s+2i)^2= (s^2+4)^2$

We have
$$(s-2i)^2 (s+2i)^2=[(s-2i) (s+2i)]^2$$
and recall that $(A-B)(A+B)=A^2-B^2$.
