<s>[INST] 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 $(s-2i)^2 (s+2i)^2= (s^2+4)^2$
 [/INST] notice:
$$(a+b)(a-b)=a^2-b^2$$
so:
$$(a+bi)(a-bi)=a^2-(bi)^2=a^2+b^2$$
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