Fast way to solve $(s-2i)^2 (s+2i)^2$

Question

$(s-2i)^2 (s+2i)^2=$

$(s^2-4si-4) (s^2+4si-4)=$

$s^4+4s^3i-4s^2-4s^3i+16s^2+16si-4s^2-16si+16 =$

$(s^4-8s^2+16) =$

$(s^2+4)^2$

Is there a quicker way to see that $(s-2i)^2 (s+2i)^2= (s^2+4)^2$

Answer 1

We have

$$(s-2i)^2 (s+2i)^2=[(s-2i) (s+2i)]^2$$

and recall that $(A-B)(A+B)=A^2-B^2$.

Answer 2

$$(s-2i)^2(s+2i)^2=\left[(s-2i)(s+2i)\right]^2=\left[s^2-4i^2\right]^2=\ldots$$

Answer 3

notice: $$(a+b)(a-b)=a^2-b^2$$ so: $$(a+bi)(a-bi)=a^2-(bi)^2=a^2+b^2$$