# 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 $$(A-B)(A+B)=A^2-B^2$$.

• I don't see why $(s-2i)^2 (s+2i)^2=[(s-2i) (s+2i)]^2$ – roy212 Nov 17 '18 at 21:46
• @roy212 We have in general (not only for complex number) $$A^2\cdot B^2 =(A\cdot B)^2$$ – user Nov 17 '18 at 21:47
• And $A^2*B^2*C^2*D^2 = (A*B*C*D)^2$? – roy212 Nov 17 '18 at 21:56
• @roy212 Yes of course! It is also true. – user Nov 17 '18 at 21:57
• As long as you are dealing with commutative things. So, okay for real and complex numbers but not quaternions or matrices. – badjohn Nov 18 '18 at 15:57

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

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