1/(s⁴+4)≡A/(s²+2i)+B/(s²-2i)⇒
1≡A(s²-2i)+B(s²+2i)⇒
1≡s²(A+B)+2i(B-A)⇒
A=-B, B-A=1/2i=-i/2.
2B=-i/2, B=-i/4, A=i/4.
1/(s⁴+4)≡¼i/(s²+2i)-¼i/(s²-2i).
ℒ⁻¹(/(s²+²))=sin(t).
If ²=2i, =√(2i).
Let =a+ib, then a²-b²+2abi=2i.
a²=b², so a=±b, and ab=1, a²=1, a=±1; a²=-1, a=±i.
Therefore, =1+i is a solution.
ℒ⁻¹((1+i)/(s²+2i))=sin(t+it).
ℒ⁻¹(¼i/(s²+2i))=sin(t+it)(¼i/(1+i)).
1/(1+i)=½(1-i), ¼i/(1+i)=⅛(1+i).
ℒ⁻¹(¼i/(s²+2i))=⅛(1+i)sin(t+it).
Now consider ℒ⁻¹(¼i/(s²-2i)).
Let ²=-2i=a+ib.
a²-b²+2abi=-2i,
a=b, ab=-1, a²=-1, a=i=b.
=1-i is a solution, =-1+i is another.
ℒ⁻¹((1-i)/(s²-2i))=sin(t-it).
ℒ⁻¹(¼i/(s²-2i))=sin(t-it)(¼i/(1-i)).
1/(1-i)=½(1+i), -¼i/(1-i)=⅛(1-i).
ℒ⁻¹(¼i/(s²+2i)-¼i/(s²-2i))=⅛(1+i)sin(t+it)+⅛(1-i)sin(t-it).
ℒ⁻¹(1/(s⁴+4)=ℒ⁻¹(¼i/(s²+2i)-¼i/(s²-2i))=
⅛(1+i)sin(t+it)+⅛(1-i)sin(t-it).
This simplifies to:
¼(sin(t)cos(it)+icos(t)sin(it)).