Complex partial fractions

Thank you Joseph O'keefe from Colyton Grammar School for this solution. Andrei Lazanu from Rumania and Laura Hannick, Townley Grammar School for Girls, also sent in good solutions.

Part (1)

$${10x^2-2x+4\over x^3 + x} = {A\over x} + {Bx+C\over x^2+1}$$ In order to work out the constants $A,B,C$, I put the RHS of the identity over a common denominator: $${10x^2-2x+4\over x^3 + x} = {A(x^2+1)+ x(Bx+C)\over x(x^2+1)}$$ Since the denominators are equal, it follows that the numerators of the fraction must also be equal: $$\eqalign{ 10x^2-2x+4 &= A(x^2+1)+ x(Bx+C)\cr &= Ax^2+Bx^2+Cx+A \cr &= (A+B)x^2+Cx+A}$$ By comparing coefficients, it follows that $A+B=10$, $C=-2$ and $A=4$ so $B=6$.

Part (2)

$${A\over x} + {Bx+C\over x^2+1}={A\over x} + {D\over (x-i)} + {E\over (x+i)}.$$ By taking $A/x$ away and replacing $B$ and $C$ by their respective values, and putting $D$ and $E$ over a common denominator: $${6x-2\over (x^2+1)}= {D(x+i)+E(x-i)\over (x-i)(x+i)}.$$ Again, since the denominators are equal, it follows that the numerators are equal so $6x-2=D(x+i)+E(x-i)$.

By comparing coefficients we have $D+E=6$ and $(D-E)i=-2$. Then $6i-2=D(2i)$ so $D=(6i-2)/2i$. Multiply both the numerator and the denominator by $i$ to get a real denominator: $$D={6i^2-2i\over -2} = {-6-2i\over -2} = 3+i$$ Then $E(-2i)= -6i-2$, so $E=3-i$. So $A=4$, $B=6$, $C=-2$, $D=3+i$ and $E=3-i$ $${10x^2-2x+4\over x^3 + x}= {4\over x} + {6x-2\over x^2+1}={4\over x} + {3+i\over (x-i)} + {3-i\over (x+i)}$$