# Complex Partial Fractions

##### Age 16 to 18

Challenge Level

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 $E=3-i$ $${10x^2-2x+4\over x^3 +
x}= {4\over x} + {6x-2\over x^2+1}={4\over x} + {3+i\over (x-i)} +
{3-i\over (x+i)}$$