Complex partial fractions
To break down an algebraic fraction into partial fractions in which all the denominators are linear and all the numerators are constants you sometimes need complex numbers.
Problem
Find real constants $A, B$ and $C$ and complex constants $D$ and $E$ such that $${10x^2-2x+4\over x^3 + x} = {A\over x} +{Bx+C\over x^2+1} = {A\over x} + {D\over x-i} + {E\over x+i}.$$
NOTES AND BACKGROUND
This problem gives an example where a rational function can be reduced to a sum of linear partial fractions IF we allow ourselves to use complex numbers. It turns out that this is always possible! This is of use in more advanced university-level applications of integration and analysis of series.
Getting Started
Just use the usual methods for partial fractions and factorise
$x^2+1$ into $(x+i)(x-i)$.
Student Solutions
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 $E=3-i$ $${10x^2-2x+4\over x^3 + x}= {4\over x} + {6x-2\over x^2+1}={4\over x} + {3+i\over (x-i)} + {3-i\over (x+i)}$$
Teachers' Resources
The 'pay off' for factorising the denominator fully is that all the
numerators then become constants.