Partial fraction decomposition for complex polynomials

$\begingroup$

I'm trying to find the partial fraction decomposition of the complex function $\frac{1}{z(z+1/2)(z-2)} $.

So I have that $\frac{1}{z(z+1/2)(z-2)} = \frac{A}{z} + \frac{B}{z+1/2} + \frac{C}{z-2}$, where

$\ 1= A(z+1/2)(z-2) + B(z)(z-2) + C(z)(z+1/2)$, which is equivalent to

$\ 1= z^2(A+B+C) + z (\frac{-3}{2}A -2B + \frac{1}{2}C) - A$ .

But here is where I am confused. How do I solve for the A, B, C values when these can be complex? I thought that we could just set

$\ A+B+C = 0$,

$\frac{-3}{2}A -2B + \frac{1}{2}C = 0$, and

$\ -A = 1$,

since none of the z-terms appear on the left-hand side of the equation, but I don't think this is right since z and A, B, C are complex numbers. So how do I go about solving this? How do I take into account both the real and imaginary parts to solve for A, B and C?

$\endgroup$ 1

2 Answers

$\begingroup$

$\ 1= z^2(A+B+C) + z (\frac{-3}{2}A -2B + \frac{1}{2}C) - A$

By the FTA, the two sides can only be identically equal iff the coefficients are pairwise equal (which is just another way to say that a polynomial is identically $0$ iff all its coefficients are $0\,$).

But here is where I am confused. How do I solve for the A, B, C values when these can be complex? I thought that we could just set

$\ A+B+C = 0$,

$\frac{-3}{2}A -2B + \frac{1}{2}C = 0$, and

$\ -A = 1$,

Yes you can do that, and that's in fact precisely how it's done.

but I don't think this is right since z and A, B, C are complex numbers.

It is right, see the first paragraph.

$\endgroup$ $\begingroup$

Hint: A+B+C may not be equal to zero because $z^{2}$ contain both real part or imagainary part.(eg z=i).so put z=x+iy and then solve.

$\endgroup$ 1

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy