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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?

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  • $\begingroup$ What you have is correct. Since the roots of the denominator are all real, $A, B, C$ will be real numbers $\endgroup$ – Doug M Apr 10 '18 at 4:28
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$\ 1= z^2(A+B+C) + z (\frac{-3}{2}A -2B + \frac{1}{2}C) - A$

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.

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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.

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    $\begingroup$ Sorry, but you are plain wrong. $A+B+C$ must be $0$ in order for the equality to hold as a polynomial identity, regardless of whether you consider it over $\,\mathbb{C}\,$ or $\,\mathbb{R}\,$. $\endgroup$ – dxiv Apr 10 '18 at 4:44

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