Question

The residue of \( f(z) = \frac{z^2}{(z-1)^3(z-2)(z-3)} \) at the pole \(z=1\) is

• A. \( \frac{23}{8} \)
• B. \( \frac{101}{16} \)
• C. \( \frac{27}{16} \)
• D. \( -8 \)

Hint:

Residue at a pole \(z_0\) of order \(m\): \( \text{Res} = \frac{1}{(m-1)!} \lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}} [(z-z_0)^m f(z)] \).
Carefully apply differentiation rules (quotient rule, chain rule).

Answer 1

The Correct Option is A

The function is \( f(z) = \frac{z^2}{(z-1)^3(z-2)(z-3)} \). The pole at \(z=1\) is of order \(m=3\). The residue at a pole \(z_0\) of order \(m\) is given by: \[ \text{Res}(f, z_0) = \frac{1}{(m-1)!} \lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}} [(z-z_0)^m f(z)] \] Here \(z_0=1, m=3\). So we need \(\frac{1}{2!} \lim_{z \to 1} \frac{d^2}{dz^2} [(z-1)^3 f(z)]\). Let \(g(z) = (z-1)^3 f(z) = \frac{z^2}{(z-2)(z-3)} = \frac{z^2}{z^2-5z+6}\). \(g'(z) = \frac{2z(z^2-5z+6) - z^2(2z-5)}{(z^2-5z+6)^2} = \frac{2z^3-10z^2+12z - 2z^3+5z^2}{(z^2-5z+6)^2} = \frac{-5z^2+12z}{(z^2-5z+6)^2}\). For \(g''(z)\), let \(N = -5z^2+12z\) and \(D_0 = z^2-5z+6\). So \(g'(z) = N/D_0^2\). \(g''(z) = \frac{N'D_0^2 - N(2D_0D_0')}{D_0^4} = \frac{N'D_0 - 2ND_0'}{D_0^3}\). \(N' = -10z+12\). \(D_0' = 2z-5\). At \(z=1\): \(D_0(1) = 1-5+6 = 2\). \(N(1) = -5+12 = 7\). \(N'(1) = -10+12 = 2\). \(D_0'(1) = 2-5 = -3\). \(g''(1) = \frac{(2)(2) - 2(7)(-3)}{(2)^3} = \frac{4 - (-42)}{8} = \frac{4+42}{8} = \frac{46}{8} = \frac{23}{4}\). Residue = \(\frac{1}{2!} g''(1) = \frac{1}{2} \times \frac{23}{4} = \frac{23}{8}\). \[ \boxed{\frac{23}{8}} \]