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: 

\[ \text{Res}(f, z_0) = \frac{1}{(m-1)!} \lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}} \left[(z-z_0)^m f(z)\right] \]

Here \(z_0=1, m=3\). So we need:

\[ \frac{1}{2!} \lim_{z \to 1} \frac{d^2}{dz^2} \left[(z-1)^3 f(z)\right] \]

Let \(g(z) = (z-1)^3 f(z) = \frac{z^2}{(z-2)(z-3)} = \frac{z^2}{z^2-5z+6}\).

First, calculate \(g'(z)\):

\[ g'(z) = \frac{2z(z^2-5z+6) - z^2(2z-5)}{(z^2-5z+6)^2} = \frac{2z^3-10z^2+12z - 2z^3+5z^2}{(z^2-5z+6)^2} = \frac{-5z^2+12z}{(z^2-5z+6)^2} \]

For \(g''(z)\), let \(N = -5z^2 + 12z\) and \(D_0 = z^2 - 5z + 6\). So \(g'(z) = \frac{N}{D_0^2}\).

\[ g''(z) = \frac{N'D_0^2 - N(2D_0 D_0')}{D_0^4} = \frac{N'D_0 - 2ND_0'}{D_0^3} \]

Now, calculate \(N'\) and \(D_0'\):

\(N' = -10z + 12\), \(D_0' = 2z - 5\).

At \(z = 1\):

\(D_0(1) = 1 - 5 + 6 = 2\), \(N(1) = -5 + 12 = 7\), \(N'(1) = -10 + 12 = 2\), \(D_0'(1) = 2 - 5 = -3\).

Now calculate \(g''(1)\):

\[ g''(1) = \frac{(2)(2) - 2(7)(-3)}{(2)^3} = \frac{4 - (-42)}{8} = \frac{4 + 42}{8} = \frac{46}{8} = \frac{23}{4} \]

The residue is:

\[ \text{Residue} = \frac{1}{2!} g''(1) = \frac{1}{2} \times \frac{23}{4} = \frac{23}{8} \]

Final Answer: \[ \boxed{\frac{23}{8}} \]