Question

The residue of the function $f(z) = \frac{ze^z}{(z-1)^3}$, at its pole is

• A. \( \frac{3e}{2} \)
• B. \( \frac{4e}{3} \)
• C. \( \frac{e}{2} \)
• D. \( e \)

Hint:

For a simple pole (m=1), the residue is $\lim_{z \to a} (z-a)f(z)$. For a pole of order $m \ge 2$, use the general formula involving derivatives. A common mistake is to forget the factorial term $(m-1)!$ in the denominator. Always ensure you differentiate the entire term $(z-a)^m f(z)$ before taking the limit, which often simplifies to just $\phi(z)$ when $f(z)=\frac{\phi(z)}{(z-a)^m}$.

Answer 1

The Correct Option is A

The given function is $f(z) = \frac{ze^z}{(z-1)^3}$. The function has a pole at $z=1$. The order of the pole is $m=3$. For a function $f(z) = \frac{\phi(z)}{(z-a)^m}$ where $\phi(a) \neq 0$, the residue at a pole of order $m$ at $z=a$ is given by the formula: $$ \text{Res}_{z=a} f(z) = \frac{1}{(m-1)!} \lim_{z \to a} \frac{d^{m-1}}{dz^{m-1}} [(z-a)^m f(z)] $$ In this case, $a=1$, $m=3$, and $\phi(z) = ze^z$. 
Step 1: Identify $\phi(z)$ and its derivatives
Here, $\phi(z) = ze^z$. We need to find the $(3-1) = 2$nd derivative of $\phi(z)$. $$ \phi'(z) = \frac{d}{dz}(ze^z) $$ Using the product rule $(uv)' = u'v + uv'$: Let $u=z$, $v=e^z$. Then $u'=1$, $v'=e^z$. $$ \phi'(z) = 1 \cdot e^z + z \cdot e^z = e^z(1+z) $$ Now, find the second derivative $\phi"(z)$: $$ \phi"(z) = \frac{d}{dz}[e^z(1+z)] $$ Again, using the product rule. Let $u=e^z$, $v=1+z$. Then $u'=e^z$, $v'=1$. $$ \phi"(z) = e^z(1) + (1+z)e^z = e^z + e^z + ze^z = 2e^z + ze^z = e^z(2+z) $$ 
Step 2: Apply the residue formula
$$ \text{Res}_{z=1} f(z) = \frac{1}{(3-1)!} \lim_{z \to 1} \frac{d^2}{dz^2} [(z-1)^3 \frac{ze^z}{(z-1)^3}] $$ $$ \text{Res}_{z=1} f(z) = \frac{1}{2!} \lim_{z \to 1} \frac{d^2}{dz^2} (ze^z) $$ We found that $\frac{d^2}{dz^2} (ze^z) = e^z(2+z)$. 
Step 3: Evaluate the limit
$$ \text{Res}_{z=1} f(z) = \frac{1}{2} \lim_{z \to 1} [e^z(2+z)] $$ Substitute $z=1$: $$ \text{Res}_{z=1} f(z) = \frac{1}{2} [e^1(2+1)] $$ $$ \text{Res}_{z=1} f(z) = \frac{1}{2} (e \cdot 3) $$ $$ \text{Res}_{z=1} f(z) = \frac{3e}{2} $$