AMU 2026 — science PYQ

1. Identify Singularities and the Contour

The given integrand is:

$$f(z)=\frac{e^z}{z(z-1)}$$

To find the singularities (poles), we set the denominator equal to zero:

$$z(z-1)=0⟹z=0\text{and}z=1$$

Both are simple poles (poles of order 1).

The path of integration $C$ is a circle centered at the origin with a radius of $2$, defined by $|z|=2$.

• For $z=0$: |0| = 0 < 2 (lies inside the contour $C$)

• For $z=1$: |1| = 1 < 2 (lies inside the contour $C$)

Since both poles lie within the region bounded by the circle, both will contribute to the integral.

2. Calculate the Residues

According to Cauchy's Residue Theorem, the value of the contour integral is given by:

$${\oint }_{C}f(z)dz=2\pi i\sum (\text{Residues inside }C)$$

Let's compute the residue at each pole:

Residue at $z=0$:

$$\text{Res}(f,0)=\underset{z\to 0}{\lim }(z-0)\cdot \frac{e^z}{z(z-1)}$$

$$\text{Res}(f,0)=\underset{z\to 0}{\lim }\frac{e^z}{z-1}=\frac{e^0}{0-1}=\frac{1}{-1}=-1$$

Residue at $z=1$:

$$\text{Res}(f,1)=\underset{z\to 1}{\lim }(z-1)\cdot \frac{e^z}{z(z-1)}$$

$$\text{Res}(f,1)=\underset{z\to 1}{\lim }\frac{e^z}{z}=\frac{e^1}{1}=e$$

3. Evaluate the Contour Integral

Sum the residues found inside the contour:

$$\sum \text{Residues}=\text{Res}(f,0)+\text{Res}(f,1)=-1+e=e-1$$

Now substitute this sum back into Cauchy's Residue formula:

$${\oint }_C\frac{e^z}{z(z-1)}dz=2\pi i(e-1)$$

Correct Answer

The correct option is (d) $2\pi i(e-1)$.