Complex Variables and Partial Differential Equations (3130005)

Question-3a-OR

BE | Semester-3 Winter-2019 | 26-11-2019

Q3) (a)

Expand Laurent series of $f (z) = \frac{1 - e^{z}}{z}$ at $z = 0$ and identify the singularity.

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

\Rightarrow f (z) = \frac{1}{z} [1 - (1 + z + \frac{z^{2}}{2!} + \frac{z^{3}}{3!} + . . .)]

\Rightarrow f (z) = \frac{1}{z} [- z - \frac{z^{2}}{2!} - \frac{z^{3}}{3!} - . . .]

\Rightarrow f (z) = - 1 - \frac{z}{2} - \frac{z^{2}}{6} - . . .

⇒ There is no principal part.

\Rightarrow f (z)

has removable singularity at

z = 0

.

Questions

Go to Question Paper