Complex Variables and Partial Differential Equations (3130005)

Question-2c-OR

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

Q2) (c)

Find the image of the infinite strips $(i) \frac{1}{4} \leq y \leq \frac{1}{2} (i) 0 \leq y \leq \frac{1}{2}$ under the transformation $w = \frac{1}{z}$ . Show the region graphically.

Let,

w = \frac{1}{z}

.

\Rightarrow z = \frac{1}{w}

\Rightarrow x + i y = \frac{1}{u + i v}

\Rightarrow x + i y = \frac{u - i v}{u^{2} - i^{2} v^{2}}

\Rightarrow x + i y = \frac{u - i v}{u^{2} + v^{2}}

\Rightarrow x + i y = \frac{u}{u^{2} + v^{2}} + i \frac{- v}{u^{2} + v^{2}}

\Rightarrow x = \frac{u}{u^{2} + v^{2}} & y = \frac{- v}{u^{2} + v^{2}}

(i) \frac{1}{4} \leq y \leq \frac{1}{2}

\Rightarrow \frac{1}{4} \leq \frac{- v}{u^{2} + v^{2}} \leq \frac{1}{2}

\Rightarrow \frac{1}{4} \leq \frac{- v}{u^{2} + v^{2}} & \frac{- v}{u^{2} + v^{2}} \leq \frac{1}{2}

\Rightarrow u^{2} + v^{2} \leq - 4 v & - 2 v \leq u^{2} + v^{2}

\Rightarrow u^{2} + v^{2} + 4 v \leq 0 & 0 \leq u^{2} + v^{2} + 2 v

\Rightarrow u^{2} + v^{2} + 4 v + 4 \leq 4 & u^{2} + v^{2} + 2 v + 1 \geq 1

\Rightarrow {(u - 0)}^{2} + {(v + 2)}^{2} \leq 4 & {(u - 0)}^{2} + {(v + 1)}^{2} \geq 1

(ii) 0 \leq y \leq \frac{1}{2}

\Rightarrow 0 \leq \frac{- v}{u^{2} + v^{2}} \leq \frac{1}{2}

\Rightarrow 0 \leq \frac{- v}{u^{2} + v^{2}} & \frac{- v}{u^{2} + v^{2}} \leq \frac{1}{2}

\Rightarrow 0 \leq - v & - 2 v \leq u^{2} + v^{2}

\Rightarrow v \leq 0 & 0 \leq u^{2} + v^{2} + 2 v

\Rightarrow v \leq 0 & u^{2} + v^{2} + 2 v + 1 \geq 1

\Rightarrow v \leq 0 & {(u - 0)}^{2} + {(v + 1)}^{2} \geq 1

Questions

Go to Question Paper