Q5P For each of the following functi... [FREE SOLUTION]

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Chapter 1: Q5P (page 1)

For each of the following functions w = f(z) = u+iv , find u and v as functions of x and y. Sketch the graph in (x,y) plane of the images of u = const. and v = const. for several values of and several values of as was done for in Figure 9.3. The curves u = const. should be orthogonal to the curves v = const.
$w = \frac{z - i}{z + i}$

Short Answer

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The solutions is,

$u = \frac{x^{2} + y^{2} - 1}{x^{2} + {(1 + y)}^{2}}, v = \frac{- 2 x}{x^{2} + {(1 + y)}^{2}}$ .

The graph is shown in the below image:

Step by step solution

To find the solution

The given function is, . $w = \frac{z - i}{z + i}$

Now, solve the given function by using z = x + iy.

$w = \frac{z - i}{z + i} = \frac{(x + i y) - i}{(x + i y) + i} = \frac{(x + i y) - i}{(x + i y) + i} . \frac{(x + i y) - i}{(x + i y) - i} = \frac{{((x + i y) - i)}^{2}}{{(x + i y)}^{2} - i^{2}}$

Next, further simplify as follow:

$\begin{array}{rcl} w & = & \frac{x^{2} + 2 x y i + y^{2} - 2 (x + i y) i + 1}{x^{2} + 2 x y i + y^{2} + 1} \\ = & \frac{x^{2} + 2 x y i + y^{2} - 2 x + 2 y + 1}{x^{2} + 2 x y i + y^{2} + 1} \\ = & \frac{x^{2} + y^{2} - 1 - 2 x y i}{x^{2} + {(1 + y)}^{2}} \\ = & u + i v \end{array}$

Hence $\begin{array}{rcl} u & = & \frac{x^{2} + y^{2} - 1}{x^{2} + {(1 + y)}^{2}}, v = \frac{- 2 x}{x^{2} + {(1 + y)}^{2}} \end{array}$ , where,u is the real function and v is an imaginary function.

Step 2: To sketch the graph of solution

If u = const., and v = const. for example 0,1,2,-1,-2 , the graph of the function is given as follows:

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Short Answer

Step by step solution

To find the solution

Step 2: To sketch the graph of solution

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