Math Problem Statement

  1. Under the transformation w = 1 z , find the image of the region (i) x > c, where c > 0, and (ii) y > c, where c < 0.

Solution

Given the transformation w=1zw = \frac{1}{z}, where z=x+iyz = x + iy is a complex number with real part xx and imaginary part yy, we need to find the images of the regions under this transformation.

Step 1: Analyze the Transformation

For w=1zw = \frac{1}{z}, let's express ww in terms of u+ivu + iv, where uu and vv are the real and imaginary parts of ww, respectively:

w=1z=1x+iyw = \frac{1}{z} = \frac{1}{x + iy}

Multiply the numerator and the denominator by the conjugate of the denominator:

w=xiy(x+iy)(xiy)=xiyx2+y2w = \frac{x - iy}{(x + iy)(x - iy)} = \frac{x - iy}{x^2 + y^2}

So, the real and imaginary parts of ww are:

u=xx2+y2,v=yx2+y2u = \frac{x}{x^2 + y^2}, \quad v = \frac{-y}{x^2 + y^2}

Step 2: Find the Image of the Region x>cx > c where c>0c > 0

The region x>cx > c represents the half-plane to the right of the line x=cx = c. We need to find how this region transforms under w=1zw = \frac{1}{z}.

  1. Consider u=xx2+y2u = \frac{x}{x^2 + y^2}. Since x>cx > c and x2+y2>0x^2 + y^2 > 0, it follows that u>cx2+y2u > \frac{c}{x^2 + y^2}.

  2. As xx increases, uu increases, and as xx approaches cc, uu approaches cc2+y2\frac{c}{c^2 + y^2}.

  3. Therefore, the image of x>cx > c under this transformation will be the region u>cx2+y2u > \frac{c}{x^2 + y^2}.

Step 3: Find the Image of the Region y>cy > c where c<0c < 0

The region y>cy > c represents the half-plane above the line y=cy = c. Since c<0c < 0, this region is below the real axis. We again analyze the transformation:

  1. For v=yx2+y2v = \frac{-y}{x^2 + y^2}, since y>cy > c and c<0c < 0, y<c-y < -c implies v>cx2+y2v > \frac{-c}{x^2 + y^2}.

  2. As yy approaches cc, vv approaches cx2+c2\frac{-c}{x^2 + c^2}.

  3. Therefore, the image of y>cy > c under this transformation will be the region v>cx2+y2v > \frac{-c}{x^2 + y^2}.

Summary of Results:

  1. The image of the region x>cx > c where c>0c > 0 under the transformation w=1zw = \frac{1}{z} is the region u>cx2+y2u > \frac{c}{x^2 + y^2}.

  2. The image of the region y>cy > c where c<0c < 0 under the transformation w=1zw = \frac{1}{z} is the region v>cx2+y2v > \frac{-c}{x^2 + y^2}.

Would you like any more details or explanations on this problem?

Here are five related questions you might find interesting:

  1. How would the transformation w=1zw = \frac{1}{z} map the region x<cx < c where c<0c < 0?
  2. What is the image of the entire right half-plane under the transformation w=1zw = \frac{1}{z}?
  3. How does the transformation w=1zw = \frac{1}{z} affect the unit circle z=1|z| = 1?
  4. What happens to the imaginary axis under the transformation w=1zw = \frac{1}{z}?
  5. How does the transformation w=1zw = \frac{1}{z} map the first quadrant of the complex plane?

Tip: When dealing with complex transformations, it can be helpful to break down the transformation into its real and imaginary components to better understand how regions are mapped in the complex plane.

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Math Problem Analysis

Mathematical Concepts

Complex Analysis
Transformation of Complex Variables

Formulas

Transformation formula w = 1/z

Theorems

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Suitable Grade Level

Advanced Undergraduate