MHB Locus in the complex plane.

Click For Summary
The discussion focuses on finding the area of the region bounded by the locus of complex numbers \( z \) that satisfy the equation \( \arg \left(\frac{z+5i}{z-5i}\right) = \pm \frac{\pi}{4} \). The relationship can be rewritten as \( \arg(z + 5) - \arg(z - 5) = \pm \frac{\pi}{4} \), indicating a geometric interpretation involving rays and angles. The analysis involves considering the line segment between \( z = 5 \) and \( z = -5 \) as a chord of a circle, with the rays creating a constant angle at their intersection. The resulting shape is identified as a circle with "holes" at the points \( z = 5 \) and \( z = -5 \). The final step is to determine the circle's radius to calculate the area of the bounded region.
juantheron
Messages
243
Reaction score
1
Area of Region Bounded by the locus of $z$ which satisfy the equation \displaystyle \arg \left(\frac{z+5i}{z-5i}\right) = \pm \frac{\pi}{4} is
 
Mathematics news on Phys.org
jacks said:
Area of Region Bounded by the locus of $z$ which satisfy the equation \displaystyle \arg \left(\frac{z+5i}{z-5i}\right) = \pm \frac{\pi}{4} is

What have you tried?
 
jacks said:
Area of Region Bounded by the locus of $z$ which satisfy the equation \displaystyle \arg \left(\frac{z+5i}{z-5i}\right) = \pm \frac{\pi}{4} is

You can take a geometric approach.

Your relation can be written \arg(z + 5) - \arg(z - 5) = \pm \frac{\pi}{4}, that is, \alpha - \beta =\pm \frac{\pi}{4}.

Consider the line segment joining z = 5 and z = -5 as the chord on a circle and consider the rays \arg(z +5) = \alpha and \arg(z - 5) = \beta subject to the restriction \alpha - \beta =\pm \frac{\pi}{4}. Consider the intersection of these rays and the angle between them at their intersection point. The angle is constant ... Now think of a circle theorem involving angles subtended by the same arc at the circumference ...

It's not hard to see you that have a circle with 'holes' at z = 5 and z = -5 (why?).

Now your job is to determine the radius of this circle and use it to get the area.
 
Last edited:

Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
View full post »

Similar threads

MHB Effie's question via email about Complex Numbers
  • · Replies 2 ·
Replies
2
Views
5K
MHB Ap1.3.51 are complex numbers, show that
  • · Replies 4 ·
Replies
4
Views
2K
High School Principal square root of a complex number, why is it unique?
  • · Replies 13 ·
Replies
13
Views
6K
Graduate Help needed with derivation: solving a complex double integral
  • · Replies 1 ·
Replies
1
Views
2K
MHB Find Intersection of Complex Number Loci Given w
  • · Replies 5 ·
Replies
5
Views
4K
Can the contour integral of z⁷ be simplified using a parameterized expression?
  • · Replies 1 ·
Replies
1
Views
1K
MHB S10.03.25 Write complex number in rectangular form
  • · Replies 3 ·
Replies
3
Views
2K
MHB Question via email about complex numbers
  • · Replies 1 ·
Replies
1
Views
6K
MHB What is the value of z if its argument is equal to its conjugate?
  • · Replies 23 ·
Replies
23
Views
4K
##\int \frac{dz}{z} ## along a line on the complex plane
  • · Replies 10 ·
Replies
10
Views
2K