Open in App
Log In Start studying!

Select your language

Suggested languages for you:

Show that a circle on the sphere that does not pass through the north pole corresponds to a circle in the complex plane.

Short Answer

Expert verified
Stereographic projection is used to establish the correspondence between a circle on a sphere not passing through the North pole, and a circle in the complex plane. Key to this proof is the invariant angle property of the stereographic projection.
See the step by step solution

Step by step solution

Unlock all solutions

Get unlimited access to millions of textbook solutions with Vaia Premium

Over 22 million students worldwide already upgrade their learning with Vaia!

Step 1 - Construct the stereographic projection

A circle on the sphere can be stereographically projected onto the complex plane. Let's take the north pole of the sphere as \(N\) and let \(O\) be a point on the circle on the sphere that doesn't pass through \(N\). Draw a line from \(N\) through \(O\) to the complex plane. The point where it intersects the plane is the stereographic projection corresponding to \(O\).

Step 2 - Establish connection between points

The key point here is to realize that all points \(O\) on the circle (which do not pass through north pole) will maintain the same angle with respect to \(N\). Hence, the radii from the origin of the complex plane to projected points are all at the same angle. This automatically gives us a circle in the complex plane.

Step 3 - Define the circle in the complex plane

Keeping the above properties in mind, define a circle in the complex plane that mirrors the circular structure in spherical coordinates. We can state that for any point \(O\) on the circle on the sphere, its projection will always lie on the defined circle in the complex plane, hence proving the correspondence between them.

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Access millions of textbook solutions in one place

  • Access over 3 million high quality textbook solutions
  • Access our popular flashcard, quiz, mock-exam and notes features
  • Access our smart AI features to upgrade your learning
Get Vaia Premium now
Access millions of textbook solutions in one place

Most popular questions from this chapter

Chapter 2

Suppose that \(J: \mathbb{C}_{\infty} \rightarrow \mathbb{C}_{\infty}\) is defined by \(J(z)=1 / z, z \in \mathbb{C}_{\infty} .\) Do our conventions imply \(J(0)=\infty\) and \(J(\infty)=\infty\) ? Does $$ \chi(J(z), J(w))=\chi(z, w) $$ hold in \(\mathbb{C}_{\infty} ?\)

Chapter 2

Which of the following sequences are convergent? (a) \(\left\\{i^{n}\right\\}\) (b) \(\left\\{z_{0}^{n}\right\\}\), where \(\left|z_{0}\right|<1\) (c) \(\left\\{\frac{\cos n+i \sin n}{n}\right\\}\) (d) \(\left\\{\frac{1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}}{n}\right\\}\) (f) $\left\\{e^{n \pi i / 3}+\left(-\frac{1}{2}-i \frac{\sqrt{3}}{2}\right)^{n}\right\\}$ (g) $\left\\{e^{n \pi i / 6}+\left(-\frac{1}{2}+i \frac{\sqrt{3}}{2}\right)^{n}\right\\}$ (f) \(\left\\{\frac{n \cos (n \pi)}{2 n+1}\right\\}\) (i) \(\left\\{\sin \left(\frac{n \pi}{8}\right)\right\\}\).

Chapter 2

If \(\lim _{z \rightarrow \infty} f(z)=a\), and \(f(z)\) is defined for every positive integer \(n\), prove that \(\lim _{n \rightarrow \infty} f(n)=a\). Give an example to show that the converse is false.

Chapter 2

Give an example of a sequence that (a) does not converge, but has exactly one limit point; (b) has \(n\) limit points, for any given integer \(n\); (c) has infinitely many limit points.

Chapter 2

Find the following limits: (a) \(\lim _{z \rightarrow 0} f(z)\), where $f(z)=\frac{x y}{x^{2}+y^{2}}+2 x i$, (b) \(\lim _{z \rightarrow 0} f(z)\), where $f(z)=\frac{x y}{x^{2}+y}+2 \frac{x}{y} i$, (c) \(\lim _{z \rightarrow 0} f(z)\), where $f(z)=\frac{x y^{3}}{x^{3}+y^{3}}+\frac{x^{8}}{y^{2}+1} i$.

Join over 22 million students in learning with our Vaia App

The first learning app that truly has everything you need to ace your exams in one place.

  • Flashcards & Quizzes
  • AI Study Assistant
  • Smart Note-Taking
  • Mock-Exams
  • Study Planner
Join over 22 million students in learning with our Vaia App Join over 22 million students in learning with our Vaia App

Recommended explanations on Math Textbooks