# Geometric intepretation

We were given a complex number problem to work on: if $\mid s\mid =1$, then show that $\Re(\frac{1}{s+1})=\frac{1}{2}$ I can prove this algebraically, but am having a little trouble with proving it geometrically. Can someone help me?

Note by Edward Jiang
5 years, 2 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

The transformation $$s\mapsto \tfrac{1}{s+1}$$ is a Mobius transformation, and Mobius transformations map circles/straight lines onto circles/straight lines. Thus the image of $$|s|=1$$ is a circle/straight line $$L$$ which passes through $$\infty$$ (the image of the point $$-1$$ on the original circle). Thus $$L$$ is a straight line. Since the point $$2$$ on the circle maps to the point $$\tfrac12$$, $$L$$ passes through $$\tfrac12$$.

Mobius transformations also preserve the angles between lines. The circle $$|s|=1$$ meets the real axis at right angles at $$1$$. Hence $$L$$ meets the real axis (the image of the real axis) at right angles at $$\tfrac12$$. Thus $$L$$ must be parallel to the imaginary axis, and we are done.

- 5 years, 2 months ago

Thanks, appreciate it. :)

- 5 years, 2 months ago

The set $$A = \{s+1 \ : \ |s| = 1\}$$ is a circle with center $$1$$ and radius $$1$$. This circle passes through the origin. The set $$A' = \{\tfrac{1}{s+1} \ : \ |s| = 1\}$$ is the inversion of $$A$$. Under inversion, circles passing through the origin map to lines that do not pass through the origin. Now, you just need to show that the line is given by $$\mathfrak{R}(z) = \tfrac{1}{2}$$.

- 5 years, 2 months ago