Find where the Mobius transformation f(z)=1+z/1-z maps the unit circle , the real axis and the imaginary axis 

To find where the Möbius transformation 

\[
f(z) = \frac{1 + z}{1 - z}
\]

maps the unit circle, the real axis, and the imaginary axis, we need to analyze the image of each of these sets under the given transformation.

### 1. Mapping the Unit Circle 

Let \( z = e^{i\theta} \) represent points on the unit circle, where \( \theta \) ranges from \( 0 \) to \( 2\pi \). Substituting \( z = e^{i\theta} \) into the transformation, we get:

\[
f(e^{i\theta}) = \frac{1 + e^{i\theta}}{1 - e^{i\theta}}.
\]

Simplify this expression:

\[
f(e^{i\theta}) = \frac{(1 + e^{i\theta})(1 - e^{-i\theta})}{(1 - e^{i\theta})(1 - e^{-i\theta})} = \frac{1 - e^{-i\theta} + e^{i\theta} - 1}{1 - e^{i\theta} - e^{-i\theta} + 1}.
\]

Simplify further:

\[
f(e^{i\theta}) = \frac{e^{i\theta} - e^{-i\theta}}{-(e^{i\theta} + e^{-i\theta})}.
\]

Using \( e^{i\theta} - e^{-i\theta} = 2i\sin\theta \) and \( e^{i\theta} + e^{-i\theta} = 2\cos\theta \), the above simplifies to:

\[
f(e^{i\theta}) = \frac{2i\sin\theta}{-2\cos\theta} = -i\tan\theta.
\]

Thus, the unit circle maps to the imaginary axis.

### 2. Mapping the Real Axis

Let \( z = x \), where \( x \) is a real number. Substituting \( z = x \) into the transformation, we get:

\[
f(x) = \frac{1 + x}{1 - x}.
\]

This is a real number as long as \( x \) is real. Therefore, the real axis maps to the real axis but stretched according to the transformation.

### 3. Mapping the Imaginary Axis

Let \( z = iy \), where \( y \) is a real number. Substituting \( z = iy \) into the transformation, we have:

\[
f(iy) = \frac{1 + iy}{1 - iy}.
\]

Multiply numerator and denominator by the conjugate of the denominator:

\[
f(iy) = \frac{(1 + iy)(1 + iy)}{(1 - iy)(1 + iy)} = \frac{1 - y^2 + 2iy}{1 + y^2}.
\]

This can be rewritten as:

\[
f(iy) = \frac{1 - y^2}{1 + y^2} + i\frac{2y}{1 + y^2}.
\]

This describes a transformation from the imaginary axis to the unit circle in the complex plane, parameterized by \( y \).

### Summary

- **Unit Circle**: Maps to the imaginary axis.
- **Real Axis**: Maps to the real axis.
- **Imaginary Axis**: Maps to the unit circle in the complex plane.

Would you like further details on any part of this transformation? 

Here are some related questions that might help deepen your understanding:

1. How can we find the fixed points of a Möbius transformation?
2. What are the implications of a Möbius transformation preserving angles?
3. Can Möbius transformations map a circle to a line? If so, how?
4. How do Möbius transformations behave near their singularities?
5. What are the properties of Möbius transformations in terms of conformal mappings?

**Tip:** Möbius transformations are powerful tools in complex analysis that preserve angles but not necessarily distances; understanding their geometric implications can greatly aid in visualizing complex mappings.

Find where the Möbius transformation f(z) = (1 + z)/(1 - z) maps the unit circle, the real axis, and the imaginary axis.

Explore the mapping of the unit circle, real axis, and imaginary axis under the Möbius transformation f(z) = (1 + z)/(1 - z). This solution uses complex analysis concepts, including geometric mappings and trigonometric identities, to illustrate how the unit circle maps to the imaginary axis, the real axis maps to itself, and the imaginary axis maps to the unit circle. Suitable for undergraduate-level students.

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