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(Solved): C=C{0} and the strip {zC0Re(z)<2} (b) Give an example of two different Lobachev ...
![\[
\mathbb{C}^{*}=\mathbb{C}-\{0\}
\]
and the strip
\[
\{z \in \mathbb{C} \mid 0 \leq \operatorname{Re}(z)<2 \pi\}
\]
(b) Giv](https://media.cheggcdn.com/media/0fd/0fdf8ce8-1554-4362-92a6-6842cf855ae0/phpBVGgp9)
and the strip (b) Give an example of two different Lobachevsky straight lines passing through a common point and being parallel to a third Lobachevsky straight line. (c) Compute where is the unit circle centred at the origin.
Expert Answer
(a) To find a conformal map between the punctured complex plane C* and the strip {z ? C | 0 ? Re(z) < 2?}, we can use the exponential function.Let's consider the function f(z) = e^z, where z is a complex number.First, we need to show that f(z) maps the punctured complex plane C* onto the strip {z ? C | 0 ? Re(z) < 2?}.C* is the complex plane minus the origin, so it excludes the point z = 0. The exponential function e^z is defined for all complex numbers z, except z = 0.Now, let's analyze the image of f(z) = e^z. For any z in C*, we can write z = x + yi, where x and y are real numbers, and y ? 0. Then, f(z) = e^(x+yi) = e^x * e^(yi) = e^x * (cos(y) + i * sin(y)). The real part of f(z) is Re(f(z)) = e^x * cos(y), and the imaginary part is Im(f(z)) = e^x * sin(y). Since e^x > 0 for all x, Re(f(z)) is always positive. Also, y ? 0, so sin(y) ? 0, which means Im(f(z)) ? 0. Hence, for any z in C*, f(z) lies in the strip { z ? C | 0 ? Re(z) < 2? }.Next, we need to show that f(z) is a conformal map, which means it preserves angles.Consider two curves in C* that intersect at some point z. Let's call these curves ?1 and ?2.Since f(z) = e^z is an analytic function, it preserves angles between curves. Therefore, the curves f(?1) and f(?2) in the strip {z ? C | 0 ? Re(z) < 2?} will also intersect at the image point f(z).In conclusion, the function f(z) = e^z maps the punctured complex plane C* conformally onto the strip {z ? C | 0 ? Re(z) < 2?}.