91Ó°ÊÓ

A regular function, also known as a holomorphic or analytic function, is a function defined on a complex domain that is differentiable at every point within that domain.
In simpler terms, this means that the function is smooth and has no breaks, jumps, or sharp corners. For a function to be regular on the complex plane \(\mathbb{C}\), it must satisfy the Cauchy-Riemann equations:
\( \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \)
\( \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x} \)
Where \(g(z) = u(x,y) + iv(x,y)\) and \(z = x + iy\).
In this exercise, \(g(z)\) is such a function, making it subject to the properties of regular functions, including smoothness and differentiability everywhere on \(\mathbb{C}\).

The exponential function \(e^z\) is a fundamental mathematical function with significant properties in complex analysis. For a complex number \(z = x + iy\), the exponential function is given by:
\(e^z = e^x (\cos y + i \sin y)\).
This formula arises from Euler's formula, which elaborates the relationship between trigonometric functions and the exponential function in the complex plane. It helps in decomposing a complex exponentiation into its real and imaginary parts. For instance, in this problem, when analyzing \(e^{\ln 3 - i \frac{\pi}{4}}\), we have:
\(e^{\ln 3} = 3\)
and
\(e^{- i \frac{\pi}{4}} = \cos \left( \frac{\pi}{4} \right) - i \sin \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}\).
Thus, combining these, we get:
\(e^{\ln 3 - i \frac{\pi}{4}} = 3 \left( \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} \right) = \frac{3}{\sqrt{2}} - i \frac{3}{\sqrt{2}}\).

Complex number inequalities involve determining the bounds on the magnitude of a complex function. In this problem, the given inequality is:
\(|g(z)| \leq 2 \cdot |e^z|\)\.
This tells us that the magnitude of the function \(g(z)\) is bounded above by twice the magnitude of the exponential function \(e^z\). When evaluating the specific point \(z = \ln 3 - i \frac{\pi}{4}\), we need to simplify \(e^z\) to determine the bound:
\(| e^{\ln 3 - i \frac{\pi}{4}} | = | \frac{3}{\sqrt{2}} - i \frac{3}{\sqrt{2}} | = \sqrt{ \left( \frac{3}{\sqrt{2}} \right) ^2 + \left( \frac{3}{\sqrt{2}} \right) ^2 } = 3\).
Therefore,
\(|g\left( \ln 3 - i \frac{\pi}{4} \right) | \leq 2 \cdot 3 = 6\)\.

Function evaluation involves substituting a specific value into a function to determine its output. To find \(g(\ln 3 - i \frac{\pi}{4})\) given the properties of \(g(z)\) and the inequality:
1. We identify the known conditions: \(g(0) = \frac{1+i}{2}\).
2. We use the inequality \(|g(z)| \leq 2 \cdot |e^z|\) to bound the magnitude.
3. Substitute \(z = \ln 3 - i \frac{\pi}{4}\):\[|g(\ln 3 - i \frac{\pi}{4})| \leq 2 \cdot |e^{\ln 3 - i \frac{\pi}{4}}| = 6\].
Given the regularity and the form of \(g(z)\), it can be shown that:
\[g(\ln 3 - i \frac{\pi}{4}) = \frac{6}{4}(1+i) = \frac{3}{2}(1+i)\].
This respects the bounds prescribed by the original inequality and the specific nature of \(g(z)\).