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Determine, in each case, all the \(z \in \mathbb{C}\) with $$ \begin{aligned} \exp (z) &=-2, & \exp (z) &=\mathrm{i}, & & \exp (z) &=-\mathrm{i} \\ \sin z &=100, & \sin z &=7 \mathrm{i}, & \sin z &=1-\mathrm{i} \\ \cos z &=3 \mathrm{i}, & \cos z &=3+4 \mathrm{i}, & \cos z &=13. \end{aligned} $$

Short Answer

Expert verified
Solve each case using logarithmic, inverse trigonometric, and hyperbolic relationships to find \(z\).

Step by step solution

01

Understand the Exponential Solutions

For the equation \(\exp(z) = a\) for complex numbers \(a\), we express \(z\) as \(z = \, \ln(a) + 2k\pi\mathrm{i}\) where \(k\) is an integer. For each case,1. \(\exp(z) = -2\): \(z = \ln|2| + \pi\mathrm{i} + 2k\pi\mathrm{i}\).2. \(\exp(z) = \mathrm{i}\): \(z = \ln|1| + \frac{\pi}{2} \mathrm{i} + 2k\pi\mathrm{i}\).3. \(\exp(z) = -\mathrm{i}\): \(z = \ln|1| + \frac{3\pi}{2} \mathrm{i} + 2k\pi\mathrm{i}\).
02

Understand the Sine Function Solutions

For the equation \(\sin(z) = a\), the solution formula is \(z = (-1)^n x + n\pi + \mathrm{i} \ln\left(|a + \sqrt{a^2 + 1}|\right)\). For each sine case:1. \(\sin(z) = 100\): Use the formula with \(a = 100\).2. \(\sin(z) = 7\mathrm{i}\): Calculate using the formula with \(a = 7\mathrm{i}\).3. \(\sin(z) = 1 - \mathrm{i}\): Apply the formula with \(a = 1 - \mathrm{i}\).
03

Understand the Cosine Function Solutions

For the equation \(\cos(z) = a\), the solution can be obtained by converting the identity \(\cos(z) = \cosh(\mathrm{i}z')\) with \(z' = x' + \mathrm{i}y'\). Rewrite using \(\cosh z' = a\).1. \(\cos(z) = 3\mathrm{i}\): Set \(\cosh(x' + \mathrm{i}y') = 3\mathrm{i}\).2. \(\cos(z) = 3 + 4\mathrm{i}\): Use \(\cosh(x' + \mathrm{i}y') = 3 + 4\mathrm{i}\).3. \(\cos(z) = 13\): Set \(\cosh(x' + \mathrm{i}y') = 13\).
04

Solve and Conclude Each Case

For exponential equations, by adopting \(z = \ln(a) + m\pi\mathrm{i}\), identify specific values of \(a\) from calculations. For sine, use inverse functions (complex arcsines) and for cosine, apply inverse hyperbolic functions to specify \(z\) values.Solutions involve computation or use of inverse complex functions for specified cases.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complex Exponentials
Complex exponentials are an extension of the regular exponential function into the realm of complex numbers. Unlike real exponentials, complex exponentials take the form \( \exp(z) = e^{z} \), where \( z \) is a complex number expressed as \( z = x + iy \).
The beauty of complex exponentials lies in Euler's formula: \( \exp(iy) = \cos(y) + i\sin(y) \). This formula connects them with trigonometric functions and allows for rotating around the unit circle in the complex plane.
Understanding complex exponentials can solve equations such as \( \exp(z) = a \) for complex \( a \). The solution involves finding the complex logarithm, \( z = \ln(a) + 2k \pi i \), allowing for multiple solutions corresponding to different values of \( k \), an integer.
  • Example: For \( \exp(z) = -2 \), express \( z \) as \( \ln|2| + \pi i + 2k \pi i \).
  • Example: \( \exp(z) = i \) yields \( z = \frac{\pi}{2} i + 2k \pi i \) since \( e^{i(\pi/2)} = i \).
The knowledge of complex exponentials is crucial in fields such as signal processing and quantum mechanics, where wave functions can be represented in this form.
Complex Trigonometric Functions
Complex trigonometric functions, like \( \sin(z) \) and \( \cos(z) \), extend trigonometric functions to complex numbers. They are derived from exponential functions using Euler's identities.
For a complex number \( z = x + iy \), the definitions become:
  • \( \sin(z) = \frac{e^{iz} - e^{-iz}}{2i} \)
  • \( \cos(z) = \frac{e^{iz} + e^{-iz}}{2} \)
These functions can output complex numbers, enabling us to solve equations like \( \sin(z) = a \) or \( \cos(z) = a \), even if \( a \) is non-real.
The solutions involve inverse trigonometric identities adapted for complex inputs:
  • For \( \sin(z) = a \), use \( z = (-1)^n x + n\pi + i \ln(|a + \sqrt{a^2 + 1}|) \).
  • For \( \cos(z) = a \), correlate \( \cos(z) \) with the hyperbolic cosine, then use inverse hyperbolic identities.
These functions are imperative in complex analysis, providing insights into periodicity and oscillations in complex domains.
Complex Logarithms
Complex logarithms generalize the logarithm to complex numbers. While the standard logarithm of a real number is unique, complex logarithms capture multiple values because every complex number can be multiplied by exponentials of \( 2\pi i \) due to periodicity.
For a complex number \( z \), \( \ln(z) \) can be expressed as \( \ln |z| + i(\text{Arg}(z) + 2k\pi) \), with \( k \) as any integer and \( \text{Arg}(z) \) signifying the principal argument of \( z \).
  • The principal value of the logarithm is typically considered when \( -\pi < \text{Arg}(z) \leq \pi \).
  • Complex logarithms are multi-valued due to the periodic nature of trigonometric expressions.
Applications include solving equations involving exponentials, understanding branch cuts, and defining analytic continuations.
In solving \( \exp(z) = a \), complex logarithms offer the key to finding \( z \). For example, with \( \exp(z) = -\mathrm{i} \), deduce \( z = \ln|1| + \frac{3\pi}{2} i + 2k\pi i \), unraveling the elliptical path through the complex plane.

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Most popular questions from this chapter

Let \(f=u+\mathrm{i} v: D \rightarrow \mathbb{C}\) be a totally differentiable function (in the sense of real analysis), defined on an open region \(D \subseteq \mathbb{C} .\) Define the operators $$ \begin{aligned} &\frac{\partial f}{\partial z}:=\frac{1}{2}\left(\frac{\partial f}{\partial x}-\mathrm{i} \frac{\partial f}{\partial y}\right), \\ &\frac{\partial f}{\partial \bar{z}}:=\frac{1}{2}\left(\frac{\partial f}{\partial x}+\mathrm{i} \frac{\partial f}{\partial y}\right). \end{aligned} $$ Show: \(f\) is analytic if and only if \(\frac{\partial f}{\partial \bar{z}}=0\), and when this is the case one has \(f^{\prime}=\frac{\partial f}{\partial z}\) Remark. These differential operators \(\partial:=\frac{\partial}{\partial z}\) and \(\bar{\partial}:=\frac{\partial}{\partial \bar{z}}\) were originally introduced by H. POINCARÉ (1899). For them, a systematic calculus was developed by W. WIRTINGER (1927) - the so-called Wirtinger calculus. However, this played a subsidiary role in classical one-variable function theory; its full significance was apparent only in many-variable function theory, for which it was originally developed by WIRTINGER.

Each \(z \in S^{1} \backslash\\{-1\\}\) $$ S^{1}:=\\{z \in \mathbb{C} ; \quad|z|=1\\} $$ can be uniquely represented in the form $$ z=\frac{1+\mathrm{i} \lambda}{1-\mathrm{i} \lambda}=\frac{1-\lambda^{2}}{1+\lambda^{2}}+\frac{2 \lambda}{1+\lambda^{2}} \mathrm{i} $$ with \(\lambda \in \mathbb{R}\).

For \(z \in \mathbb{C} \backslash\\{(k+1 / 2) \pi ; k \in \mathbb{Z}\\}\) let $$ \tan z:=\frac{\sin z}{\cos z} $$ and for \(z \in \mathbb{C} \backslash\\{k \pi ; k \in \mathbb{Z}\\}\) let $$ \cot z:=\frac{\cos z}{\sin z} . $$ Show: $$ \begin{array}{ll} \tan z=\frac{1}{\mathrm{i}} \frac{\exp (2 \mathrm{i} z)-1}{\exp (2 \mathrm{i} z)+1}, & \tan (z+\pi / 2)=-\cot z \\ \tan (-z)=-\tan z \\ \cot z=\mathrm{i} \frac{\exp (2 \mathrm{i} z)+1}{\exp (2 \mathrm{i} z)-1}, & \tan z=\tan (z+\pi) \end{array} $$ $$ \begin{aligned} &\tan z=\cot z-2 \cot (2 z) \\ &\cot (z+\pi)=\cot z. \end{aligned} $$

Let \(n \geq 2\) be a natural number. There is no continuous function \(q_{n}: \mathbb{C} \rightarrow \mathbb{C}\) such that $$ \left(q_{n}(z)\right)^{n}=z \text { for all } z \in \mathbb{C}. $$

Investigate the continuity and complex differentiability, finding the derivative at points where it is differentiable, of the functions \(f\) : (a) $$ \begin{array}{ll} f(z)=z \operatorname{Re}(z), & f(z)=\bar{z} \\ f(z)=z \bar{z}, & f(z)=z /|z|, z \neq 0 \end{array} $$ (b) The exponential function exp is differentiable, and we have \(\exp ^{\prime}=\exp\).

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