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Introduction
The equation \(\sin z = 2\) may seem straightforward at first glance, but it opens the door to a rich exploration of complex analysis, transcendental functions, and the nature of solutions in the complex plane. Unlike real numbers, where the sine function is limited to the interval \([-1, 1]\), the extension to complex numbers allows the sine function to take on any complex value, including values outside this range. This tutorial aims to provide a thorough understanding of how to solve \(\sin z = 2\), including the derivation of solutions, the role of complex logarithms, the structure of solutions in the complex plane, and related concepts.
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Understanding the Sine Function in the Complex Plane
The Sine Function: Real vs. Complex
In real analysis, the sine function, \(\sin x\), is well-understood within the interval \([-1, 1]\). Its graph oscillates between these bounds, with zeros at integer multiples of \(\pi\), and is periodic with period \(2\pi\).
However, when extended to the complex plane, \(z = x + iy\), the sine function exhibits a much richer behavior. The complex sine function is defined via its exponential form:
\[
\sin z = \frac{e^{i z} - e^{-i z}}{2i}
\]
This expression allows for solutions where \(\sin z\) exceeds the real bounds of \([-1, 1]\). Specifically, \(\sin z\) can take on any complex value, including real numbers greater than 1, such as 2.
Basic Properties
- Periodicity: \(\sin z\) is periodic with period \(2\pi\).
- Complex Values: For complex \(z\), \(\sin z\) can be unbounded.
- Zeroes: \(\sin z = 0\) at \(z = n\pi\), where \(n\) is an integer.
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Solving \(\sin z = 2\)
Recognizing the Nature of the Equation
Since \(\sin z = 2\) involves a complex variable, the solutions are complex numbers \(z\) satisfying the equation. Given the definition:
\[
\sin z = \frac{e^{i z} - e^{-i z}}{2i}
\]
we aim to find all \(z\) such that:
\[
\frac{e^{i z} - e^{-i z}}{2i} = 2
\]
Multiplying both sides by \(2i\):
\[
e^{i z} - e^{-i z} = 4i
\]
This equation involves exponential functions, which suggests solving using logarithmic methods and properties of complex analysis.
Deriving the Solutions
Step 1: Express \(e^{i z}\) in terms of a single variable.
Let:
\[
w = e^{i z}
\]
Then:
\[
e^{-i z} = \frac{1}{w}
\]
Our transformed equation becomes:
\[
w - \frac{1}{w} = 4i
\]
Step 2: Multiply through by \(w\):
\[
w^2 - 1 = 4i w
\]
Rearranged as a quadratic in \(w\):
\[
w^2 - 4i w - 1 = 0
\]
Step 3: Solve the quadratic:
\[
w = \frac{4i \pm \sqrt{(4i)^2 - 4 \times 1 \times (-1)}}{2}
\]
Calculate the discriminant:
\[
(4i)^2 - 4 \times 1 \times (-1) = 16 i^2 + 4 = 16(-1) + 4 = -16 + 4 = -12
\]
Thus:
\[
w = \frac{4i \pm \sqrt{-12}}{2}
\]
Express \(\sqrt{-12}\):
\[
\sqrt{-12} = \sqrt{12} \times i = 2 \sqrt{3} \, i
\]
Therefore:
\[
w = \frac{4i \pm 2 \sqrt{3} i}{2} = 2i \pm \sqrt{3} i
\]
Factor out \(i\):
\[
w = i (2 \pm \sqrt{3})
\]
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Expressing \(z\) in terms of \(w\)
Recall that:
\[
w = e^{i z}
\]
Taking the natural logarithm:
\[
i z = \ln w + 2 \pi i n, \quad n \in \mathbb{Z}
\]
The addition of \(2 \pi i n\) accounts for the multi-valued nature of the complex logarithm.
Thus:
\[
z = -i \ln w + 2 \pi n
\]
Substitute \(w\):
\[
z = -i \ln \left(i (2 \pm \sqrt{3})\right) + 2 \pi n
\]
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Computing the Logarithm
The logarithm of a complex number \(w = r e^{i \theta}\) is:
\[
\ln w = \ln r + i \theta
\]
where:
- \(r = |w|\), the modulus of \(w\),
- \(\theta = \arg(w)\), the argument of \(w\).
Calculate the modulus:
\[
|w| = |i (2 \pm \sqrt{3})| = |i| \times |2 \pm \sqrt{3}| = 1 \times |2 \pm \sqrt{3}|
\]
Since \(2 + \sqrt{3} > 0\) and \(2 - \sqrt{3} > 0\), both are positive real numbers:
\[
|w| = 2 \pm \sqrt{3}
\]
Calculate the arguments:
\[
\arg(w) = \arg(i (2 \pm \sqrt{3})) = \arg(i) + \arg(2 \pm \sqrt{3})
\]
- \(\arg(i) = \frac{\pi}{2}\),
- \(2 \pm \sqrt{3}\) are positive real numbers, so:
\[
\arg(2 \pm \sqrt{3}) = 0
\]
Therefore:
\[
\arg(w) = \frac{\pi}{2} + 0 = \frac{\pi}{2}
\]
Now, the logarithm:
\[
\ln w = \ln |w| + i \arg(w) = \ln (2 \pm \sqrt{3}) + i \frac{\pi}{2}
\]
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Final Expression for \(z\)
Recall:
\[
z = -i \left( \ln (2 \pm \sqrt{3}) + i \frac{\pi}{2} \right) + 2 \pi n
\]
Distribute \(-i\):
\[
z = -i \ln (2 \pm \sqrt{3}) - i \times i \frac{\pi}{2} + 2 \pi n
\]
Since \(i \times i = -1\):
\[
z = -i \ln (2 \pm \sqrt{3}) + \frac{\pi}{2} + 2 \pi n
\]
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Summary of Solutions
The solutions to \(\sin z = 2\) are:
\[
\boxed{
z = \frac{\pi}{2} - i \ln (2 \pm \sqrt{3}) + 2 \pi n, \quad n \in \mathbb{Z}
}
\]
where the \(\pm\) accounts for the two roots derived from the quadratic solution.
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Geometric Interpretation and Properties of Solutions
Distribution in the Complex Plane
- The solutions are infinitely many, forming a discrete set in the complex plane.
- Each solution differs by a multiple of \(2\pi\) along the real axis, reflecting the periodicity of sine.
- The imaginary part involves the logarithm of real constants, resulting in fixed shifts vertically in the complex plane.
Nature of the Solutions
- These solutions are complex numbers with non-zero imaginary parts, indicating that solutions are not on the real axis.
- The imaginary parts involve \(\ln(2 \pm \sqrt{3})\), which are positive real numbers, ensuring the solutions are located away from the real axis.
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Related Concepts and Extensions
Sine Function Outside the Range \([-1, 1]\)
In real analysis, \(\sin x = 2\) has no solutions because \(\sin x\) is bounded. In complex analysis, the sine function is unbounded, allowing solutions for any complex value.
Multi-Valued Nature of Complex Logarithm
The logarithm function in complex analysis is multi-valued due to the periodic nature of the complex exponential:
\[
\ln w = \ln r + i (\theta + 2 \pi n)
\]
Hence, solutions involve an infinite set parameterized by \(n \in \mathbb{Z}\).
General Solution for \(\sin z = c\)
For any complex constant
Frequently Asked Questions
What does solving sin z = 2 entail in complex analysis?
Solving sin z = 2 involves finding all complex numbers z such that their sine equals 2, which extends beyond real numbers since sine of real numbers ranges between -1 and 1. This requires using complex analysis techniques and the inverse sine function in the complex plane.
Is there a real solution to sin z = 2?
No, there is no real solution because the sine of any real number is always between -1 and 1. Therefore, sin z = 2 has no solutions in real numbers.
How do you find solutions to sin z = 2 in the complex plane?
To solve sin z = 2 in the complex plane, use the complex inverse sine function: z = arcsin(2) + 2π n or z = π - arcsin(2) + 2π n, where arcsin(2) is a complex number computed via logarithmic expressions, and n is any integer.
What is the value of arcsin(2) in complex analysis?
In complex analysis, arcsin(2) can be expressed as arcsin(2) = -i ln( i2 + √(1 - 4) ), which simplifies to complex logarithmic forms involving square roots of negative numbers.
Can you provide the general solution formula for sin z = 2?
Yes, the general solutions are z = arcsin(2) + 2π n and z = π - arcsin(2) + 2π n, where n is any integer and arcsin(2) is calculated using complex logarithms.
What are the implications of solving sin z = 2 in physics or engineering?
Solving sin z = 2 in physics or engineering often relates to analyzing wave phenomena or oscillations extended into the complex domain, which can help in understanding resonance, signal processing, or stability analysis involving complex variables.
Are solutions to sin z = 2 unique in the complex plane?
No, solutions are not unique; due to the periodic nature of sine in the complex plane, there are infinitely many solutions differing by multiples of 2π added to the principal solutions.
What tools are commonly used to solve equations like sin z = 2?
Complex logarithms, inverse sine functions, and branch cut analysis are commonly used tools to solve equations like sin z = 2 in the complex plane.
