The equation sin z = 2, where z is a complex number, presents a fascinating challenge. Unlike its real-number counterpart, which is bounded between -1 and 1, the sine function in the complex plane exhibits a far richer behavior, allowing for solutions to equations like this. This article will delve into the methods for solving sin z = 2, exploring the intricacies of complex numbers and the exponential form of trigonometric functions. We will uncover the infinite set of solutions and demonstrate how to find them.
Understanding the Complex Sine Function
To tackle sin z = 2, we need to understand the definition of the sine function for complex numbers. We can express the sine function in its exponential form using Euler's formula:
e^(ix) = cos x + i sin x
From this, we can derive the expressions for sine and cosine:
sin x = (e^(ix) - e^(-ix)) / (2i)
cos x = (e^(ix) + e^(-ix)) / 2
Replacing the real variable 'x' with the complex variable 'z = x + iy', we get:
This equation reveals that the complex sine function involves both exponential and imaginary components. This is crucial in understanding how sin z can achieve values outside the [-1, 1] range restricted to real numbers.
Solving sin z = 2 using Exponential Form
Now, let's solve sin z = 2 using the exponential form:
To find z, we take the natural logarithm of both sides:
iz = ln(i(2 ± √3))
Using the properties of logarithms and Euler's formula, we can express the logarithm of a complex number in its polar form. The magnitude of i(2±√3) is 2 ± √3, and the argument is π/2. Therefore:
iz = ln(2 ± √3) + i(π/2 + 2kπ), where k is an integer.
Finally, multiplying by -i:
z = -i ln(2 ± √3) + (π/2 + 2kπ)
This equation provides an infinite set of solutions for z, indexed by the integer k. Each value of k gives a different solution in the complex plane.
Practical Example
Let's find one specific solution for k = 0:
z = -i ln(2 + √3) + π/2
We can approximate this as:
z ≈ -i(1.317) + 1.571 ≈ 1.571 - 1.317i
This is just one solution; there are infinitely many others obtained by varying k.
Conclusion
Solving sin z = 2 in the complex plane demonstrates the power of complex analysis and the elegant interplay between exponential and trigonometric functions. By utilizing Euler's formula and the properties of complex logarithms, we were able to uncover the infinite set of solutions that lie within the complex plane. This exploration highlights the difference between the real and complex domains of trigonometric functions, showcasing the richer mathematical landscape offered by complex numbers.
FAQs
1. Why is sin z = 2 solvable in the complex plane but not in the real plane? The real sine function is bounded between -1 and 1. However, in the complex plane, the sine function is extended, allowing for values beyond this range.
2. Are the solutions z periodic? Yes, the solutions are periodic with a period of 2π along the real axis.
3. How many solutions are there? There are infinitely many solutions, one for each integer value of k.
4. Can we visualize these solutions? Yes, they form a lattice-like structure in the complex plane, with solutions spaced regularly along a line parallel to the real axis.
5. What are the applications of solving equations like sin z = 2? These types of equations appear in various branches of physics and engineering, particularly in the study of wave phenomena and differential equations with complex solutions.
Note: Conversion is based on the latest values and formulas.
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