Question

Evaluate each of the following in \(x+i y\) form, and compare with a computer solution. $$\sin \left[i \ln \left(\frac{\sqrt{3}+i}{2}\right)\right]$$

Short answer

-1/2

Step-by-step solution

Understand the expression

The given expression is \ \( \sin \[i \ln \left(\frac{\sqrt{3} + i}{2}\right)\] \). We need to evaluate this expression in terms of \( x + iy \) form, where \( x \) and \( y \) are real numbers.

Simplify the logarithmic term

Rewrite the complex number \( \frac{\sqrt{3} + i}{2} \) in polar form as \( re^{i \theta} \). Here, \( r \) is the magnitude and \( \theta \) is the argument of the complex number. \ \( r = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{i}{2}\right)^2} = 1 \) and \( \theta = \tan^{-1}(\frac{1/2}{\sqrt{3}/2}) = \tan^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6} \). Thus, \[ \frac{\sqrt{3} + i}{2} = e^{i \frac{\pi}{6}} \]

Apply logarithmic function

Next, apply the natural logarithm to \( e^{i \frac{\pi}{6}} \): \[ \ln(e^{i \frac{\pi}{6}}) = i \frac{\pi}{6} \]

Multiply by imaginary unit

Now, multiply the result by \( i \): \[ i \cdot ( i \frac{\pi}{6} ) = -\frac{\pi}{6} \]

Evaluate sine function

Finally, evaluate the sine function: \[ \sin( -\frac{\pi}{6} ) = -\sin(\frac{\pi}{6}) = -\frac{1}{2} \]

Key concepts

polar form

Polar form is a way to represent complex numbers using their magnitude and angle. Instead of writing a complex number as \(a + bi\), where \(a\) and \(b\) are real numbers, we use the magnitude \(r\) and the angle \(\theta\). Here, \(r\) is the distance from the origin to the point, and \(\theta\) is the angle from the positive real axis.

For example, if we have a complex number like \(z = \frac{\sqrt{3} + i}{2}\), it has a magnitude and an argument. We can calculate the magnitude as \(r = \sqrt{(\frac{\sqrt{3}}{2})^2 + (\frac{i}{2})^2}\), which simplifies to 1. The argument \(\theta\) is found using the inverse tangent function, resulting in \(\frac{\pi}{6}\). Thus, we can write the complex number in polar form as:

\[ \frac{\sqrt{3} + i}{2} = e^{i \pi/6} \]
Polar form makes it easier to handle operations like multiplication, division, and exponentiation of complex numbers.

logarithm of complex numbers

Logarithms can also be applied to complex numbers. When you take the logarithm of a complex number in polar form, it simplifies the process greatly.

For a complex number \(z = re^{i \theta}\), its natural logarithm (ln) is given by:
\[ \ln(z) = \ln(re^{i \theta}) = \ln(r) + i \theta \]
In our example problem, we applied this property to the complex number \(e^{i \pi/6}\). Taking the logarithm of this expression results in:
\[ \ln(e^{i \pi/6}) = i \pi/6 \]

sine function in complex numbers

The sine function can be extended to complex arguments using Euler's formula. When you compute the sine of a complex number, you can use the following identity:
\[ \sin(z) = \frac{e^{iz} - e^{-iz}}{2i} \]
In our exercise, we needed to find \(\sin(-i \ln(e^{i \pi/6}))\). After we've simplified the expression and substituted \(-i \pi/6\) into the sine formula, we can calculate:
\[ \sin(-\pi/6) = - \sin(\pi/6) \] which solves to:
\[ -\frac{1}{2} \]
Essentially, the sine function's extension to complex numbers maintains similar periodic properties as with real numbers.

Euler's formula

Euler's formula is a fundamental bridge between complex exponentials and trigonometric functions. It states:
\[ e^{i\theta} = \cos(\theta) + i \sin(\theta) \]
This formula is incredibly powerful in expressing complex numbers in exponential form and simplifies the computations of many complex expressions.

In our problem, Euler's formula allowed us to convert \( \frac{\sqrt{3} + i}{2} \) to \( e^{i \pi/6} \), making it straightforward to apply the logarithm and subsequent sine operations. Euler's formula is not just helpful but essential for working with complex numbers in various mathematical and engineering fields.