Finding the Imaginary Part of log sin(x + iy)

The problem asks for the imaginary part of the complex logarithm of the sine of a complex number, specifically $\log \sin(x + iy)$. To find the imaginary part of a complex logarithm $\log z$, we need to determine the argument of the complex number $z$. The formula for the logarithm of a complex number $z$ is $\log z = \log |z| + i \arg(z)$, where $\log |z|$ is the natural logarithm of the magnitude of $z$ and $\arg(z)$ is the argument (or phase) of $z$. The imaginary part of $\log z$ is simply $\arg(z)$.

Understanding sin(x + iy)

First, let's express $\sin(x + iy)$ in the standard form $A + iB$. We use the formula for the sine of a sum of two angles, where one angle is real and the other is purely imaginary:

\(\sin(x + iy) = \sin x \cos(iy) + \cos x \sin(iy)\)

Now, we use the relationship between trigonometric functions of purely imaginary arguments and hyperbolic functions:

• \(\cos(iy) = \cosh y\)
• \(\sin(iy) = i \sinh y\)

Substituting these into the expression for \(\sin(x + iy)\):

\(\sin(x + iy) = \sin x (\cosh y) + \cos x (i \sinh y)\)

\(\sin(x + iy) = \sin x \cosh y + i \cos x \sinh y\)

So, we have $\sin(x + iy)$ in the form $A + iB$, where:

• \(A = \sin x \cosh y\)
• \(B = \cos x \sinh y\)

Calculating the Imaginary Part (Argument)

The imaginary part of $\log \sin(x + iy)$ is the argument of $\sin(x + iy)$. For a complex number $Z = A + iB$, the principal argument is typically given by \(\arg(Z) = \tan^{-1} \left( \frac{B}{A} \right)\), considering the quadrant of $(A, B)$. Assuming $A \neq 0$, we can find the argument:

\(\text{Imaginary Part} = \arg(\sin(x + iy)) = \tan^{-1} \left( \frac{\cos x \sinh y}{\sin x \cosh y} \right)\)

We can simplify the expression inside the $\tan^{-1}$ using the definitions of cotangent and tangent/hyperbolic tangent:

\(\frac{\cos x \sinh y}{\sin x \cosh y} = \left( \frac{\cos x}{\sin x} \right) \cdot \left( \frac{\sinh y}{\cosh y} \right)\)

Since \(\frac{\cos x}{\sin x} = \cot x\) and \(\frac{\sinh y}{\cosh y} = \tanh y\), the expression becomes:

\(\frac{\cos x \sinh y}{\sin x \cosh y} = \cot x \tanh y\)

Therefore, the imaginary part of $\log \sin(x + iy)$ is:

\(\text{Imaginary Part} = \tan^{-1}(\cot x \tanh y)\)

Summary of the Imaginary Part Calculation

To summarize the steps for finding the imaginary part of $\log \sin(x + iy)$:

1. Expand \(\sin(x + iy)\) using complex trigonometric functions and hyperbolic functions to get it into the form \(A + iB\).
2. Identify the real part \(A = \sin x \cosh y\) and the imaginary part \(B = \cos x \sinh y\) of \(\sin(x + iy)\).
3. The imaginary part of \(\log \sin(x + iy)\) is the argument of \(\sin(x + iy)\), which is \(\arg(\sin(x + iy))\).
4. Calculate the argument using the formula \(\tan^{-1}(B/A)\).
5. Simplify the expression \(B/A\) to arrive at the final form.

Following these steps, we found the imaginary part to be \(\tan^{-1}(\cot x \tanh y)\).