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The problem from the image asks to "Determine \( \coth \left( \frac{3\pi i}{4} \right) \)".

To solve this:

We start with the definition of the hyperbolic cotangent function for complex arguments:
\[
\coth(z) = \frac{\cosh(z)}{\sinh(z)}
\]
where \( z = \frac{3\pi i}{4} \).

### Step 1: Calculate \( \cosh \left( \frac{3\pi i}{4} \right) \)
Recall the definition of the hyperbolic cosine:
\[
\cosh(z) = \frac{e^z + e^{-z}}{2}
\]
For \( z = \frac{3\pi i}{4} \), this becomes:
\[
\cosh\left( \frac{3\pi i}{4} \right) = \frac{e^{\frac{3\pi i}{4}} + e^{-\frac{3\pi i}{4}}}{2}
\]
Using Euler's formula \( e^{ix} = \cos(x) + i\sin(x) \), we get:
\[
e^{\frac{3\pi i}{4}} = \cos\left( \frac{3\pi}{4} \right) + i \sin\left( \frac{3\pi}{4} \right) = -\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}
\]
\[
e^{-\frac{3\pi i}{4}} = \cos\left( -\frac{3\pi}{4} \right) + i \sin\left( -\frac{3\pi}{4} \right) = -\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}
\]
Adding these together:
\[
\cosh\left( \frac{3\pi i}{4} \right) = \frac{-\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}} + -\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}}{2} = \frac{-\sqrt{2}}{2} = -\frac{1}{\sqrt{2}}
\]

### Step 2: Calculate \( \sinh \left( \frac{3\pi i}{4} \right) \)
The hyperbolic sine function is defined as:
\[
\sinh(z) = \frac{e^z - e^{-z}}{2}
\]
Substituting \( z = \frac{3\pi i}{4} \):
\[
\sinh\left( \frac{3\pi i}{4} \right) = \frac{e^{\frac{3\pi i}{4}} - e^{-\frac{3\pi i}{4}}}{2}
\]
Using the same values for \( e^{\frac{3\pi i}{4}} \) and \( e^{-\frac{3\pi i}{4}} \) as above:
\[
\sinh\left( \frac{3\pi i}{4} \right) = \frac{\left( -\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}} \right) - \left( -\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}} \right)}{2}
\]
\[
= \frac{2i\frac{1}{\sqrt{2}}}{2} = i \frac{1}{\sqrt{2}}
\]

### Step 3: Calculate \( \coth \left( \frac{3\pi i}{4} \right) \)
Now that we have both \( \cosh\left( \frac{3\pi i}{4} \right) = -\frac{1}{\sqrt{2}} \) and \( \sinh\left( \frac{3\pi i}{4} \right) = i \frac{1}{\sqrt{2}} \), we can compute \( \coth \):
\[
\coth\left( \frac{3\pi i}{4} \right) = \frac{\cosh\left( \frac{3\pi i}{4} \right)}{\sinh\left( \frac{3\pi i}{4} \right)} = \frac{-\frac{1}{\sqrt{2}}}{i \frac{1}{\sqrt{2}}} = \frac{-1}{i} = i
\]

Thus, the value of \( \coth \left( \frac{3\pi i}{4} \right) \) is \( i \).

Would you like further details or clarifications? Here are 5 related questions you might find interesting:

1. What is the relationship between hyperbolic and trigonometric functions?
2. How does Euler's formula help in solving hyperbolic functions for complex arguments?
3. Can you explain the derivation of the formula for \( \cosh(z) \) and \( \sinh(z) \)?
4. How do hyperbolic functions behave for purely imaginary arguments?
5. How does the cotangent function differ from the hyperbolic cotangent function?

**Tip:** When solving hyperbolic functions with imaginary arguments, Euler's formula is key to simplifying the calculations!

Learn how to solve coth(3πi/4) using Euler's formula and hyperbolic function identities. Step-by-step solution for understanding hyperbolic cotangent and complex number calculations.

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