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To solve these problems, we will use **Euler's formula**, which is given by:

\[
e^{ix} = \cos(x) + i \sin(x)
\]

### a. \(\exp(2 + 3i)\)

This expression is equivalent to \(e^{2 + 3i}\), and can be separated into two parts:

\[
e^{2 + 3i} = e^2 \cdot e^{3i}
\]

Using Euler's formula for \(e^{3i}\):

\[
e^{3i} = \cos(3) + i \sin(3)
\]

Thus, the full expression becomes:

\[
e^{2 + 3i} = e^2 \cdot (\cos(3) + i \sin(3))
\]

Now, we calculate the real and imaginary parts:

\[
\text{Real part: } e^2 \cos(3)
\]
\[
\text{Imaginary part: } e^2 \sin(3)
\]

Therefore, \( \exp(2 + 3i) = e^2 \cos(3) + e^2 \sin(3) i \).

---

### b. \(e^{\left(\frac{3\pi}{2}i\right)}\)

Using Euler's formula directly for \(e^{ix}\), we have:

\[
e^{\left(\frac{3\pi}{2}i\right)} = \cos\left(\frac{3\pi}{2}\right) + i \sin\left(\frac{3\pi}{2}\right)
\]

From trigonometric values:

\[
\cos\left(\frac{3\pi}{2}\right) = 0, \quad \sin\left(\frac{3\pi}{2}\right) = -1
\]

Thus:

\[
e^{\left(\frac{3\pi}{2}i\right)} = 0 - i
\]

So the expression becomes \( 0 - i \).

---

### c. \(9^{6 - 5i}\)

We can rewrite \(9^{6 - 5i}\) as:

\[
9^{6 - 5i} = (9^6) \cdot 9^{-5i}
\]

Since \(9 = e^{\ln(9)}\), we convert \(9^{-5i}\) to an exponential form:

\[
9^{-5i} = e^{-5i \ln(9)}
\]

Using Euler's formula:

\[
e^{-5i \ln(9)} = \cos(5 \ln(9)) + i \sin(5 \ln(9))
\]

Thus, the full expression becomes:

\[
9^{6 - 5i} = 9^6 \cdot (\cos(5 \ln(9)) + i \sin(5 \ln(9)))
\]

Finally, the real part is \(9^6 \cos(5 \ln(9))\) and the imaginary part is \(9^6 \sin(5 \ln(9)) i\).

---

Would you like a more detailed breakdown or further clarifications on any part of the solution?

### Related questions:
1. What is the general form of Euler's formula?
2. How do we split complex exponentials into real and imaginary parts?
3. What are the trigonometric values of angles like \(3\pi/2\)?
4. How do logarithms help in expressing powers of numbers in complex exponentials?
5. Can Euler's formula be extended to solve differential equations involving complex numbers?

### Tip:
For exponentials of complex numbers, always try to break the problem into the product of a real exponential and a complex exponential using Euler's formula.

Use Euler's formula to write each of the following expressions in the form a + bi. a) exp(2 + 3i), b) e^((3π/2)i), c) 9^(6-5i).

Learn how to apply Euler's formula to solve complex exponential expressions and rewrite them in the form a + bi. This detailed guide covers problems such as exp(2 + 3i), e^((3π/2)i), and 9^(6-5i). Ideal for advanced high school or college-level students.

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