Suppose  z
is a complex number satisfying  z−1z=4+3i
. Find the length (or modulus) of  z+1z
.

Express your solution in closed form.

Note : If the answer is  12345−−−−−√3
then input   1234513
or  12345^(1/3).

To solve for the modulus of \( z + \frac{1}{z} \), given that \( z \) is a complex number satisfying \( z - \frac{1}{z} = 4 + 3i \), we can follow these steps:

### Step 1: Express \( z \) in polar form
Let \( z = r e^{i\theta} \), where \( r \) is the modulus of \( z \) and \( \theta \) is the argument.

### Step 2: Solve the equation \( z - \frac{1}{z} = 4 + 3i \)
First, express \( \frac{1}{z} \):
\[
\frac{1}{z} = \frac{1}{r e^{i\theta}} = \frac{1}{r} e^{-i\theta}
\]
Then, the given equation becomes:
\[
r e^{i\theta} - \frac{1}{r} e^{-i\theta} = 4 + 3i
\]
This can be rewritten as:
\[
r e^{i\theta} - \frac{1}{r} e^{-i\theta} = \left( r - \frac{1}{r} \right) \cos(\theta) + i \left( r + \frac{1}{r} \right) \sin(\theta) = 4 + 3i
\]
This implies that:
\[
\left( r - \frac{1}{r} \right) \cos(\theta) = 4 \quad \text{and} \quad \left( r + \frac{1}{r} \right) \sin(\theta) = 3
\]

### Step 3: Find \( r \) and \( \theta \)
Square both sides of the equations and add them:
\[
\left( \left( r - \frac{1}{r} \right) \cos(\theta) \right)^2 + \left( \left( r + \frac{1}{r} \right) \sin(\theta) \right)^2 = 4^2 + 3^2 = 25
\]
Expanding the squared terms:
\[
\left( r - \frac{1}{r} \right)^2 \cos^2(\theta) + \left( r + \frac{1}{r} \right)^2 \sin^2(\theta) = 25
\]
This equation is quite complex to solve directly, but we proceed with a different method to simplify the calculations.

### Step 4: Use properties of moduli
We can find the modulus of \( z + \frac{1}{z} \) using the equation:
\[
\left| z + \frac{1}{z} \right|^2 = \left( z + \frac{1}{z} \right) \left( \overline{z} + \frac{1}{\overline{z}} \right)
\]
Given \( z - \frac{1}{z} = 4 + 3i \), the conjugate equation is \( \overline{z} - \frac{1}{\overline{z}} = 4 - 3i \).

Now, multiply \( (z + \frac{1}{z}) \) by its conjugate:
\[
\left| z + \frac{1}{z} \right|^2 = \left( z + \frac{1}{z} \right) \left( \overline{z} + \frac{1}{\overline{z}} \right) = |z|^2 + 2 + \frac{1}{|z|^2}
\]

We already know \( |z|^2 - \frac{1}{|z|^2} = 4 + 3i \), so:
\[
\text{Re}\left(z + \frac{1}{z}\right) = 4 \quad \text{and} \quad \text{Im}\left(z + \frac{1}{z}\right) = 3
\]

Finally, the modulus of \( z + \frac{1}{z} \) is:
\[
\left| z + \frac{1}{z} \right| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5
\]

Thus, the closed-form solution for the modulus is **5**. 

If you have any questions, or want further details, feel free to ask!

Here are some related questions you might explore:

1. How do you compute the modulus of a complex number?
2. What is the geometric interpretation of \( z + \frac{1}{z} \) for complex \( z \)?
3. How does the argument of \( z \) affect \( z - \frac{1}{z} \)?
4. Can you find \( z \) given \( z - \frac{1}{z} = 4 + 3i \) explicitly?
5. How do you interpret \( \left( z + \frac{1}{z} \right) \) in terms of the unit circle?

**Tip:** When dealing with complex numbers, it's often useful to work in polar form, as it simplifies multiplication and division.

Explore how to find the modulus of z + 1/z for a complex number z satisfying z - 1/z = 4 + 3i. Learn the step-by-step solution using polar form and modulus formulas. Suitable for advanced high school students and anyone interested in complex numbers.

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