Let a and b be two positive real numbers and `z_(1)` and `z_(2)` be two non-zero complex numbers such that `a|z_(1)|=b|z_(2)|`. If `z=(az_(1))/(bz_(2))+(bz_(2))/(az_(1))`, then

Options

A. Re(z)=0
B. Im(z)=0
C. |z|`=a/b`
D. `|z| gt 2`

Explanation

To solve the problem, we need to analyze the expression given for \( z \) and the conditions provided. Let's break it down step by step.

### Step 1: Define the expression for \( z \)
We start with the expression given in the problem:
\[ z = \frac{a z_1}{b z_2} + \frac{b z_2}{a z_1} \]

### Step 2: Introduce a new variable \( \omega \)
Let’s define:
\[ \omega = \frac{a z_1}{b z_2} \]
This allows us to rewrite \( z \) as:
\[ z = \omega + \frac{1}{\omega} \]

### Step 3: Analyze the modulus condition
From the problem, we know that:
\[ a |z_1| = b |z_2| \]
This implies:
\[ \left| \frac{a z_1}{b z_2} \right| = 1 \]
Thus, we have:
\[ |\omega| = 1 \]

### Step 4: Express \( \omega \) in exponential form
Since \( |\omega| = 1 \), we can express \( \omega \) as:
\[ \omega = e^{i\theta} \]
for some angle \( \theta \).

### Step 5: Substitute back into the expression for \( z \)
Now substituting \( \omega \) back into the expression for \( z \):
\[ z = e^{i\theta} + \frac{1}{e^{i\theta}} = e^{i\theta} + e^{-i\theta} \]

### Step 6: Simplify using Euler's formula
Using Euler's formula, we know:
\[ e^{i\theta} + e^{-i\theta} = 2\cos(\theta) \]
Thus, we have:
\[ z = 2\cos(\theta) \]

### Step 7: Conclude about the nature of \( z \)
Since \( \cos(\theta) \) is a real number, it follows that \( z \) is purely real. Therefore, the imaginary part of \( z \) is:
\[ \text{Im}(z) = 0 \]

### Final Answer
Thus, we conclude that \( z \) is a purely real number.

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