Let `z_(1), z_(2)` be two complex numbers such that `z_(1) ne 0 and z_(2)//z_(1)` is purely real, then `|(2i z_(1) + 5 z_(2))/(2iz_(1)-5z_(2))|` is equal to

Options

A. 3
B. 2
C. 1
D. 0

Answer

To solve the problem, we need to find the modulus of the expression \(\frac{2i z_1 + 5 z_2}{2i z_1 - 5 z_2}\) given that \(z_2/z_1\) is purely real. Let's denote \(z_2 = k z_1\) where \(k\) is a real number.

### Step 1: Rewrite the expression
We start with the expression:
\[ \frac{2i z_1 + 5 z_2}{2i z_1 - 5 z_2} \]
Substituting \(z_2 = k z_1\):
\[ \frac{2i z_1 + 5(k z_1)}{2i z_1 - 5(k z_1)} = \frac{2i z_1 + 5k z_1}{2i z_1 - 5k z_1} \]

### Step 2: Factor out \(z_1\)
Since \(z_1 \neq 0\), we can factor \(z_1\) out of the numerator and the denominator:
\[ = \frac{z_1(2i + 5k)}{z_1(2i - 5k)} = \frac{2i + 5k}{2i - 5k} \]

### Step 3: Find the modulus
Now we need to find the modulus of the simplified expression:
\[ \left| \frac{2i + 5k}{2i - 5k} \right| \]
Using the property of modulus:
\[ \left| \frac{a}{b} \right| = \frac{|a|}{|b|} \]
we have:
\[ = \frac{|2i + 5k|}{|2i - 5k|} \]

### Step 4: Calculate the modulus of the numerator and denominator
1. **Numerator**:
\[ |2i + 5k| = \sqrt{(0)^2 + (2 + 5k)^2} = \sqrt{(2 + 5k)^2} = |2 + 5k| \]

2. **Denominator**:
\[ |2i - 5k| = \sqrt{(0)^2 + (2 - 5k)^2} = \sqrt{(2 - 5k)^2} = |2 - 5k| \]

### Step 5: Combine the results
Now we can express the modulus as:
\[ \left| \frac{2i + 5k}{2i - 5k} \right| = \frac{|2 + 5k|}{|2 - 5k|} \]

### Step 6: Analyze the result
Notice that if \(k\) is a real number, the expression \(\frac{|2 + 5k|}{|2 - 5k|}\) simplifies under certain conditions. Specifically, if \(k\) is such that \(2 + 5k\) and \(2 - 5k\) are equal in magnitude but opposite in sign, the modulus will equal 1.

Thus, we conclude:
\[ \left| \frac{2i z_1 + 5 z_2}{2i z_1 - 5 z_2} \right| = 1 \]

### Final Answer
The value of \(\left| \frac{2i z_1 + 5 z_2}{2i z_1 - 5 z_2} \right|\) is \(1\).

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