If `z_(1)=2 + 7i and z_(2)=1- 5i`, then verify that
`|(z_(1))/(z_(2))|= (|z_(1)|)/(|z_(2)|)`

Answer

To verify that \(\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}\) for the complex numbers \(z_1 = 2 + 7i\) and \(z_2 = 1 - 5i\), we will follow these steps:

### Step 1: Find \(z_1\) and \(z_2\)
Given:
- \(z_1 = 2 + 7i\)
- \(z_2 = 1 - 5i\)

### Step 2: Calculate \(\frac{z_1}{z_2}\)
To find \(\frac{z_1}{z_2}\), we will multiply the numerator and denominator by the conjugate of the denominator:

\[ \frac{z_1}{z_2} = \frac{2 + 7i}{1 - 5i} \cdot \frac{1 + 5i}{1 + 5i} \]

Calculating the numerator:

\[ (2 + 7i)(1 + 5i) = 2 \cdot 1 + 2 \cdot 5i + 7i \cdot 1 + 7i \cdot 5i \]
\[ = 2 + 10i + 7i + 35i^2 \]
Since \(i^2 = -1\):
\[ = 2 + 17i - 35 = -33 + 17i \]

Calculating the denominator:

\[ (1 - 5i)(1 + 5i) = 1^2 - (5i)^2 = 1 - 25(-1) = 1 + 25 = 26 \]

Thus, we have:

\[ \frac{z_1}{z_2} = \frac{-33 + 17i}{26} \]

### Step 3: Find the modulus \(\left|\frac{z_1}{z_2}\right|\)
The modulus of a complex number \(\frac{a + bi}{c}\) is given by:

\[ \left|\frac{a + bi}{c}\right| = \frac{\sqrt{a^2 + b^2}}{|c|} \]

Here, \(a = -33\), \(b = 17\), and \(c = 26\).

Calculating the modulus:

\[ \left|\frac{z_1}{z_2}\right| = \frac{\sqrt{(-33)^2 + (17)^2}}{26} \]
\[ = \frac{\sqrt{1089 + 289}}{26} \]
\[ = \frac{\sqrt{1378}}{26} \]

### Step 4: Calculate \(|z_1|\) and \(|z_2|\)
Now, we will find the moduli of \(z_1\) and \(z_2\):

\[ |z_1| = \sqrt{2^2 + 7^2} = \sqrt{4 + 49} = \sqrt{53} \]

\[ |z_2| = \sqrt{1^2 + (-5)^2} = \sqrt{1 + 25} = \sqrt{26} \]

### Step 5: Calculate \(\frac{|z_1|}{|z_2|}\)
Now we can find \(\frac{|z_1|}{|z_2|}\):

\[ \frac{|z_1|}{|z_2|} = \frac{\sqrt{53}}{\sqrt{26}} = \frac{\sqrt{53}}{\sqrt{26}} = \sqrt{\frac{53}{26}} \]

### Step 6: Compare the two results
We have:

\[ \left|\frac{z_1}{z_2}\right| = \frac{\sqrt{1378}}{26} \]
and
\[ \frac{|z_1|}{|z_2|} = \sqrt{\frac{53}{26}} \]

Since both expressions simplify to the same value, we have verified that:

\[ \left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|} \]