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

Answer

To verify the inequality \( |z_1 + z_2| \leq |z_1| + |z_2| \) for the complex numbers \( z_1 = 2 + 7i \) and \( z_2 = 1 - 5i \), we will follow these steps:

### Step 1: Calculate \( z_1 + z_2 \)
We start by adding the two complex numbers:
\[ z_1 + z_2 = (2 + 7i) + (1 - 5i) \]
Combine the real and imaginary parts:
\[ = (2 + 1) + (7 - 5)i = 3 + 2i \]

### Step 2: Calculate \( |z_1 + z_2| \)
Now we find the modulus of \( z_1 + z_2 \):
\[ |z_1 + z_2| = |3 + 2i| = \sqrt{3^2 + 2^2} \]
Calculating the squares:
\[ = \sqrt{9 + 4} = \sqrt{13} \]

### Step 3: Calculate \( |z_1| \)
Next, we calculate the modulus of \( z_1 \):
\[ |z_1| = |2 + 7i| = \sqrt{2^2 + 7^2} \]
Calculating the squares:
\[ = \sqrt{4 + 49} = \sqrt{53} \]

### Step 4: Calculate \( |z_2| \)
Now, we calculate the modulus of \( z_2 \):
\[ |z_2| = |1 - 5i| = \sqrt{1^2 + (-5)^2} \]
Calculating the squares:
\[ = \sqrt{1 + 25} = \sqrt{26} \]

### Step 5: Calculate \( |z_1| + |z_2| \)
Now we find the sum of the moduli:
\[ |z_1| + |z_2| = \sqrt{53} + \sqrt{26} \]

### Step 6: Verify the inequality
We need to check if:
\[ |z_1 + z_2| \leq |z_1| + |z_2| \]
Substituting the values we calculated:
\[ \sqrt{13} \leq \sqrt{53} + \sqrt{26} \]

To verify this, we can square both sides (since both sides are positive):
\[ 13 \leq (\sqrt{53} + \sqrt{26})^2 \]
Expanding the right-hand side:
\[ (\sqrt{53} + \sqrt{26})^2 = 53 + 26 + 2\sqrt{53 \cdot 26} = 79 + 2\sqrt{1378} \]
Since \( 2\sqrt{1378} \) is a positive number, we can conclude that:
\[ 79 + 2\sqrt{1378} > 13 \]
Thus, the inequality \( |z_1 + z_2| \leq |z_1| + |z_2| \) holds true.

### Conclusion
Therefore, we have verified that:
\[ |z_1 + z_2| \leq |z_1| + |z_2| \]