If `z=4 + 3i`, then verify that
`-|z| le R e (z) lt |z|`

Answer

To verify the inequality \(-|z| \leq \text{Re}(z) < |z|\) for the complex number \(z = 4 + 3i\), we will follow these steps:

### Step 1: Identify the complex number
Given:
\[ z = 4 + 3i \]

### Step 2: Calculate the modulus of \(z\)
The modulus of a complex number \(z = a + bi\) is given by:
\[ |z| = \sqrt{a^2 + b^2} \]
For our complex number:
\[ |z| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

### Step 3: Determine the real part of \(z\)
The real part of \(z\) is:
\[ \text{Re}(z) = 4 \]

### Step 4: Substitute values into the inequality
Now we need to check the inequality:
\[ -|z| \leq \text{Re}(z) < |z| \]
Substituting the values we calculated:
\[ -|5| \leq 4 < |5| \]
This simplifies to:
\[ -5 \leq 4 < 5 \]

### Step 5: Verify the inequalities
1. Check the left part of the inequality:
\(-5 \leq 4\) is **true**.
2. Check the right part of the inequality:
\(4 < 5\) is also **true**.

### Conclusion
Since both parts of the inequality hold true, we have verified that:
\[ -|z| \leq \text{Re}(z) < |z| \]
is indeed valid for \(z = 4 + 3i\).

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