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

Answer

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

### Step 1: Calculate the magnitude of \(z\)
The magnitude of a complex number \(z = a + bi\) is given by the formula:
\[ |z| = \sqrt{a^2 + b^2} \]
For \(z = 4 + 3i\), we have \(a = 4\) and \(b = 3\).
\[ |z| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

### Step 2: Identify the imaginary part of \(z\)
The imaginary part of the complex number \(z = a + bi\) is simply \(b\). Therefore, for \(z = 4 + 3i\):
\[ \text{Im}(z) = 3 \]

### Step 3: Verify the inequality \(-|z| < \text{Im}(z)\)
We need to check if:
\[ -|z| < \text{Im}(z) \]
Substituting the values we found:
\[ -5 < 3 \]
This inequality is true.

### Step 4: Verify the inequality \(\text{Im}(z) \leq |z|\)
Next, we check if:
\[ \text{Im}(z) \leq |z| \]
Substituting the values:
\[ 3 \leq 5 \]
This inequality is also true.

### Conclusion
Since both inequalities \(-|z| < \text{Im}(z)\) and \(\text{Im}(z) \leq |z|\) hold true, we have verified that:
\[ -|z| < \text{Im}(z) \leq |z| \]