If log `sqrt(3)((|z|^(2)-|z|+1)/(2+|z|))gt2`, then the locus of z is

Options

A. |z|=5
B. |z|lt 5
C. |z| gt5
D. |z|=0

Answer

To solve the inequality \( \log_{\sqrt{3}}\left(\frac{|z|^2 - |z| + 1}{2 + |z|}\right) > 2 \), we will follow these steps:

### Step 1: Rewrite the logarithmic inequality
We start with the given inequality:
\[ \log_{\sqrt{3}}\left(\frac{|z|^2 - |z| + 1}{2 + |z|}\right) > 2 \]
To eliminate the logarithm, we can use the property of logarithms:
\[ \frac{|z|^2 - |z| + 1}{2 + |z|} > (\sqrt{3})^2 \]
Since \( (\sqrt{3})^2 = 3 \), we rewrite the inequality as:
\[ \frac{|z|^2 - |z| + 1}{2 + |z|} > 3 \]

### Step 2: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:
\[ |z|^2 - |z| + 1 > 3(2 + |z|) \]
This simplifies to:
\[ |z|^2 - |z| + 1 > 6 + 3|z| \]

### Step 3: Rearrange the inequality
Rearranging the terms leads to:
\[ |z|^2 - |z| - 3|z| + 1 - 6 > 0 \]
This simplifies to:
\[ |z|^2 - 4|z| - 5 > 0 \]

### Step 4: Factor the quadratic expression
Now we can factor the quadratic:
\[ |z|^2 - 4|z| - 5 = (|z| - 5)(|z| + 1) \]
Thus, the inequality becomes:
\[ (|z| - 5)(|z| + 1) > 0 \]

### Step 5: Determine the intervals
To find the intervals where this product is positive, we need to consider the roots:
1. \( |z| - 5 = 0 \) gives \( |z| = 5 \)
2. \( |z| + 1 = 0 \) gives \( |z| = -1 \) (not applicable since modulus cannot be negative)

The critical points divide the number line into intervals:
- \( |z| < -1 \) (not applicable)
- \( -1 < |z| < 5 \)
- \( |z| > 5 \)

### Step 6: Test the intervals
We test the intervals:
1. For \( |z| < 5 \): Choose \( |z| = 0 \) → \( (0 - 5)(0 + 1) < 0 \) (not valid)
2. For \( |z| > 5 \): Choose \( |z| = 6 \) → \( (6 - 5)(6 + 1) > 0 \) (valid)

### Conclusion
The solution to the inequality is:
\[ |z| > 5 \]
This means the locus of \( z \) is outside the circle of radius 5 centered at the origin in the complex plane.

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