What is the value if `z^2 + 1/z^2 ` when z=`5+2sqrt6` ?

Options

A. 49
B. 50
C. 98
D. 100

Answer

To find the value of \( z^2 + \frac{1}{z^2} \) when \( z = 5 + 2\sqrt{6} \), we can follow these steps:

### Step 1: Calculate \( \frac{1}{z} \)
To find \( \frac{1}{z} \), we rationalize the denominator. We multiply the numerator and denominator by the conjugate of the denominator:

\[ \frac{1}{z} = \frac{1}{5 + 2\sqrt{6}} \cdot \frac{5 - 2\sqrt{6}}{5 - 2\sqrt{6}} = \frac{5 - 2\sqrt{6}}{(5 + 2\sqrt{6})(5 - 2\sqrt{6})} \]

### Step 2: Simplify the Denominator
Now, we simplify the denominator using the difference of squares:

\[ (5 + 2\sqrt{6})(5 - 2\sqrt{6}) = 5^2 - (2\sqrt{6})^2 = 25 - 24 = 1 \]

Thus,

\[ \frac{1}{z} = 5 - 2\sqrt{6} \]

### Step 3: Calculate \( z + \frac{1}{z} \)
Now we can find \( z + \frac{1}{z} \):

\[ z + \frac{1}{z} = (5 + 2\sqrt{6}) + (5 - 2\sqrt{6}) = 5 + 5 = 10 \]

### Step 4: Calculate \( z^2 + \frac{1}{z^2} \)
Using the identity \( z^2 + \frac{1}{z^2} = (z + \frac{1}{z})^2 - 2 \):

\[ z^2 + \frac{1}{z^2} = (10)^2 - 2 = 100 - 2 = 98 \]

### Final Answer
Thus, the value of \( z^2 + \frac{1}{z^2} \) is \( \boxed{98} \).

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