What is the conjugate of `((1+2i)/(2+i))^(2)`?

Options

A. `7/25+i24/25`
B. `-7/25-i24/25`
C. `-7/25+i24/25`
D. `7/25-i24/25`

Explanation

To find the conjugate of \(\left(\frac{1+2i}{2+i}\right)^{2}\), we will follow these steps:

### Step 1: Simplify the expression \(\frac{1+2i}{2+i}\)
To simplify this expression, we will rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator.

\[ \frac{1+2i}{2+i} \cdot \frac{2-i}{2-i} = \frac{(1+2i)(2-i)}{(2+i)(2-i)} \]

### Step 2: Calculate the denominator
The denominator is calculated as follows:

\[ (2+i)(2-i) = 2^2 - i^2 = 4 - (-1) = 4 + 1 = 5 \]

### Step 3: Calculate the numerator
Now calculate the numerator:

\[ (1+2i)(2-i) = 1 \cdot 2 + 1 \cdot (-i) + 2i \cdot 2 + 2i \cdot (-i) = 2 - i + 4i - 2i^2 \]
Since \(i^2 = -1\), we have:

\[ = 2 - i + 4i + 2 = 4 + 3i \]

### Step 4: Combine the results
Now we can combine the results from the numerator and denominator:

\[ \frac{1+2i}{2+i} = \frac{4 + 3i}{5} = \frac{4}{5} + \frac{3}{5}i \]

### Step 5: Square the result
Now we need to square the simplified expression:

\[ \left(\frac{4}{5} + \frac{3}{5}i\right)^{2} \]

Using the formula \((a + bi)^2 = a^2 + 2abi + (bi)^2\):

\[ = \left(\frac{4}{5}\right)^{2} + 2 \cdot \frac{4}{5} \cdot \frac{3}{5}i + \left(\frac{3}{5}i\right)^{2} \]
\[ = \frac{16}{25} + \frac{24}{25}i + \frac{9}{25}(-1) \]
\[ = \frac{16}{25} - \frac{9}{25} + \frac{24}{25}i \]
\[ = \frac{7}{25} + \frac{24}{25}i \]

### Step 6: Find the conjugate
The conjugate of a complex number \(a + bi\) is \(a - bi\). Therefore, the conjugate of \(\frac{7}{25} + \frac{24}{25}i\) is:

\[ \frac{7}{25} - \frac{24}{25}i \]

### Final Answer
The conjugate of \(\left(\frac{1+2i}{2+i}\right)^{2}\) is:

\[ \frac{7}{25} - \frac{24}{25}i \]

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