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

Options

A. `(7)/(25)+i(24)/(25)`
B. `-(7)/(25)-i(24)/(25)`
C. `-(7)/(25)+i(24)/(25)`
D. `(7)/(25)-i(24)/(25)`

Answer

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

### Step 1: Simplify the expression inside the parentheses
We start with the expression \(\frac{1 + 2i}{2 + i}\).

To simplify this, we multiply the numerator and the denominator by the conjugate of the denominator, which is \(2 - i\):

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

### Step 2: Calculate the denominator
Now, calculate the denominator:

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

### Step 3: Calculate the numerator
Next, 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:

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

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

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

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

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

Calculating each term:

1. \(\left(\frac{4}{5}\right)^{2} = \frac{16}{25}\)
2. \(2\left(\frac{4}{5}\right)\left(\frac{3}{5}i\right) = \frac{24}{25}i\)
3. \(\left(\frac{3}{5}i\right)^{2} = \frac{9}{25}i^{2} = -\frac{9}{25}\)

Combining these results:

\[ \frac{16}{25} - \frac{9}{25} + \frac{24}{25}i = \frac{7}{25} + \frac{24}{25}i \]

### 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
Thus, the conjugate of \(\left(\frac{1 + 2i}{2 + i}\right)^{2}\) is:

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

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