If \(\frac{(3-i)^2}{2+i} = A + iB\), where \(A\) and \(B\) are real numbers, then \(A\) and \(B\) are equal to

Question

The number of all four-digit numbers which begin with 4 and end with either zero or five is

Answer 1

The problem requires finding the real and imaginary parts of the complex fraction \(\frac{(3-i)^2}{2+i}\). To do this, we'll follow these steps:

Calculate \((3-i)^2\).
Multiply the result by the conjugate of the denominator to simplify the fraction.
Extract the real and imaginary components to determine \(A\) and \(B\).

Step 1: Calculate \((3-i)^2\)

To find the square of the complex number \(3-i\):

\((3-i)^2 = 3^2 - 2 \cdot 3 \cdot i + (-i)^2 = 9 - 6i + (-1) = 8 - 6i\)

Step 2: Simplify \(\frac{8-6i}{2+i}\)

To divide by a complex number, multiply by the conjugate of the denominator:

\(\frac{8-6i}{2+i} \cdot \frac{2-i}{2-i} = \frac{(8-6i)(2-i)}{(2+i)(2-i)}\)

First, calculate the denominator:

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

Now, calculate the numerator:

\((8-6i)(2-i) = 16 - 8i - 12i + 6i^2 = 16 - 20i + 6(-1) = 16 - 20i - 6 = 10 - 20i\)

So, the division results in:

\(\frac{10-20i}{5} = \frac{10}{5} - \frac{20i}{5} = 2 - 4i\)

Step 3: Identify \(A\) and \(B\)

The real part \(A\) is 2, and the imaginary part \(B\) is -4. Therefore, the values are \(A=2\) and \(B=-4\).

The correct answer is \(A = 2\), \(B = -4\).

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