Question

The real part of \(\frac{1+2i}{(2-i)^2}\) is

• A. \(-\frac{1}{5}\)
• B. \(\frac{1}{5}\)
• C. \(-\frac{2}{5}\)
• D. \(\frac{2}{5}\)

Hint:

When simplifying complex fractions, always handle the denominator first. Expand any powers and then multiply by the conjugate to make the denominator a real number. Remember the conjugate of \(a+bi\) is \(a-bi\) and their product is always a real number: \((a+bi)(a-bi) = a^2 + b^2\).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are asked to find the real part of a given complex number. The complex number is given as a fraction.

Step 2: Key Formula or Approach:
To find the real part, we first need to simplify the expression and write it in the standard form \(a+bi\).
First, expand the denominator \((2-i)^2\).
Then, multiply the numerator and the denominator by the conjugate of the resulting denominator to eliminate the imaginary part from the denominator.

Step 3: Detailed Explanation:
Let the complex number be \(z = \frac{1+2i}{(2-i)^2}\).

Step 3a: Simplify the denominator
\[ (2-i)^2 = 2^2 - 2(2)(i) + i^2 \]
\[ = 4 - 4i - 1 \quad (\text{since } i^2 = -1) \]
\[ = 3 - 4i \]
So, the expression becomes:
\[ z = \frac{1+2i}{3-4i} \]

Step 3b: Rationalize the denominator
To convert this into the form \(a+bi\), multiply the numerator and denominator by the conjugate of the denominator, which is \(3+4i\).
\[ z = \frac{1+2i}{3-4i} \times \frac{3+4i}{3+4i} \]
\[ z = \frac{(1+2i)(3+4i)}{(3-4i)(3+4i)} \]
Numerator:
\[ (1)(3) + (1)(4i) + (2i)(3) + (2i)(4i) = 3 + 4i + 6i + 8i^2 = 3 + 10i - 8 = -5 + 10i \]
Denominator:
\[ (3)^2 - (4i)^2 = 9 - 16i^2 = 9 - 16(-1) = 9 + 16 = 25 \]
So, the complex number is:
\[ z = \frac{-5 + 10i}{25} \]

Step 3c: Write in standard form
\[ z = \frac{-5}{25} + \frac{10i}{25} = -\frac{1}{5} + \frac{2}{5}i \]
The standard form is \(a+bi\), where \(a\) is the real part and \(b\) is the imaginary part.
Here, the real part is \(a = -\frac{1}{5}\).

Step 4: Final Answer:
The real part of the complex number is \(-\frac{1}{5}\).