Question

If \( z = \frac{3+i}{2-i} \), then \( z^{-1} \) is equal to

• A. \(\frac{1+i}{2}\)
• B. \(\frac{1+i}{2}\)
• C. \(\frac{1-i}{2}\)
• D. \(2(1-i)\)
• E. \(1-i\)

Hint:

To simplify a complex fraction, multiply numerator and denominator by the conjugate of the denominator. Also, remember that \(z^{-1}=\frac{1}{z}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This question asks for the multiplicative inverse of a complex number, denoted as \(z^{-1}\). The inverse \(z^{-1}\) is defined as \(\frac{1}{z}\). We can solve this by first finding \(z^{-1}\) in terms of the given fraction and then simplifying it to the standard form \(a+bi\).
Step 2: Key Formula or Approach:
Given \(z = \frac{3+i}{2-i}\), its inverse is:
\[ z^{-1} = \frac{1}{z} = \frac{1}{\frac{3+i}{2-i}} = \frac{2-i}{3+i} \] To simplify a fraction with a complex denominator, we multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of \(a+bi\) is \(a-bi\).
Step 3: Detailed Explanation:
We need to simplify the expression \(z^{-1} = \frac{2-i}{3+i}\).
The conjugate of the denominator \((3+i)\) is \((3-i)\).
Multiply the numerator and denominator by this conjugate:
\[ z^{-1} = \frac{2-i}{3+i} \times \frac{3-i}{3-i} \] \[ z^{-1} = \frac{(2-i)(3-i)}{(3+i)(3-i)} \] Expand the numerator using FOIL (First, Outer, Inner, Last):
\[ (2)(3) + (2)(-i) + (-i)(3) + (-i)(-i) = 6 - 2i - 3i + i^2 \] Since \(i^2 = -1\), the numerator becomes:
\[ 6 - 5i - 1 = 5 - 5i \] Expand the denominator using the difference of squares formula \((a+b)(a-b) = a^2 - b^2\):
\[ (3)^2 - (i)^2 = 9 - (-1) = 9 + 1 = 10 \] Combine the simplified numerator and denominator:
\[ z^{-1} = \frac{5 - 5i}{10} \] Factor out 5 from the numerator and simplify the fraction:
\[ z^{-1} = \frac{5(1 - i)}{10} = \frac{1 - i}{2} \] Step 4: Final Answer:
The inverse \(z^{-1}\) is \(\frac{1-i}{2}\). Therefore, option (C) is the correct answer.