KCET · Maths · Complex Number
The conjugate of the complex number \(\frac{(1+i)^{2}}{1-i}\) is
- A \(1-i\)
- B \(1+i\)
- C \(-1+i\)
- D \(-1-i\)
Answer & Solution
Correct Answer
(B) \(1+i\)
Step-by-step Solution
Detailed explanation
Given complex number is \(\frac{(1+i)^{2}}{1-i}\)
\[
=\frac{\left(1+i^{2}+2 i\right)}{1+i} \times \frac{1+i}{1+i}=\frac{2 i(1+i)}{1-i^{2}}
\]
\[
=\frac{2 i+2 i^{2}}{1-i^{2}}=\frac{2 i-2}{2}=i-1
\]
\(\therefore\) Required conjugate is \(1-i\)
\[
=\frac{\left(1+i^{2}+2 i\right)}{1+i} \times \frac{1+i}{1+i}=\frac{2 i(1+i)}{1-i^{2}}
\]
\[
=\frac{2 i+2 i^{2}}{1-i^{2}}=\frac{2 i-2}{2}=i-1
\]
\(\therefore\) Required conjugate is \(1-i\)
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