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Question

Mathematics Question on conjugate of a complex number

The conjugate of the complex number (1+i)21−i \frac {(1+i)^2}{1-i} is

A

1−i1 - i

B

1+i1 + i

C

−1+i-1+ i

D

−1−i-1 - i

Answer

−1−i-1 - i

Explanation

Solution

Given complex number is (1+i)21−i\frac{(1 + i)^2}{1 - i}
=(1+i2+2i)1−i×1+i1+i= \frac{\left(1+ i^{2} +2 i\right)}{1-i} \times \frac{1+i}{1+i}
=2i(1+i)1−i2= \frac{2i\left(1+i\right)}{1-i^{2}}
=2i+2i21+1=2i−22= \frac{2 i +2i^{2}}{1+1} =\frac{2i-2}{2}
= i - 1
∴\therefore Required conjugate is −1−i -1-i