Q68 / 180 · KCET 2007 · Mathematics

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

A
$1-i$
Correct answer
B
$1+i$
C
$-1+i$
D
$-1-i$

Solution

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 $1-i$

Hence, the correct answer is option: 'B'.

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