A) \[1-i\]
B) \[1+i\]
C) \[-1+i\]
D) \[-1-i\]
Correct Answer: D
Solution :
Given complex number is \[\frac{{{(1+i)}^{2}}}{1-i}\] \[=\frac{(1+{{i}^{2}}+2i)}{1-i}\times \frac{1+i}{1+i}\] \[=\frac{2i(1+i)}{1-{{i}^{2}}}\] \[=\frac{2i+2{{i}^{2}}}{1+1}\] \[=\frac{2i-2}{2}\] \[=i-1\] \[\therefore \] Required conjugate is \[-i-1\]You need to login to perform this action.
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