A) \[-\frac{2}{5}\]
B) \[\frac{6}{5}\]
C) \[\frac{2}{5}\]
D) \[-\frac{6}{5}\]
Correct Answer: C
Solution :
Given that, \[\frac{{{(1+i)}^{2}}}{2-i}=x+iy\] \[\Rightarrow \] \[\frac{1+{{i}^{2}}+2i}{2-i}=x+iy\] \[\Rightarrow \] \[\frac{1-1+2i}{2-i}=x+iy\] \[\Rightarrow \] \[\frac{2i}{2-i}\times \frac{2+i}{2+i}=x+iy\] \[\Rightarrow \] \[\frac{2i(2+i)}{{{(2)}^{2}}+{{(i)}^{2}}}=x+iy\] \[\Rightarrow \] \[\frac{4i-2}{5}=x+iy\] \[\Rightarrow \] \[x+iy=-\frac{2}{5}+\frac{4}{5}i\] Now, \[x+y=-\frac{2}{5}+\frac{4}{5}=\frac{2}{5}\]
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