A) \[\frac{3}{2}+2i\]
B) \[\frac{3}{2}-2i\]
C) \[2-\frac{3}{2}i\]
D) \[2+\frac{3}{2}i\]
Correct Answer: B
Solution :
We have, \[\left| \,z\, \right|-z=1+2i\] If \[z=x+iy,\]then this equation reduces to \[\left| \,x+i\,y\, \right|-(x+i\,y)=1+2i\] \[\Rightarrow \] \[\left( \sqrt{{{x}^{2}}+{{y}^{2}}}-x \right)+(-iy)=1+2i\] On comparing real and imaginary parts of both sides of this equation, we get \[\sqrt{{{x}^{2}}+{{y}^{2}}}-x=1\] \[\Rightarrow \] \[\sqrt{{{x}^{2}}+{{y}^{2}}}=1+x\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}={{(1+x)}^{2}}\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}=1+{{x}^{2}}+2x\] \[\Rightarrow \] \[{{y}^{2}}=1+2x\] ?(i) and \[-y=2\] \[\Rightarrow \] \[y=-2\] Putting this value in Eq. (i), we get \[{{(-2)}^{2}}=1+2x\] \[\Rightarrow \] \[2x=3\] \[\Rightarrow \] \[x=\frac{3}{2}\] Hence, \[z=x+iy\] \[=\frac{3}{2}-2i\]
You need to login to perform this action.
You will be redirected in
3 sec
