A) 1
B) 2
C) 9
D) 4
E) 3
Correct Answer: E
Solution :
Let\[z=x+iy\] \[\therefore \] \[|z+3-i|=1\] \[\Rightarrow \] \[|x+iy+3-i|=1\] \[\Rightarrow \] \[|x+3+i(y-1)|=1\] \[\Rightarrow \] \[{{(x+3)}^{2}}+{{(y-1)}^{2}}=1\] ...(i) But \[{{\tan }^{-1}}\frac{y}{x}=\pi \] \[\Rightarrow \] \[y=0\] ...(ii) \[\therefore \]From Eqs. (i) and (ii), we get \[{{(x+3)}^{2}}=0\] \[\Rightarrow \] \[x=-3,-3\] \[\therefore \]Required complex number is \[x+iy=-3\] \[\therefore \] \[|x+iy|=|-3|=3\]
You need to login to perform this action.
You will be redirected in
3 sec
