A) 1
B) 2
C) \[\sqrt{3}\]
D) \[\sqrt{5}\]
Correct Answer: D
Solution :
Put \[z=x+iy,\] we get \[\left( x+\sqrt{2}.\sqrt{{{(x+1)}^{2}}+y} \right)+i(y+1)=0+i0\] \[\therefore \]On equating real and imaginary parts, we get \[x+\sqrt{2}.\sqrt{{{(x+1)}^{2}}+{{y}^{2}}}=0\] and \[y+1=0\] so, on solving (i) and (ii), we get \[x=-2,\,\,y=-1\] \[\therefore z=-2-i\Rightarrow |z|\,\,=\sqrt{5}\]You need to login to perform this action.
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