A) \[2\sqrt{3}\]
B) \[\frac{7}{2}\]
C) \[\sqrt{10}\]
D) \[\frac{15}{4}\]
Correct Answer: B
Solution :
[b] \[\therefore \,\,\,\,\,\,\,|z-i|\,\,=\,\,|z+2i|\] is perpendicular bisector of line segment joining \[(0,1)\] and \[(0,-2)\] that is \[y=-\frac{1}{2}\] ...(i) \[|z|=\frac{5}{2}\] represents a circle having equation \[{{x}^{2}}+{{y}^{2}}=\frac{25}{4}\] ...(ii) From (i) and (ii) \[x=\pm \sqrt{6},\,\,\,y=-\frac{1}{2}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,z=\pm \sqrt{6}-\frac{1}{2}i\] So \[\left| z+3i \right|\,\,=\,\,\sqrt{{{\left( \pm \sqrt{6} \right)}^{2}}+{{\left( \frac{5}{2} \right)}^{2}}}=\frac{7}{2}\]
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