A) \[\frac{\sqrt{34}}{3}\]
B) \[\frac{5}{4}\]
C) \[\frac{\sqrt{41}}{4}\]
D) \[\frac{5}{3}\]
Correct Answer: D
Solution :
Let \[z=x+iy\] \[\therefore \]\[|z|=\sqrt{{{x}^{2}}+{{y}^{2}}}\] Now\[|z|+z=3+i\] \[\therefore \]\[\sqrt{{{x}^{2}}+{{y}^{2}}}+x+iy=3+i\] \[\Rightarrow \]\[x+\sqrt{{{x}^{2}}+{{y}^{2}}}=3\]and\[y=1\] \[\Rightarrow \]\[x+\sqrt{{{x}^{2}}+}1=3\]and\[\Rightarrow \]\[\sqrt{{{x}^{2}}+}1=3-x\] \[\Rightarrow \]\[{{x}^{2}}+1={{(3-x)}^{2}}\]\[\Rightarrow \]\[x=4/3\] \[\therefore \]\[|z|=\sqrt{\frac{16}{9}+1}=\frac{5}{3}\]
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