A) quadrant I
B) quadrant II
C) quadrant III
D) quadrant IV
Correct Answer: D
Solution :
Given, \[\frac{(1-i\sqrt{3})(1+i)}{(\sqrt{3}+i)}=\frac{(1-i\sqrt{3}+i+\sqrt{3})(\sqrt{3}-i)}{(3+1)}\] \[(\because \,\,{{i}^{2}}=-1)\] \[=\frac{1}{4}.\{(1+\sqrt{3})+i\,(1-\sqrt{3})\}.(\sqrt{3}-i)\] \[=\frac{1}{4}.\{\sqrt{3}(1+\sqrt{3})+i\,(1-\sqrt{3})\sqrt{3}\] \[-(1+\sqrt{3})\,i+(1-\sqrt{3})\}\] \[=\frac{1}{4}.\{\sqrt{3}+3+1-\sqrt{3})+(\sqrt{3}-3-1-\sqrt{3})i\}\] \[=\frac{1}{4}.\{4-4i\}=1-i\] The point \[(1-i)\] in Arg and plane is \[(1,-1)\] which lies in IV th quadrant.
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