A) \[-1250\sqrt{3}\,i\]
B) \[1250\sqrt{3}\,i\]
C) \[-1350\sqrt{3}\,i\]
D) \[1365\sqrt{3}\,i\]
Correct Answer: C
Solution :
\[z=1+ai,\] \[a>0\] \[{{z}^{3}}=1-3{{a}^{2}}+(3a-{{a}^{3}})i\] is a real number \[\Rightarrow \,\,3a-{{a}^{3}}=0\] \[\Rightarrow \,\,{{a}^{2}}=3\] \[\Rightarrow \,\,a=\sqrt{3},\] \[\Rightarrow \,\,a=\sqrt{3},\] \[a\,>\,0\] \[\Rightarrow \,\,z=1+\sqrt{3}i\] \[=2\left( \cos \frac{\pi }{3}+i\,\sin \frac{\pi }{3} \right)\] Now \[1+z+{{z}^{2}}+\,......\,+{{z}^{11}}=\frac{1\left( 1-{{z}^{12}} \right)}{1-z}=\frac{1-{{12}^{12}}\left( \cos 4\pi +i\,\sin 4\pi \right)}{1-\left( 1+i\sqrt{3} \right)}\]\[=\frac{1-{{2}^{12}}}{-i\sqrt{3}}=\frac{4095}{i\sqrt{3}}=-1365\,\,\sqrt{3}\,i\]
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