A) \[\frac{25\sqrt{3}}{2}\]
B) \[\frac{25\sqrt{3}}{4}\]
C) \[\frac{75\sqrt{3}}{2}\]
D) \[\frac{75\sqrt{3}}{4}\]
Correct Answer: B
Solution :
Given, \[|z|\,=5,\,\,|\beta z|\,=5\] \[(\because \,\,|\beta |\,=1)\] and \[|z+\beta z+{{\beta }^{2}}z+{{\beta }^{3}}z|\] \[=\,|z||1+\beta +{{\beta }^{2}}+{{\beta }^{3}}|\] \[=\,|z||-{{\beta }^{4}}|\] \[(\because \,1+\beta +{{\beta }^{2}}+{{\beta }^{3}}+{{\beta }^{4}}=0)\] = 5 So, trinagle will be an equilateral triangle. \[\therefore \] Area \[=\frac{\sqrt{3}}{4}|z{{|}^{2}}=\frac{\sqrt{3}}{4}\times 25=\frac{25\sqrt{3}}{4}\]
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