A) \[a=b=2+\sqrt{3}\]
B) \[a=b=2-\sqrt{3}\]
C) \[a=2-\sqrt{3},b=2+\sqrt{3}\]
D) None of these
Correct Answer: B
Solution :
Since the triangle with vertices \[{{z}_{1}}=a+i,{{z}_{2}}=1+bi\] and \[{{z}_{3}}=0\] is equilateral, we have \[z_{1}^{2}+z_{2}^{2}+z_{3}^{2}={{z}_{1}}{{z}_{2}}+{{z}_{2}}{{z}_{3}}+{{z}_{3}}{{z}_{1}}\] Þ \[{{(a+i)}^{2}}+{{(1+ib)}^{2}}+0=(a+i)(1+ib)+0+0\] Þ \[{{a}^{2}}-{{b}^{2}}+2i(a+b)=a-b+i(1+ab)\] Equating real and imaginary parts, \[{{a}^{2}}-{{b}^{2}}=a-b\] ??(i) and \[2(a+b)=1+ab\] ..... (ii) From (i), \[(a-b)[(a+b)-1]=0\] Þ either \[a=b\]or \[a+b=1\] Taking \[a=b\], we get from (ii) \[4a=1+{{a}^{2}}\]or \[{{a}^{2}}-4a+1=0\] \[\therefore \,\,\,\,a=\frac{4\pm \sqrt{16-4}}{2}=2\pm \sqrt{3}\] Since \[0<a<1\]and \[0<b<1,\]we have \[a=b=2-\sqrt{3}\] Taking \[a+b=1\]or \[b=1-a,\]we get from (ii) \[2=1+a(1-a)\]or \[{{a}^{2}}-a+1=0\], which gives imaginary values of \[a\]. Hence \[a=b=2-\sqrt{3}\] is the required value of \[a\] and \[b\].
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