question_answer

If \[{{z}_{1}},{{z}_{2}},{{z}_{3}}\in C,\,i{{z}^{3}}+{{z}^{2}}-z+i=0\], where \[i=\sqrt{-1}\] then the area of triangle formed by \[{{z}_{1}},{{z}_{2}}\], and \[{{z}_{3}}\] is equal to

A)  \[\frac{1}{\sqrt{2}}\]                                    

B)  \[\sqrt{2}\]

C)  \[2\sqrt{2}\]                                     

D)  \[\frac{1}{2\sqrt{2}}\]

Correct Answer: A

Solution :

\[i{{z}^{3}}+{{z}^{2}}+{{i}^{2}}z+i=0\] \[\Rightarrow ({{z}^{2}}+i)\,\,(iz+1)=0\] \[\therefore z=\frac{-1}{i}\times \frac{i}{i}=i\] or \[{{z}^{2}}=-i\Rightarrow z={{(-i)}^{1/2}}\] As, \[-i=\frac{-2i}{2}=\frac{{{(1-i)}^{2}}}{2}\Rightarrow \sqrt{-i}=\frac{\pm (1-i)}{2}\] So, area \[(\Delta ABC)=\frac{1}{\sqrt{2}}\]