A) equilateral
B) right angled
C) isosceles
D) None of these
Correct Answer: A
Solution :
\[z_{1}^{2}+z_{2}^{2}+2{{z}_{1}}{{z}_{2}}\cos \theta +1=0\] \[\Rightarrow \] \[{{\left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)}^{2}}+2\left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)\cos \theta +1=0\] \[\Rightarrow \] \[{{\left( \frac{{{z}_{1}}}{{{z}_{2}}}+\cos \theta \right)}^{2}}=-(1-{{\cos }^{2}}\theta )=-{{\sin }^{2}}\theta \] \[\Rightarrow \] \[\frac{{{z}_{1}}}{{{z}_{2}}}=-\cos \theta \pm i\sin \theta \] \[\Rightarrow \] \[\left| \frac{{{z}_{1}}}{{{z}_{2}}} \right|=\sqrt{{{(-\cos \theta )}^{2}}+{{\sin }^{2}}\theta }=1\] \[\Rightarrow \] \[|{{z}_{1}}|\,\,=\,\,|{{z}_{2}}|\] \[\Rightarrow \] \[|{{z}_{1}}-0|\,\,=\,\,|{{z}_{2}}-0|\] Thus, triangle with vertices\[{{z}_{1}},\,\,{{z}_{2}}\]is isosceles.
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