Exam

JEE Advanced

Year

2013

Paper

Paper 2

Subject

Mathematics

Chapter

Complex Number

JEE Advanced 2013 · Paper 2 · Mathematics · Complex Number

Let w = 3 + i 2 and P = \ w ^ n : n = 1, 2, 3, \ .Further H _ 1 = \...

Let $w = \dfrac{\sqrt{3} + \text{i}}{2}$ and $\text{P} = \left\{{w }^{\text{n}} \text{:} \text{n} = \text{1, 2, 3,} \ldots\right\}$.Further ${\text{H}}_{1} = \left\{\text{z} \in \text{C} \text{:} \text{Re z} > \dfrac{1}{2}\right\}$ and ${\text{H}}_{2} = \left\{\text{z} \in \text{C} \text{:} \text{Re z} < \dfrac{- 1}{2}\right\}$, where C is the set of all complex numbers. If ${\text{z}}_{1} \in \text{P} \cap {\text{H}}_{1} \text{,} {\text{z}}_{2} \in \text{P} \cap {\text{H}}_{2}$ and O represents the origin, then $∠ {\text{z}}_{1} \text{O} {\text{z}}_{2} =$

Step-by-step solution

\text{P} = {&omega;}^{\text{n}} = \text\cos \dfrac{\text{n} &pi;}{6} + \text{i} \text\sin \dfrac{\text{n} &pi;}{6} \text{,} {\text{H}}_{1} = \text{Rez} > 1 / 2
{\text{z}}_{1} = \text{P} &cap; {\text{H}}_{1} = \dfrac{\sqrt{3} + \text{i}}{2} \text{,} \dfrac{\sqrt{3} - \text{i}}{2}
{\text{z}}_{2} = \text{P} &cap; {\text{H}}_{2} = - 1 \text{,} \dfrac{- \sqrt{3} + \text{i}}{2} \text{,} \dfrac{- \sqrt{3} - \text{i}}{2}
&ang; {\text{z}}_{1} \text{O} {\text{z}}_{2} = \dfrac{2 &pi;}{3} \text{,} \dfrac{5 &pi;}{6} \text{,} &pi;