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Question

Mathematics Question on Complex Numbers and Quadratic Equations

Let w=3+i2w = \frac{\sqrt 3 + i }{2} and P=Wn:n=1,2,3,....P = \\{W^n:n=1,2,3,....\\} Further H_1 = \bigg \\{z \in C: Re \, z >\frac{1}{2}\bigg \\} and $H_2=\bigg[ z\in C: Re ,z

A

π2\frac{\pi}{2}

B

π6\frac{\pi}{6}

C

2π3\frac{2\pi}{3}

D

1π6\frac{1\pi}{6}

Answer

1π6\frac{1\pi}{6}

Explanation

Solution

PLAN It is the simple representation of points on Argand plane and
to find the angle between the points
Here, P = Wn=(cosπ6+isinπ6)n=cosnπ6+isinnπ6W^n=\bigg(cos\frac{\pi}{6}+i \, sin\frac{\pi}{6}\bigg)^n=cos\frac{n\pi}{6}+i \, sin\frac{n\pi}{6}
H_1=\bigg \\{z \in \, C: Re(z)>\frac{1}{2}\bigg \\}
PH1\therefore \, P \cap H_1 represents those points for which cosnπ6is+ve\frac{n\pi}{6} is +ve
Hence, it belongs to I or IV quadrant
z1=PH=cosπ6+isinπ6\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, z_1= P \cap H =cos\frac{\pi}{6}+ i sin\frac{\pi}{6}
or cos11π6+isin11π6 \, \, \, \, \, \, \, \, \, \, \, \, \, \, cos\frac{11 \pi}{6}+ i \, sin\frac{11\pi}{6}
z1=32+i2or32i2\therefore \, \, \, \, \, \, \, \, \, \, \, z_1=\frac{\sqrt 3}{2}+\frac{i}{2} \, or \, \frac{\sqrt 3}{2}-\frac{i}{2} \, \, \, \, ....(i)
Similarly
z2=PH2z_2=P \, \cap \, H_2 i.e. those points for which
cosnπ6<0\, \, \, \, \, \, \, \, \, \, \, cos\frac{n\pi}{6}<0
z2=cosπ+isinπ,cos5π6,cos7π6\therefore \, \, z_2=cos\pi+i sin \pi ,cos \frac{5\pi}{6},\frac{cos 7\pi}{6}
\hspace45mm +i sin \frac{7\pi}{6}
z2=1.32+i2,32i2\Rightarrow \, \, \, \, \, \, \, z_2=-1.\frac{-\sqrt 3}{2}+\frac{i}{2},\frac{-\sqrt 3}{2}-\frac{i}{2}
Thus,z1Oz2=2π3,1π6Thus, \, \, \angle z_1Oz_2=\frac{2\pi}{3},\frac{1\pi}{6}