Question

# The centre of regular polygon of $$\mathrm{n}$$ sides is located at $$Z=0$$ and one of its vertices is $$Z_{1}$$ lf $$Z_{2}$$ is the vertex adjacent to $$Z_{1}$$ then $$Z_{2}=$$

A
Z1(cos2πn±isin2πn)
B
Z1(cosπn±isinπn)
C
Z1(cosπ2n±isinπ2n)
D
Z1(cosπ3n±isinπ3n)

Solution

## The correct option is A $$Z_{1}(\displaystyle \cos\frac{2\pi}{n}\pm i\sin\frac{2\pi}{n})$$Polygon of nsides$$\Rightarrow$$ subsequent vertices subtend angle of $$\frac{2\pi }{n}$$ atorigin.$$[\because 2 \pi$$  is total angle at centre and there are n sides$$\Rightarrow$$ each side subtends $$\frac{2\pi }{n}]$$So,$$z_{2}$$  is adjacent to $$z_{1}$$ $$\Rightarrow$$ $$z_{2}$$   can either be clockwise (cw) or anticlockwise(acw) adjacent.So,$$z _{2}=z_{1}e^{\tfrac{i2\pi }{2}}$$ or $$z _{2}=z _{1}e^{\tfrac{-i2\pi }{2}}$$$$\Rightarrow$$ $$z _{2}=z _{1}(cos\tfrac{2\pi}{n}+isin\tfrac{2\pi}{n})$$ or$$z _{2}=z_{1}(cos\tfrac{2\pi}{n}-isin\tfrac{2\pi}{n})$$$$= z _{1}(cos\tfrac{2\pi}{n}\pm+isin\tfrac{2\pi}{n})$$Maths

Suggest Corrections

0

Similar questions
View More

People also searched for
View More