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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)
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B
Z1(cosπn±isinπn)
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C
Z1(cosπ2n±isinπ2n)
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D
Z1(cosπ3n±isinπ3n)
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Solution

The correct option is A $$Z_{1}(\displaystyle \cos\frac{2\pi}{n}\pm i\sin\frac{2\pi}{n})$$
Polygon of n


sides


$$\Rightarrow $$ subsequent vertices subtend angle of $$ \frac{2\pi }{n}$$ at
origin.


$$[\because 2 \pi$$  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

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