Area of an n -gon on the Complex Plane
When the roots of x 9 + x 8 + x 7 + x 6 + x 5 + x 4 + x 3 + x 2 + x + 1 and 1 are simultaneously graphed on the complex plane, a regular n -gon is formed with area q . Find n + q rounded to the nearest integer.
The answer is 13.
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2 solutions
Yup!! Did it the same way!!
Perfect!
can you please express that little trigonometry
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The radius of the 10-sided regular polygon is 1 . Divide it into 10 triangles and you'll notice that they are isosceles, and name the different side L . Take one triangle and bisect it by the different angle. Let's call the angle bisector a . Now, you'll get a right triangle with sides a , 2 L and hypotenuse 1 . But, the angle opposite to 2 L is 1 0 π . So, by the sine function, we get L = 2 sin ( 5 π ) and by the cosine function a = cos ( 5 π ) . With the formula of the area of a polygon: A = 2 a L n , substitute: A = 2 cos ( 5 π ) ( 2 sin ( 5 π ) ) ( 1 0 ) , use a trigonometric identity to get: A = 5 sin ( 5 π )
The above graph plots the complex roots of given equation, and by adding plot of 1+0i, i.e. (1,0) in the complex plane in the above graph, we get a regular polygon of 10 sides, and we can approximate it's area to the area of unit circle. Thus, the answer is :- n+q=floor(10+3.14....)=13.
If we multiply that equation by ( x − 1 ) , we obtain x 1 0 − 1 , which also contains the point ( 1 , 0 ) . Now, that equation represents the 10th roots of unity, and we know if that is plotted, there will be a regular polygon of 10 sides, so n = 1 0 . And the area of that polygon, after a little trigonometry, will be q = 5 sin ( 5 π ) . So, p + q ≈ 1 2 . 9 3 , and rounded to the nearest integer, 1 3 .