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Precalculus Mathematics Homework Help
Find the integer that is nearest to the area of complex plane A
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[QUOTE="Greychu, post: 3832702, member: 160521"] Consider the region A in the complex plane that consists of all points z such that both [itex]\frac{z}{40}[/itex] and [itex]\frac{40}{\overline{z}}[/itex] have real and imaginary parts between 0 and 1, inclusive. What is the integer that is nearest the area of A? Let z = a + bi and [itex]\overline{z}[/itex] = a - bi a = real part of z, b = imaginary part of z substitute z and [itex]\overline{z}[/itex], we get [itex]\frac{z}{40}[/itex] = [itex]\frac{a + bi}{40}[/itex] [itex]\frac{40}{\overline{z}}[/itex] = [itex]\frac{40(a+bi)}{a^{2} + b^{2}}[/itex] Since both both [itex]\frac{z}{40}[/itex] and [itex]\frac{40}{\overline{z}}[/itex] have real and imaginary parts between 0 and 1, inclusive: Hence For [itex]\frac{z}{40}[/itex]: 0 < [itex]\frac{a}{40}[/itex] < 1, 0 < [itex]\frac{b}{40}[/itex] < 1 For [itex]\frac{40}{\overline{z}}[/itex]: 0 < [itex]\frac{40(a)}{a^{2} + b^{2}}[/itex] < 1, 0 < [itex]\frac{40(b)}{a^{2} + b^{2}}[/itex] < 1 From these 4 equations, if 0 < a < 40 and 0 < b < 40 are fulfilled, then [itex]a^{2}[/itex] + [itex]b^{2}[/itex] = 1600 Well the problem is I don't know how to find out the area A, since the points z can be randomly distributed in the range of 0 < a < 40 and 0 < b < 40. Any ideas? If assuming the complex plane is an square with a = x = 40 and b = y = 40, the area is 1600, which is differ from the answer. Source: [url]http://en.wikipedia.org/wiki/International_Mathematical_Olympiad[/url] [/QUOTE]
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Precalculus Mathematics Homework Help
Find the integer that is nearest to the area of complex plane A
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