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?(adsbygoogle = window.adsbygoogle || []).push({});

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:

http://en.wikipedia.org/wiki/International_Mathematical_Olympiad

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# Homework Help: Find the integer that is nearest to the area of complex plane A

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