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

Consider the region A in the complex plane that consists of all points z such that both $\frac{z}{40}$ and $\frac{40}{\overline{z}}$ 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 $\overline{z}$ = a - bi
a = real part of z, b = imaginary part of z

substitute z and $\overline{z}$, we get

$\frac{z}{40}$ = $\frac{a + bi}{40}$

$\frac{40}{\overline{z}}$ = $\frac{40(a+bi)}{a^{2} + b^{2}}$

Since both both $\frac{z}{40}$ and $\frac{40}{\overline{z}}$ have real and imaginary parts between 0 and 1, inclusive:
Hence
For $\frac{z}{40}$:
0 < $\frac{a}{40}$ < 1,
0 < $\frac{b}{40}$ < 1

For $\frac{40}{\overline{z}}$:
0 < $\frac{40(a)}{a^{2} + b^{2}}$ < 1,
0 < $\frac{40(b)}{a^{2} + b^{2}}$ < 1

From these 4 equations, if 0 < a < 40 and 0 < b < 40 are fulfilled, then

$a^{2}$ + $b^{2}$ = 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.

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