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

1. Mar 25, 2012

### Greychu

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.

Source:

Last edited: Mar 25, 2012
2. Nov 12, 2016

### haruspex

It helps to know about geometrical inverses. If you 'invert' a straight line with respect to a point off the line you get a circle.
The $\frac 1{\bar z}$ is equivalent to such an inversion. Inverting the two lines x=1/40, y=1/40 thus gives two intersecting circles as shown in the diagram at the link provided. The region of z for which $\frac {40}{\bar z}$ has real and imaginary parts between 0 and 1 is therefore the grey area to the right and above those circles, but extending off to infinity in the rest of the first quadrant. Combining with the other constraint leaves just the grey area.
Finding its area is easy. The white area can be cut into two quadrants and a square.