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

• Greychu
The square has area 40²=1600. Each 'quarter' of the circles are congruent by symmetry so they have equal areas. These four areas together are thus 1600*4=6400.In summary, the region A in the complex plane 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. The integer nearest the area of A is 6400. This can be found by inverting two lines x=1/40 and y=1/40, which creates two intersecting circles and the grey area above and to the right of these circles
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:

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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.

## 1. What is the definition of the "area of complex plane A"?

The area of complex plane A refers to the total space or region covered by the points plotted on the complex plane A, which is a two-dimensional graph used to visualize complex numbers.

## 2. How do you calculate the area of complex plane A?

The area of complex plane A can be calculated by finding the product of the length and width of the plane. In other words, it is equal to the length of the horizontal axis (real numbers) multiplied by the length of the vertical axis (imaginary numbers).

## 3. What is the purpose of finding the integer nearest to the area of complex plane A?

Finding the integer nearest to the area of complex plane A is a way to simplify the complex number and make it easier to work with. It can also help in visualizing the location of the complex number on the plane.

## 4. Can there be more than one integer nearest to the area of complex plane A?

Yes, there can be more than one integer nearest to the area of complex plane A. This is because the area of the complex plane is continuous and infinite, so there are multiple integers that can be considered the nearest.

## 5. How do you determine the integer that is nearest to the area of complex plane A?

To determine the integer nearest to the area of complex plane A, you can use the rounding method. If the area of the complex plane is a decimal, you can round it up or down to the nearest integer. If the area is already an integer, then that number is already the nearest integer.

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