Question

1*2.4/2.6consider 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. find the area of $a.$

Answer 1

First, note that

`(40)/(\bar z) = (40z)/(z\bar z) = (40z)/(|z|^2)`

Let
`z=x+i\,y`. Then we have

`\mathrm{Re}\left(\frac z{40}\right) = \frac x{40} \\\\ ~~~~ 0 < \mathrm{Re}\left(\frac z{40}\right) < 1 \implies 0 < \frac x{40} < 1 \implies 0 < x < 40`

`\mathrm{Im}\left(\frac z{40}\right) = \frac y{40} \\\\~~~~ 0 < \mathrm{Im}\left(\frac z{40}\right) < 1 \implies 0 < y < 40`

`\mathrm{Re}\left((40)/(\bar z)\right) = \mathrm{Re}\left((40z)/(|z|^2)\right) = (40x)/(√(x^2+y^2)) \\\\ ~~~~ 0 < \mathrm{Re}\left((40)/(\bar z)\right) < 1 \implies 0 < (40x)/(x^2+y^2) < 1 \implies 0 < 40x < x^2+y^2`

`\mathrm{Im}\left((40)/(\bar z)\right) = (40y)/(x^2+y^2) \\\\ ~~~~ 0 < \mathrm{Im}\left((40)/(\bar z)\right) < 1 \implies 0 < 40y < x^2+y^2`

n the complex plane, the first two inequalities represent a square with vertices at 0, 40, 40 + 40i, and 40i.

The other two inequalities are regions outside disks. The boundaries are

`40x = x^2 + y^2 \implies x^2 - 40x + y^2 = 0 \implies (x-20)^2 + y^2 = 20^2`

and

`40y = x^2 + y^2 \implies x^2 + (y-20)^2 = 20^2`

which are circles with radius 20 and centered at either (20, 0) or (0, 20).

See the attached plot of the region
`A`. We can use basic geometry or calculus to find its area.

Geometry:

The square has area 40² = 1600. From this we subtract the area of a 20-by-20 square, whose area is 20² = 400, as well as the area of a semicircle with radius 20, whose area is π/2•20² = 200π. Then the area of
`A` is 1600 - 400 - 200π = 1200 - 200π.

Calculus:

In polar coordinates, and taking advantage of symmetry, the area is given by the double integral

`\displaystyle 2 \int_0^(\pi/4) \int_(40\cos(\theta))^(40\sec(\theta)) r \, dr \, d\theta`

where
`x=r\cos(\theta)` and
`y=r\sin(\theta)` give the polar functions

`(x-20)^2 + y^2 = 20^2 \implies r^2 - 40r\cos(\theta) = 0 \implies r = 40r\cos(\theta)`

and

`x=40 \implies r\cos(\theta) = 40 \implies r = 40\sec(\theta)`

Then in the integral, we have

`\displaystyle 2 \int_0^(\pi/4) \int_(40\cos(\theta))^(40\sec(\theta)) r \, dr \, d\theta = \int_0^(\pi/4) \bigg((40\sec(\theta))^2 - (40\cos(\theta))^2\bigg) \, d\theta \\\\ ~~~~~~~~ = 1600 \int_0^(\pi/4) \bigg(\sec^2(\theta) - \cos^2(\theta)\bigg) \, d\theta \\\\ ~~~~~~~~ = 1600 \tan(\theta)\bigg|_0^(\pi/4) - 800 \int_0^(\pi/4) (1 + \cos(2\theta)) \, d\theta \\\\ ~~~~~~~~ = 1600 - 800 \bigg(\theta + \frac12 \sin(2\theta)\bigg)\bigg|_0^(\pi/4) \\\\ ~~~~~~~~ = 1600 - 800 \left(\frac\pi4 + \frac12 \sin\left(\frac\pi2\right)\right) \\\\ ~~~~~~~~ = 1600 - 200\pi - 400 = \boxed{1200-200\pi}`