# Math Help - Sketching regions in complex plane

1. ## Sketching regions in complex plane

Draw a diagram showing the region ${z \epsilon \mathbb{C} : Re(\frac{1}{z} - 2) \leq 1}$ ?

Attempt:

$Re(\frac{1}{z} - 2) \leq 1$

$Re(\frac{1 - 2z}{z}) \leq 1$

$Re(\frac{1 - 2(x + iy)}{x + iy} ) \leq 1$

$Re(\frac{1 - 2x - i2y}{x + iy} \frac{x - iy}{x - iy})$

$Re(\frac{x - iy - 2x^2 - 2y^2}{x^2 + y^2}) \leq 1$

$\frac{x - 2x^2 - 2y^2}{x^2 + y^2} \leq 1$

$x - 2x^2 - 2y^2 \leq x^2 + y^2$

$-3x^2 + x -3y^2 \leq 0$

$3x^2 - x + \frac{1}{4} + 3y^2 \geq 4$ (completing the square)

I am not sure what else I can do

2. Originally Posted by SyNtHeSiS
Draw a diagram showing the region ${z \epsilon \mathbb{C} : Re(\frac{1}{z} - 2) \leq 1}$ ?

Attempt:

$Re(\frac{1}{z} - 2) \leq 1$

$Re(\frac{1 - 2z}{z}) \leq 1$

$Re(\frac{1 - 2(x + iy)}{x + iy} ) \leq 1$

$Re(\frac{1 - 2x - i2y}{x + iy} \frac{x - iy}{x - iy})$

$Re(\frac{x - iy - 2x^2 - 2y^2}{x^2 + y^2}) \leq 1$

$\frac{x - 2x^2 - 2y^2}{x^2 + y^2} \leq 1$

$x - 2x^2 - 2y^2 \leq x^2 + y^2$

$-3x^2 + x -3y^2 \leq 0$ Correct up to here.

$3x^2 - x + \frac{1}{4} + 3y^2 \geq 4$ (completing the square)

I am not sure what else I can do
Complete the square a bit more carefully! First divide by –3, and get $x^2 - \frac13x + y^2\geqslant0$. Then you should get $(x-\frac16)^2 + y^2 \geqslant\frac1{36}$, which you ought to recognise as the region outside some circle.

3. Looks good until the last line. I get $(x-1/6)^2+y^2\ge (1/6)^2$ after completing the square. This is the area outside the circle with the center (1/6, 0) and radius 1/6.