Sketching regions in complex plane

[TsAmE]

Sketch the diagram showing the region \(\displaystyle {z \epsilon \mathbb{C} : |z - 2| \leq Rez}\)

Attempt:

\(\displaystyle |z - 2| \leq Rez\)

\(\displaystyle |z - 2| \leq x\)

\(\displaystyle \sqrt{x + iy -2} \leq x\)

\(\displaystyle \sqrt{(x - 2)^2 + y^2} \leq x\)

\(\displaystyle (x - 2)^2 + y^2 \leq x\)

\(\displaystyle x^2 - 4x + 4 + y^2 < x\)

\(\displaystyle x^2 - 5x + 4 + y^2 < 0\)

I thought that this would give me a circle, but when I tried to factorise the x terms I couldnt, since ther discriminant is < 0

 

[Deleted member 4993]

Sketch the diagram showing the region \(\displaystyle {z \epsilon \mathbb{C} : |z - 2| \leq Rez}\)

Attempt:

\(\displaystyle |z - 2| \leq Rez\)

\(\displaystyle |z - 2| \leq x\)

\(\displaystyle \sqrt{x + iy -2} \leq x\)

\(\displaystyle \sqrt{(x - 2)^2 + y^2} \leq x\)

\(\displaystyle (x - 2)^2 + y^2 \leq x\)

\(\displaystyle x^2 - 4x + 4 + y^2 < x\)

\(\displaystyle x^2 - 5x + 4 + y^2 < 0\)

I thought that this would give me a circle, but when I tried to factorise the x terms I couldnt, since ther discriminant is < 0

\(\displaystyle (x-2.5)^2 + y^2 - 2.25< 0\)

\(\displaystyle (x-2.5)^2 + y^2 < 1.5^2\)

 

[TsAmE]

Thanks! The answer happened to be an ellipse (refer to attachment). So I tried to change the form:

\(\displaystyle (x - \frac{5}{2})^2 + y^2 \leq \frac{9}{4}\)

\(\displaystyle \frac{4(x - \frac{5}{2})^2}{9} + \frac{4y^2}{9} \leq 1\)

but since \(\displaystyle a^2 = b^2 = 9\) for the ellipse, howcome the major and minor axis are equal? Is it cause in this case it a circle (this must mean the memo I got is wrong)?