Complex number, finding complex plane

Question

I have to find the complex plane region that is determined by the following condition:

$$|2z+3|\gt 4$$

I develop using my definition of $z$ as $a+bi$

and I get to

$$a^2+3a-b^2 \gt \frac{7}{4}$$

After this, I do not know what to do, any help would be appreciated.

Answer 1

$$|2z+3|^2 = |2a+3+2bi|^2=(2a+3)^2+4b^2>16$$ Dividing both sides by $4$, we get $$\left(a+\frac{3}{2}\right)^2+b^2>4$$

The resulting region is the exterior of a circle, not including the boundary.

(You made a mistake with the minus in front of $b^2$ )

Answer 2

Ok. Such problems are problems on the geometry of complex numbers.

Let's first use, not the complex variable $z$, but the complex variable $$ w = 2z + 3. $$ In the $w$-plane, your inequality is $$ |w| > 4, $$ which is simply the strict exterior of the disc with center $0$ and radius $4$.

What does going back to the $z$ variable do to that disc? $$ z = {1 \over 2}\left(w - 3\right), $$ so: (i) shift the disc by 3 units "to the left" along the real axis, and then (ii) downscale the entire plane by factor $1/2$. We end up with a disc centered at $z = -3/2$ and with radius $2$. The exterior of that disc is the desired region.

A good source: Part 2 of Lectures and Problems: A Gift to Young Mathematicians.

Answer 3

You can just write as $|z+3/2|>2$, i.e. $|z-(-3/2)|>2$, so the region is given by the points whose distance to $-3/2$ is bigger than $2$, i.e. the region is the plane after removing the closed disc of center $-3/2$ and radius $2$.