sketch abs(3z-2+5i)<3
i replace z with rcis(theta)
then do you move the other terms the the other side of the inequality and use them as asymptotes?
so it would become a circle radius 3r and asymptotes at 5 and -5i?
$\displaystyle |3z - 2 + 5i| < 3 \Rightarrow 3 \left|z - \left(\frac{2}{3} - \frac{5}{3} i \right) \right| < 3 \Rightarrow \left|z - \left(\frac{2}{3} - \frac{5}{3} i \right) \right| < 1$.
The boundary of the required region is a circle of radius 1 and centre at $\displaystyle z = \frac{2}{3} - \frac{5}{3} i$ (geometrically, do you know why?)
Now you have to determine whether the required region is inside or outside this circle.
Hello CarmineCortezThe goal is to have 'no' coefficient for $\displaystyle z$ - although, of course, the coefficient is really $\displaystyle 1$. (Don't forget that, will you?)
The reason this works is that geometrically (to use Mr F's word) $\displaystyle |z - a|$ is the distance between the points representing the complex numbers $\displaystyle z$ and $\displaystyle a$. So if $\displaystyle a$ is a constant, $\displaystyle |z - a|$ will be the distance of the point representing z from a fixed point, which we can call A. This in turn is why the equation $\displaystyle |z - a| = r$ (where r is real) represents a circle, centre A, radius $\displaystyle r$.
OK now?
Grandad