Math Help - sketch on complex plane

1. sketch on complex plane

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?

2. Originally Posted by CarmineCortez
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?
$|3z - 2 + 5i| < 3 \Rightarrow 3 \left|z - \left(\frac{2}{3} - \frac{5}{3} i \right) \right| < 3 \Rightarrow \left|z - \left(\frac{2}{3} - \frac{5}{3} i \right) \right| < 1$.

The boundary of the required region is a circle of radius 1 and centre at $z = \frac{2}{3} - \frac{5}{3} i$ (geometrically, do you know why?)

Now you have to determine whether the required region is inside or outside this circle.

3. Originally Posted by mr fantastic
$|3z - 2 + 5i| < 3 \Rightarrow 3 \left|z - \left(\frac{2}{3} - \frac{5}{3} i \right) \right| < 3 \Rightarrow \left|z - \left(\frac{2}{3} - \frac{5}{3} i \right) \right| < 1$.

The boundary of the required region is a circle of radius 1 and centre at $z = \frac{2}{3} - \frac{5}{3} i$ (geometrically, do you know why?)

Now you have to determine whether the required region is inside or outside this circle.

the region should be inside the circle since we have abs(z-...)<1

so the goal is to have no coefficient for z? or a 1 for the inequality. I'm a little confused here.

4. Sketch on complex plane

Hello CarmineCortez
Originally Posted by CarmineCortez
the region should be inside the circle since we have abs(z-...)<1

so the goal is to have no coefficient for z? or a 1 for the inequality. I'm a little confused here.
The goal is to have 'no' coefficient for $z$ - although, of course, the coefficient is really $1$. (Don't forget that, will you?)

The reason this works is that geometrically (to use Mr F's word) $|z - a|$ is the distance between the points representing the complex numbers $z$ and $a$. So if $a$ is a constant, $|z - a|$ 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 $|z - a| = r$ (where r is real) represents a circle, centre A, radius $r$.

OK now?