Homework Help: Sketch the region of the complex plane

1. Oct 31, 2013

SteveDC

1. The problem statement, all variables and given/known data

Sketch the region of the complex plane specified by:

|z - 4 + 3i| ≤ 5

2. Relevant equations

3. The attempt at a solution
I have tried re-writing the modulus as √[(z)^2 (- 4)^2 + (3i)^2] and from this I have managed to arrive at z ≤ 3√2

But not sure if I needed to do this or how I would take it from here in terms of sketching this

2. Oct 31, 2013

tiny-tim

Hi SteveDC!
|z - (4 - 3i)| ≤ 5 ?

3. Oct 31, 2013

SteveDC

Sorry, I might need a bigger hint then this! I still don't really understand

4. Oct 31, 2013

tiny-tim

how would you draw |z| ≤ 5 ?

5. Oct 31, 2013

SteveDC

As a line along the real axis stretching to a point that is less than or equal to 5?

6. Oct 31, 2013

pasmith

That's not how you compute $|z - 4 + 3i|$. Recall that if $w = a + ib$ then $|w|^2 = a^2 + b^2$.

Set $z = x + iy$ and see what happens.

7. Oct 31, 2013

HallsofIvy

In any set in which an absolute value is defined we can interpret |x- y| as the distance between x and y. In particular, in the complex plane, |z- a| is the distance between z and a. If $|z- b|\le r$, for z a variable, b a specific complex number, and r a real number, then z is any point on or inside the circle with center at b and radius r.

(If z is a complex number, $z\le 3\sqrt{2}$ makes no sense. The complex numbers are not an "ordered field".)

8. Oct 31, 2013

tiny-tim

(just got back :tongue2:)
ahh … that's where your misunderstandning is …

|z| ≤ 5 is a circle, the circle of all points whose distance from 0 is ≤ 5

i] do you see why that is? (or do you need an explanation?)

ii] now what does |z - i| ≤ 5 look like?

9. Nov 1, 2013

SteveDC

Think I've got this now. ii] a circle round the midpoint at i, with radius less than or equal to 5, and z will lie on that radius.

10. Nov 1, 2013

Crake

If by "midpoint" you mean "centered", then you are correct.

11. Nov 1, 2013

SteveDC

Yep, thanks everyone