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The expression is this:

$|3z+2| \leq 1$

then, I did

$|3x+2+3iy| \leq 1$

and I got stuck.

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Think of $z$ not just as a complex number here, but as a vector with a magnitude and a direction undergoing two transformations to become vector $v$; first $z$ is being dilated by length 3, and then being translated rightward by length 2. The locus of $v$ such that their magnitude is less than length of 1 is easier to visualize, and you can work backwards from there. –  Uticensis Feb 15 '11 at 14:44
Thank you. That way to analyze the problem made me think better –  Tomas Sironi Feb 15 '11 at 14:46
Try to think of it as a ball centered at a point. bring a 3 over to the other side and you will see what the center and radius are. –  AnonymousCoward Feb 15 '11 at 23:02

1 Answer 1

up vote 1 down vote accepted

Hint: How do you find the modulus of a complex number?

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"$(x^2+y^2)^{1/2}$" sory for these fool questions, I should think more before asking here. Now I know how to proceed. Thanks! –  Tomas Sironi Feb 15 '11 at 14:38
After all, I should remove the square roots so I can operate the inequation, right? –  Tomas Sironi Feb 15 '11 at 14:41
As you know the modulus is positive, you can square both sides of the inequality. –  Ross Millikan Feb 15 '11 at 14:44

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