[SOLVED] Complex analysis, determine the set in the complex plane that satisfies...

**arbolis** · June 1st 2010, 06:42 PM

I've been stuck for 3 days on this simple problem. I must graph the set of the complex plane where z satisfies $|3z+2|<1$ . I've used the way of writing $z=re^{i\theta}$ but reached a non sense condition for r. Well, the equation |3z+1|=1 would hold if $r=-\frac{1}{3 \cos \theta} \pm \frac{\sqrt {-2-\sin ^2 \theta}}{3 \cos \theta}$ . I really have no idea where I went wrong. I know there's likely an easy way to solve the problem, but still, I don't know where I'm wrong by replacing z for $re^{i\theta}$ and then calculate the modulus of $3z+2$ , etc.

**Drexel28** · June 1st 2010, 06:48 PM

Originally Posted by arbolis

I've been stuck for 3 days on this simple problem. I must graph the set of the complex plane where z satisfies $|3z+2|<1$ . I've used the way of writing $z=re^{i\theta}$ but reached a non sense condition for r. Well, the equation |3z+1|=1 would hold if $r=-\frac{1}{3 \cos \theta} \pm \frac{\sqrt {-2-\sin ^2 \theta}}{3 \cos \theta}$ . I really have no idea where I went wrong. I know there's likely an easy way to solve the problem, but still, I don't know where I'm wrong by replacing z for $re^{i\theta}$ and then calculate the modulus of $3z+2$ , etc.

In general, $\left\{z\in\mathbb{C}:|az+b|<r\right\}$ is a disk of radius $r$ centered around...what?

**arbolis** · June 1st 2010, 06:51 PM

Originally Posted by Drexel28

In general, $\left\{z\in\mathbb{C}:|az+b|<r\right\}$ is a disk of radius $r$ centered around...what?

My bad, a semester of physics totally destroyed me...
Around -b?

**tonio** · June 1st 2010, 06:53 PM

Originally Posted by arbolis

I've been stuck for 3 days on this simple problem. I must graph the set of the complex plane where z satisfies $|3z+2|<1$ . I've used the way of writing $z=re^{i\theta}$ but reached a non sense condition for r. Well, the equation |3z+1|=1 would hold if $r=-\frac{1}{3 \cos \theta} \pm \frac{\sqrt {-2-\sin ^2 \theta}}{3 \cos \theta}$ . I really have no idea where I went wrong. I know there's likely an easy way to solve the problem, but still, I don't know where I'm wrong by replacing z for $re^{i\theta}$ and then calculate the modulus of $3z+2$ , etc.

Hint - Screw the polar form: write $z=x+iy\,,\,\,x,y\in\mathbb{R}$ , substitute in $|3x+2|<1$ , square and do a little algebra there.

You must get the set of all points inside a certain circle (with center on the negative x-axis) of radius $\frac{1}{3}$ ...

A big deal of problems of this kind can easily be solved as above.

Tonio

**Drexel28** · June 1st 2010, 06:55 PM

Originally Posted by arbolis

My bad, a semester of physics totally destroyed me...
Around -b?

Note that literally $|z-z_0|<r$ is the open disk of radius $r$ around $z_0$ . So, if $a\ne 0$ then $|az+b|<r\Leftrightarrow |z-\tfrac{-b}{a}|<\frac{r}{|a|}$

**arbolis** · June 1st 2010, 07:21 PM

I had tried at first the way of writing z=x+iy but reached non sense probably due to arithmetic errors. Like now: I reached $\sqrt {9x^2 +6x +4 + 9 y^2}<1$ . So $9x^2 +6x +4 + 9 y^2<1$ . Skipping the steps (with 1 completion of square) I reach $\left ( x + \frac{1}{3} \right ) ^2 + y^2 < -\frac{2}{9}$ . So 2 positive numbers add to a negative one... Are the 2 first inequations right at least?

**tonio** · June 1st 2010, 07:34 PM

Originally Posted by arbolis

I had tried at first the way of writing z=x+iy but reached non sense probably due to arithmetic errors. Like now: I reached $\sqrt {9x^2 +6x +4 + 9 y^2}<1$ . So $9x^2 +6x +4 + 9 y^2<1$ . Skipping the steps (with 1 completion of square) I reach $\left ( x + \frac{1}{3} \right ) ^2 + y^2 < -\frac{2}{9}$ . So 2 positive numbers add to a negative one... Are the 2 first inequations right at least?

No, they aren't. Why is it that sometimes people rushes to open up parentheses? Leave that as a last resource!

$|3z+2|<1\iff |3x+2+3yi|^2<1\iff (3x+2)^2+9y^2<1$ $\iff 9\left(x+2/3\right)^2+9y^2<1\iff$ $\left(x+2/3\right)^2+y^2<\frac{1}{9}$ ...

Drexel's solution though is more elegant and short, once you know that $|z-z_0|<a\,,\,0<a\in\mathbb{R}$ , is the inside of a circle.

Tonio

**Bruno J.** · June 1st 2010, 08:19 PM

Originally Posted by Drexel28

In general, $\left\{z\in\mathbb{C}:|az+b|<r\right\}$ is a disk of radius $r$ centered around...what?

(Of radius $r/|a|$ )

**Drexel28** · June 1st 2010, 08:51 PM

Originally Posted by Bruno J.

(Of radius $r/|a|$ )

Typo, clearly fixed in my next post.

**HallsofIvy** · June 2nd 2010, 03:31 AM

The point people are trying to make to you is to do this "geometrically" not "algebraically".

In the complex numbers, |z- a| is the distance between z and a. |z- a|= r is satisfied by all z on the circle with center a and radius r. |z- a|< r is the open disk of point inside that circle.

Here, your problem was |3z+ 2|< 1 which is the same as 3|z+ 2/3|< 1 or
|z-(-2/3)|< 1/3.

**arbolis** · June 2nd 2010, 01:52 PM

Ok thanks to all, I get it. So the center of the disk is at -2/3, right?

I'm still totally puzzled why $(3x+2)^2+9y^2<1$ is true while $9x^2 +6x +4 + 9 y^2<1$ is not true. I know I don't need to solve the problem this way, but still, I'm stuck at understanding why it isn't true.

Edit: Oops, ok I get it. I forgot a factor 2. Problem solved.

**tonio** · June 2nd 2010, 02:25 PM

Originally Posted by arbolis

Ok thanks to all, I get it. So the center of the disk is at -2/3, right?

I'm still totally puzzled why $(3x+2)^2+9y^2<1$ is true while $9x^2 +6x +4 + 9 y^2<1$ is not true. I know I don't need to solve the problem this way, but still, I'm stuck at understanding why it isn't true.

Edit: Oops, ok I get it. I forgot a factor 2. Problem solved.

Indee, and the center is a point on the plane, not a number: $(-2/3,\,0)$

Tonio

**arbolis** · June 2nd 2010, 02:30 PM

Originally Posted by tonio

Indee, and the center is a point on the plane, not a number: $(-2/3,\,0)$

Tonio

Yes sorry for being imprecise. I meant the point on the real axis.

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