# Thread: Sketching a complex relation

1. ## Sketching a complex relation

So I've been working on this 1 question for about an hour trying to find some sort of insight or similar question that can help me out. I've been looking for some sort of example where there is a graph of z^2.

My problem is this:

Sketch the set, s, where s = {z| | z^2 - 1 | < 1 } ... z is a complex number

I know that if it was s = {z| | z | < 1 } then I would get a unit disc. And, if there were numbers inside the absolute value then it would a shift or some sort.

The closest thing I can think of is factoring the interior of the absolute value so I get conjugates but I don't see how that helps. Another thing I thought of was use the graph of z^2 bounded by a positive integer but I don't know what that looks like. I suppose I could made z = x + iy but then I might get bogged down with calculation.

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the next question is

Sketch the set, s, where s = {Z| | Z | > 2 | Z - 1 | } ... Z is a complex number

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Any help would be greatly appreciated.

Thank you

2. ## Re: Sketching a complex relation

Originally Posted by aande
So I've been working on this 1 question for about an hour trying to find some sort of insight or similar question that can help me out. I've been looking for some sort of example where there is a graph of z^2.

My problem is this:

Sketch the set, s, where s = {z| | z^2 - 1 | < 1 } ... z is a complex number

I know that if it was s = {z| | z | < 1 } then I would get a unit disc. And, if there were numbers inside the absolute value then it would a shift or some sort.

The closest thing I can think of is factoring the interior of the absolute value so I get conjugates but I don't see how that helps. Another thing I thought of was use the graph of z^2 bounded by a positive integer but I don't know what that looks like. I suppose I could made z = x + iy but then I might get bogged down with calculation.

-

the next question is

Sketch the set, s, where s = {Z| | Z | > 2 | Z - 1 | } ... Z is a complex number

-

Any help would be greatly appreciated.

Thank you
\displaystyle \begin{align*} |z^2 - 1| &< 1 \\ \left|(x + iy)^2 - 1\right| &< 1 \\ \left|x^2 + 2ixy + i^2y^2 - 1\right| &< 1 \\ \left|\left(x^2 - y^2 - 1\right) + i\left(2xy\right)\right| &< 1 \\ \sqrt{\left(x^2 - y^2 - 1\right)^2 + \left(2xy\right)^2} &< 1 \end{align*}

Go from here...

3. ## Re: Sketching a complex relation

Originally Posted by Prove It
\displaystyle \begin{align*} |z^2 - 1| &< 1 \\ \left|(x + iy)^2 - 1\right| &< 1 \\ \left|x^2 + 2ixy + i^2y^2 - 1\right| &< 1 \\ \left|\left(x^2 - y^2 - 1\right) + i\left(2xy\right)\right| &< 1 \\ \sqrt{\left(x^2 - y^2 - 1\right)^2 + \left(2xy\right)^2} &< 1 \end{align*}

Go from here...
Thanks.

However, this is what I meant by "getting bogged down with calculation". I knew what your wrote down myself but how am I supposed to graph this by hand? Is there some sort of insight that I'm just not seeing? Or, does this all miraculously simplify into something sketch-able if you just grind your way through it?

What I expect is something like two circles next to each other but I don't know. Something like a 8 on it's side that is placed at the origin. The only reason I think it's this is because I might be able to get
| z^2 - 1 | < 1
| (z+1) (z-1) | < 1
I know that |z+1|<1 is a unit disc centred at (-1,0) and |z-1|<1 is a unit disc centred at (1,0)

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btw, could you help me on the other question as well.