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The set $$\{z\in\Bbb C:1<|2z-6|\le 2\}$$ and the set $$\{z\in\Bbb C:|z|=|\Re(z)|+|\Im(z)|\}$$

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1 Answer

Your first step should be work out what the sets look like. I’ll leave the first one to you and get you a good start on the second one.

If $z=x+iy$, then $|z|=\sqrt{x^2+y^2}$, $|\Re(z)|=|x|$, and $|\Im(z)|=|y|$, so

$$\{z\in\Bbb C:|z|=|\Re(z)|+|\Im(z)|\}=\left\{x+iy\in\Bbb C:\sqrt{x^2+y^2}=|x|+|y|\right\}\;.$$

The defining equation on the righthand side can be squared to give $x^2+y^2=\left(|x|+|y|\right)^2$, or, after multiplying out, $x^2+y^2=x^2+2|xy|+y^2$; it should be an easy matter to use this to see what the set actually looks like, and once you’ve done that, it should be clear that the set is closed and unbounded.

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I have sketches for both of them and can see that the first one is two circle around 3 one with radius 1/2 and one with radius 1 so and the second one is just the lines y=x for Re(z)>0 and y=-x for Re(z)<0 so I'm assuming that the first one is closed and bounded and the second is open, not sure about the boundedness on this one but I don't know how to say why/prove it. I know it can be done using an epsilon delta argument or by stating something about the closed ball and all of the points being inside a closed ball or something like that –  Adam Nov 3 '12 at 18:04
    
I’ve not worked out the first one, but you definitely have the wrong picture for the second one. $|1+i|=\sqrt2\ne|1|+|i|$. Your (incorrect) set for the second one is obviously not bounded: is it completely contained in some circle centred at $0$? It is closed: it contains all of its limit points. Alternatively, its complement is open: if $z$ is not on either of those two lines, there is an open ball centred at $z$ that is disjoint from both lines. –  Brian M. Scott Nov 3 '12 at 18:10
    
Oh ok then,I don't understand how to get a picture for the second one out of your answer as surely the x^2 and y^2 cancel and you just get 2|xy|=0 I also don't understand how it is closed is the a closed ball somewhere or do I need and epsilon delta type argument –  Adam Nov 3 '12 at 18:19
    
@Adam: If $2|xy|=0$, then $xy=0$; what does that tell you about $x$ and $y$? Closed balls are closed sets, but many closed sets look nothing at all like closed balls. Exactly what definition of closed set have you been given? –  Brian M. Scott Nov 3 '12 at 18:21
    
x=-y. I haven't actually been given a definition of a closed set but I have one about open sets. A subset U of a metric space (X,delta) is called open for any x element of u, there exists an r>0 such that an open ball radius epsilon is a subset of U –  Adam Nov 3 '12 at 18:27
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