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I am given the task to sketch all the points in the complex plane satisfying $$ \mathrm{Re}(1/z)<1 $$ I am not very good at sketching, nor seeing how to draw this in the complex plane. I was thinking that since

$$ \frac{1}{z} = \frac{|z|}{z|z|} = \frac{x - iy}{x^2 + y^2} $$

then $\mathrm{Re}(1/z)=x/(x^2+y^2)$. Our inequality is therefore equivalent to

$$\mathrm{Re}(1/z)<1 \Leftrightarrow x < x^2 + y^2 \Leftrightarrow \left(\frac{1}{2}\right)^2 < \left( x - \frac{1}{2}\right)^2 + y^2$$

So the equality represents all points in $\mathbb{R}$ that lie outside a disk of radius $1/2$ and centre $(1/2,0)$. But I have not plotted anything in the complex plane??

Any help sketching and understanding this would be greatly appreciated.

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You mean points in $\Bbb R^2$, not in $\Bbb R$. But this is just the complex plane under the identification $(x,y) \leftrightarrow x+{\rm i}y$. –  Per Manne Oct 10 '12 at 9:05

1 Answer 1

up vote 4 down vote accepted

You did everything right. Just draw a circle of radius $\frac12$ around the point $(\frac12,0)$, and it's exterior is the required region. Here is the plot in WA:
enter image description here

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