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So I heard this a long time ago and I recently started thinking about it again. So I was told that the complex function $f(z)=1/z$ maps everything inside a circle to points outside the circle (the remaining part of the complex plane). Why is this?

I realize we can write $$f(z)=\frac{1}{z}=\frac{\overline{z}}{|z|^2}$$ so that there is a reflection $\overline{z}$ and a dilation $1/|z|^2$. But I don't see why this then only maps to points outside the circle. Can you explain?

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A point is inside the (unit) circle if $|z|<1$, thus $$|f(z)|=\frac{1}{|z|}>1$$ is outside the circle if $z$ was inside the circle, and vice-versa.

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  • $\begingroup$ Thank you for your quick reply! What if we had the circle centered at (0,0) and with radius $2$ (say), so that $1/|z| < 1$ for some $z$ in this set? $\endgroup$ – Numbersandsoon Dec 14 '13 at 16:12
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    $\begingroup$ For the circle of radius $2$, the relevant function is $z\mapsto4/z$. More generally, the function exchanging the inside of the circle centered at $c\in\mathbb{C}$ with radius $r>0$ with the outside is $z\mapsto r^2/(z-c)+c$. $\endgroup$ – Daniel Robert-Nicoud Dec 14 '13 at 16:16
  • $\begingroup$ Aha, I see. Because I was told that in general $1/z$ mapped the inside to the outside, and I couldn't see why that was. So I was right - it only holds for the unit circle that $z \mapsto 1/z$ maps the inner to the outer of the circle. Thanks! $\endgroup$ – Numbersandsoon Dec 14 '13 at 16:19
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We can also write the numbers $z$ in polar form as $re^{i\theta}$, in which case $$f(z)=f(re^{i\theta})=\frac {e^{-i\theta}}r$$ So the angle is negated and the radius is inverted. This means that $r\lt1\implies \frac 1r\gt 1$, and every complex point is mirrored across the real axis, since $e^{-i\theta}$ preserves the real component (due to $\cos$ mapping a negative angle to the same value as the absolute value of that angle) and negates the imaginary component (since $\sin (-x)=-\sin x$).

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$0 < |z| < 1$ iff $1 < |1/z| < \infty$

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