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Is there a way to algebraically show that a circle that contains a point at the origin is a line when mapped with $f(z)=1/z$? I tried to work through an example, such as $|z+2i|=2$, I still got a circle. Is this more of an inspection thing or...? $$|z+2i|=2$$ Since $w=f(z)=\frac{1}{z}$ we know that $\frac{1}{w}$. $$|\frac{1}{w}+2i|=2$$ $$\frac{|1+2wi|}{|w|}=4$$ $$|1+2wi|=4|w|$$

Then, let $w=u+vi$

$$(1-2v)^2+(2u)^2=4u^2+4v^2$$ Then we get $$4v^2-4v+1+4u^2=4u^2+4v^2$$ $$-4v+1=0$$ which gives a line. Thank you all :)

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    Can you show your work how you got a circle? – 2017-02-13
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    Sure. Give me a few to type it up... – 2017-02-13
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    $(2\lvert w\rvert)^2 = 4\lvert w\rvert^2 = 4u^2 + 4v^2$. – 2017-02-13
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    $g(t) = \frac{1}{r+r e^{i t}}, g'(t) = \frac{-ir e^{it}}{r^2(1+e^{it})^2} =\frac{-2i}{r\cos^2 t/2}$ so $\text{arg }g'(t)$ is constant – 2017-02-13
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    You forgot to square the $2$ on the RHS. – 2017-02-13
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    Thank you so much everyone. I see the error now. :D – 2017-02-13

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Let's begin by studying $|z+2i|=2$. This describes a circle centered at $-2i$ of radius $2$. If we were to put this into $(x,y)$ coordinates, we get $x^2+(y+2)^2=4$.

Observe that $\frac{1}{z}$ is its own inverse, so the image of the circle under the map $\frac{1}{z}$ is the same as the solutions to $$ \left|\frac{1}{z}+2i\right|=2. $$ Squaring both sides, we get $$ \left(\frac{1}{z}+2i\right)\left(\frac{1}{\overline{z}}-2i\right)=4. $$ Multiplying this out, one gets $$ \frac{1}{z\overline{z}}+\frac{2i}{\overline{z}}-\frac{2i}{z}+4=4 $$ or $$ \frac{1}{z\overline{z}}+\frac{2i}{\overline{z}}-\frac{2i}{z}=0 $$ Multiplying through by $|z|$, we get $$ 1+2zi-2\overline{z}i=0. $$ Making the substitution $z=x+iy$, we get $$ 1-4y=0 $$ which is a line.

The error in the original question is that when the equation was squared, the $2|w|$ did not become $4|w|^2$, the $2$ was not squared. When you make that change to the original question, you get $1-4v=0$ after simplifying, which agrees with this answer.

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A circle with center at $a$ containing the origin is described by $$ \lvert z - a \rvert = \lvert a \rvert \, . $$ The substitution $z = 1/w$ gives $$ \lvert \frac 1w - a \rvert = \lvert a \rvert \, . $$ or $$ \lvert a w \rvert^2 = \lvert 1 - aw \rvert^2 = 1 + \lvert a w \rvert^2 - 2 \operatorname{Re}( a w) \\ \Longleftrightarrow \operatorname{Re}( a w) = \frac 12 \\ \Longleftrightarrow w = \frac{1}{ a}(\frac 12 + iy) \text{ for some } y \in \Bbb R $$ which is a line.

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You may want to try to look at polar coordinates. A circle that contains the Origin can be described as $r=acos\theta+bsin\theta$, where $a$ and $b$ are numbers (not both zero). I'll leave it up to you to verify that such circle has some $\theta$ for which it passes through the Origin. Now reciprocating gives a new polar form $r=\frac{1}{asin\theta+bcos\theta}$ which can be written as $a(rsin\theta)+b(rcos\theta)=1$ from which we get $ay+bx=1$, a line (not through the Origin). For more information, you may want to google Mobius Transformations, that also highlight other cases.