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Let $C$ be a circle in the complex plane, and let $x$ be a fixed, non-zero complex number. Prove that $\{xz : z \in C\}$ is also a circle.

I would really appreciate any help that would get me started.

3 Answers 3

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By definition

$C=\{z\in\Bbb C\;;\;|z-a|=r\,\,,\,\,a\in\Bbb C\,,\,\,0

Now, putting $\,x=Re^{i\theta}\,$ , we get

$\,\{xz\;;\;z\in C\}=\{xa+rRe^{i(t+\theta)}\;;\;0\leq t\leq 2\pi\}$

which is a circle, too.

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Recall that multiplication of a complex number is really a rotation and a dilation. More explicitly, if you write the circle and $x$ in polar form then what do you have?

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Note that if $|\cdot|$ denotes the norm of a complex number, then for $z,w \in \mathbb{C}$ we have $|zw|=|z|\cdot|w|$. Since $x$ is fixed it has a fixed norm. Now think about the definition of a circle: every point in the circle will have the same norm, right? Do you see how the result follows?