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The circle in complex numbers has the form

$$|z-c|=r$$

or

$$\{z \in \mathbb{C}:|z-c|=r\}$$

How does one express a circle (in complex form) in polar coordinates and what is the "polar coordinate form" of a complex circle?

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  • $\begingroup$ You just asked this question. Why did you delete it and ask again? $\endgroup$ Jan 28, 2016 at 11:32

2 Answers 2

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May be what you are looking for is $$ z=c+r\,e^{i\theta},\quad 0\le\theta<2\,\pi? $$

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  • $\begingroup$ Can you explain how to prove it? $\endgroup$
    – Tapi
    Jul 26, 2021 at 13:10
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A quick look at wikipedia would have saved you the trouble;

The general equation for a circle with center at $(r_o, \gamma)$ and radius $a$ is $$r^2 - 2rr_0\cos(\varphi - \gamma) + r_0^2 = a^2$$

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  • $\begingroup$ So this applies to "complex circle" as well? $\endgroup$
    – mavavilj
    Jan 28, 2016 at 13:49
  • $\begingroup$ @mavavilj the "complex circle" and the circle in the 2d plane are exactly the same thing $\endgroup$
    – Ant
    Jan 28, 2016 at 13:53
  • $\begingroup$ What happens to $i$ when one plugs $z=|z|cos\theta+isin\theta$ to e.g. $|z-1|=1$? If $i$ is left, then it's not "really" a circle in $\mathbb{R}$? $\endgroup$
    – mavavilj
    Jan 28, 2016 at 17:44
  • $\begingroup$ @mavavilj in the above equation, $r = |z|$ and $\varphi = \arg z$. Also, $z = |z| \cos \theta + i |z| \sin \theta$. I don't really understand your comment... $\endgroup$
    – Ant
    Jan 28, 2016 at 18:41
  • $\begingroup$ In that case I probably need to see the derivation of the equation you posted. $\endgroup$
    – mavavilj
    Jan 28, 2016 at 18:54

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