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- Thread starter shen07
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- Sep 16, 2013

- 337

Have you tried taking a generalized equation for a circle in the Cartesian plane, and then converting that into Polar Coordinates...?

A comparison of that and your complex form above might help...

- Jan 30, 2012

- 2,492

Consider a second coordinate system centered at $a$. The variable $z'$ will range over points in that system while $z$ will range over points in the original system. Then $z=z'+a$ (e.g.., the center $z'=0$ of the second system is mapped to $z=a$).

The equation of the required circle in the second system is $|z'|^2=a^2$. Express $z'$ through $z$ and substitute it into this equation. Then use the following properties.

$|z|^2=z\bar{z}$

$2\operatorname{Re}z=z+\bar{z}$

$\overline{z_1z_2}=\bar{z}_1\bar{z}_2$

$\bar{\bar{z}}=z$

Last edited:

- Jan 17, 2013

- 1,667

Let \(\displaystyle z=x+iy ,

a=s+it \)

Then the equation of the circle can be written as

\(\displaystyle |z-a|=r\)

\(\displaystyle (x-s)^2+(y-t)^2=r^2\)

\(\displaystyle x^2+y^2-2xs-2yt +t^2+s^2=r^2\)

- Jan 17, 2013

- 1,667

\(\displaystyle 2 \text{Re}(z)=z+\bar{z}\)$|z|^2=z\bar{z}$

$\operatorname{Re}z=z+\bar{z}$

$\overline{z_1z_2}=\bar{z}_1\bar{z}_2$

$\bar{\bar{z}}=z$

\(\displaystyle z=\bar{z} \,\,\, z\in \mathbb{R}\)

- Feb 15, 2012

- 1,967

Note that: $\overline{a}z = (h-ik)(x+iy) = (hx+ky) + i(hy-kx)$, so that:

$\mathfrak{Re}(\overline{a}z) = hx+ky = xh+yk$.

In $\Bbb R^2$, the equation for a circle of radius $r$ centered at $(h,k)$ is:

$(x - h)^2 + (y - k)^2 = r^2$, so, expanding this, we have:

$x^2 - 2xh + h^2 + y^2 - 2yk + k^2 = r^2$

$x^2 + y^2 - 2(xh + yk) + h^2 + k^2 = r^2$

$|z|^2 - 2\mathfrak{Re}(\overline{a}z) + |a|^2 = r^2$.