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If $$\frac{1}{a_1-2i} ,\frac{1}{a_2-2i}, … ,\frac{1}{a_8 - 2i}$$ represents sides of a regular polygon in complex plane , find the area of this octagon where $a_1,…,a_8$ belongs to real numbers, what I did first I took help of inversion and said $|a_1-a_2|$ , and similarily others to be equal and then use the fact the angle between them is $\dfrac{3π}{4}$, but this doesn't lead me to good system of equations to solve is there a more better way to solve this or a idea after what I did ?

WizardMath
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    Hint: $\frac{1}{a_k - 2i}$ are the images of $a_k$ under an inversion with respect to the unit circle centered at $2i$. Now $a_k$ lies on a line (the real axes) and circle inversion map this line to a circle... – achille hui Apr 24 '21 at 23:49
  • Can u give a clear view how 1/a_k -2i are images of real numbers a_k under an inversion wrt to unit circle centered at 2i ? – WizardMath Apr 25 '21 at 00:17
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    I make a mistake in my previous comment. Strictly speaking, the map $z \mapsto \frac{1}{z-2i}$ is a composition of 3 maps: $z \mapsto z-2i$ (a translation), $z \mapsto \frac{z}{|z|^2} = \frac1{\bar{z}}$ (an inversion wrt unit circle at origin) and $z \mapsto \bar{z}$ (a reflection). This doesn't stop the fact it send the real axis to a circle. Notice the map send $0$ to $\frac{i}{2}$ and $\infty$ to $0$. If you compute the distance between $\frac{1}{x-2i}$ and the midpoint $\frac{i}{4}$ for any real number $x$, you always get $\frac14$. – achille hui Apr 25 '21 at 02:15
  • Wow thats really cool , how does it helps in calculating area ? – WizardMath Apr 25 '21 at 02:20
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    Well, the regular octagon lies on a circle of radius $\frac14$.... – achille hui Apr 25 '21 at 04:50
  • Ty got it now . – WizardMath Apr 25 '21 at 06:03

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