User:Hans G. Oberlack/Sandkiste

Elements

Base is the square of given side length $a_{0}$ with centroid at $S_{0}$
Inscribed are the largest possible circle and right triangle of the same area.

General case

Segments in the general case

0) The side length of the base square: $a_{0}$
1) The radius of the inscribed circle: $r_{1}={\frac {\sqrt {2}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\quad \approx 0.34\cdot a_{0}\quad$, see Calculation 1
2) The side length of the inscribed right triangle: $a_{2}={\sqrt {2\pi }}\cdot r_{1}={\frac {2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\quad \approx 0.847\cdot a_{0}\quad$, see Calculation 1

Perimeters in the general case

0) Perimeter of base square $P_{0}=4\cdot a_{0}\quad$
1) Perimeter of the inscribed circle: $P_{1}=2\pi \cdot r_{1}={\frac {2\pi {\sqrt {2}}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\quad \approx 2.122\cdot a_{0}\quad$
2) Perimeter of the inscribed triangle: $P_{2}=4\cdot a_{2}={\frac {4\cdot 2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\quad \approx 3.387\cdot a_{0}\quad$

Areas in the general case

0) Area of the base square $A_{0}=a_{0}^{2}$
1) Area of the inscribed circle: $A_{1}=\pi r_{1}^{2}=\pi \left({\frac {\sqrt {2}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)^{2}\cdot a_{0}^{2}\quad \approx 0.358\cdot a_{0}^{2}\quad$
2) Area of the inscribed triangle: $A_{2}={\frac {a_{2}^{2}}{2}}={\frac {1}{2}}\left({\frac {2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)^{2}\cdot a_{0}^{2}\quad \approx 0.358\cdot a_{0}^{2}\quad$

Centroids in the general case

Centroids as graphically displayed

0) Centroid position of the base square: $S_{0}=0+0i=0$
1) Centroid position of the inscribed circle: $S_{1}={\frac {1}{2}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (1+i)\cdot a_{0}\quad \approx (0.162+0.162i)\cdot a_{0}\quad$, see Calculation 2
2) Centroid position of the inscribed square: $S_{2}={\frac {1}{6}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot a_{0}\quad \approx (-0.218-0.218i)\quad$, see Calculation 3

Orientated centroids

The centroid positions of the following shapes will be expressed orientated so that the first shape n with $S_{n}\neq S_{0}$ will be of type $S_{n}=0-ki$ with $k>0$. The graphical representation does not correspond to the mathematical expression.
0) Orientated centroid position of the base square: $SO_{0}=0+0i=0$
1) Orientated centroid position of the inscribed circle: $SO_{1}={\frac {1}{\sqrt {2}}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (0-i)\cdot a_{0}\quad \approx (0-0.229i)\quad$, see calculation 4
2) Orientated centroid position of the inscribed right triangle: $SO_{2}={\frac {1}{3{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (0-i)\cdot a_{0}\quad \approx (0+0.308i)$, see calculation 5

Normalised case

File:FS Q(CR).png

In the normalised case the area of the base is set to 1.
$A_{0}=1\quad \Rightarrow a_{0}^{2}=1\quad \Rightarrow a_{0}^{2}=1$

Segments in the normalised case

0) Side length of the base square: $a_{0}=1$
1) The radius of the inscribed circle: $r_{1}={\frac {\sqrt {2}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\quad \approx 0.340\quad$
2) The side length of the inscribed right triangle: $a_{2}={\sqrt {2\pi }}\cdot r_{1}={\frac {2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\quad \approx 0.847\quad$

Calculations

Equations of given elements and relations

(0) The area of the right triangle $\triangle AKH$ has to be the same as the area of the circle with radius $r_{1}$ around center point $S_{1}\quad \Rightarrow A_{1}=A_{2}$
(1) $|AB|=|BC|=|CD|=|AD|=a_{0}\quad$since $\square ABCD$ is a square
(2) $|AC|={\sqrt {2}}\cdot a_{0}\quad$since ${\overline {AC}}$ is the diagonal of square $\square ABCD$
(3) $|AK|=|AH|=a_{2}\quad$since $\triangle AKH$ is a right triangle
(4) $|S_{1}F|=|FC|=|CG|=|GS_{1}|=r_{1}\quad$since $\square S_{1}FCG$ is a square, because $F$ and $G$ are tangent points of the circle around $S_{1}$ with the sides of square $\square ABCD$

Calculation 1

The radius $r_{1}$ is calculated:

a) First the relation between $r_{1}$ and $a_{2}$ is determined from equation (0):
$A_{1}=A_{2}\quad$ applying equation (0)
$\quad \Leftrightarrow {\frac {a_{2}^{2}}{2}}=\pi r_{1}^{2}\quad$, definitions of areas of right triangles and circles
$\quad \Leftrightarrow a_{2}^{2}=2\pi r_{1}^{2}\quad$, multiplying both sides by two
$\quad \Leftrightarrow a_{2}={\sqrt {2\pi }}\cdot r_{1}\quad$, since negative length cannot be applied

b) Then the relation between $r_{1}$ and $a_{0}$ is determined from the following identity:
$|AC|={\sqrt {2}}\cdot a_{0}\quad$ since $\square ABCD$ is a square
$\quad \Leftrightarrow |AE|+|ES_{1}|+|S_{1}C|={\sqrt {2}}\cdot a_{0}\quad$, because the three segments are forming the diagonal of the square
$\quad \Leftrightarrow {\frac {\sqrt {2}}{2}}\cdot a_{2}+|ES_{1}|+|S_{1}C|={\sqrt {2}}\cdot a_{0}\quad$, since $\triangle AKH$ is a right isosceles trinagle
$\quad \Leftrightarrow {\frac {\sqrt {2}}{2}}({\sqrt {2\pi }}\cdot r_{1})+|ES_{1}|+|S_{1}C|={\sqrt {2}}\cdot a_{0}\quad$, applying part a) of the calculation
$\quad \Leftrightarrow {\frac {{\sqrt {2}}{\sqrt {2}}}{2}}({\sqrt {\pi }}\cdot r_{1})+|ES_{1}|+|S_{1}C|={\sqrt {2}}\cdot a_{0}\quad$, rearranging
$\quad \Leftrightarrow {\sqrt {\pi }}\cdot r_{1}+|ES_{1}|+|S_{1}C|={\sqrt {2}}\cdot a_{0}\quad$, rearranging
$\quad \Leftrightarrow {\sqrt {\pi }}\cdot r_{1}+r_{1}+|S_{1}C|={\sqrt {2}}\cdot a_{0}\quad$, because E is a tangent point on the circle and a diagonal side of the triangle $\triangle AKH$
$\quad \Leftrightarrow {\sqrt {\pi }}\cdot r_{1}+r_{1}+{\sqrt {2}}\cdot r_{1}={\sqrt {2}}\cdot a_{0}\quad$, because $\square S_{1}FCG$ is a square with side length $r_{1}$
$\quad \Leftrightarrow r_{1}\cdot ({\sqrt {\pi }}+1+{\sqrt {2}})={\sqrt {2}}\cdot a_{0}\quad$, extracting $r_{1}$ out of the bracket
$\quad \Leftrightarrow r_{1}={\frac {\sqrt {2}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\quad \approx 0.338\cdot a_{0}$, rearranging

c) Eventually the relation between $a_{2}$ and $a_{0}$ is determined by using part a) of the calculation:
$a_{2}={\sqrt {2\pi }}\cdot r_{1}\quad$, from part a) of the calculation
$\quad \Leftrightarrow a_{2}={\sqrt {2\pi }}\cdot {\frac {\sqrt {2}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}$, using the result from part b) of the calculation
$\quad \Leftrightarrow a_{2}={\frac {2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\quad \approx 0.847\cdot a_{0}$, rearranging

Calculation 2

Calculating the centroid of the inscribed circle as displayed:
$S_{1}=S_{0}+{\overrightarrow {S_{0}S_{1}}}$
$\quad \Leftrightarrow S_{1}=(0+0i)+{\overrightarrow {S_{0}C}}+{\overrightarrow {CS_{1}}}$
$\quad \Leftrightarrow S_{1}=\left(|S_{0}C|\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)+{\overrightarrow {CS_{1}}}\quad$, since ${\overrightarrow {S_{0}C}}$ is collinear to the diagonal of the square
$\quad \Leftrightarrow S_{1}=\left({\frac {{\sqrt {2}}\cdot a_{0}}{2}}\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)+{\overrightarrow {CS_{1}}}\quad$, since ${\overrightarrow {S_{0}C}}$ is half the diagonal of the square
$\quad \Leftrightarrow S_{1}=\left({\frac {a_{0}}{2}}\cdot (1+i)\right)+{\overrightarrow {CS_{1}}}\quad$, rearranging
$\quad \Leftrightarrow S_{1}=\left({\frac {a_{0}}{2}}\cdot (1+i)\right)-\left(|S_{1}C|\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)\quad$, rearranging
$\quad \Leftrightarrow S_{1}=\left({\frac {a_{0}}{2}}\cdot (1+i)\right)-\left({\sqrt {2}}\cdot r_{1}\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)\quad$, since ${\overrightarrow {S_{1}C}}$ is the diagonal of the square $\square {S_{1}FCG}$ with side length $r_{1}$
$\quad \Leftrightarrow S_{1}=\left({\frac {a_{0}}{2}}\cdot (1+i)\right)-\left(r_{1}\cdot (1+i)\right)\quad$, rearranging
$\quad \Leftrightarrow S_{1}=\left({\frac {a_{0}}{2}}-r_{1}\right)\cdot (1+i)\quad$, rearranging
$\quad \Leftrightarrow S_{1}=\left({\frac {a_{0}}{2}}-{\frac {\sqrt {2}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\right)\cdot (1+i)\quad$, applying Calculation 1b
$\quad \Leftrightarrow S_{1}=\left({\frac {1}{2}}-{\frac {\sqrt {2}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot a_{0}\quad$, rearranging
$\quad \Leftrightarrow S_{1}=\left({\frac {{\sqrt {\pi }}+{\sqrt {2}}+1}{2({\sqrt {\pi }}+{\sqrt {2}}+1)}}-{\frac {2{\sqrt {2}}}{2({\sqrt {\pi }}+{\sqrt {2}}+1)}}\right)\cdot (1+i)\cdot a_{0}\quad$, rearranging
$\quad \Leftrightarrow S_{1}=\left({\frac {{\sqrt {\pi }}-{\sqrt {2}}+1}{2({\sqrt {\pi }}+{\sqrt {2}}+1)}}\right)\cdot (1+i)\cdot a_{0}\quad$, rearranging
$\quad \Leftrightarrow S_{1}={\frac {1}{2}}\cdot \left({\frac {{\sqrt {\pi }}-{\sqrt {2}}+1}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot a_{0}\quad \approx (0.162+0.162i)\cdot a_{0}$, rearranging

Calculation 3

Calculating the centroid of the right triangle as displayed:
$S_{2}=S_{0}+{\overrightarrow {S_{0}S_{2}}}$
$\quad \Leftrightarrow S_{2}=(0+0i)+{\overrightarrow {S_{0}E}}+{\overrightarrow {ES_{2}}}$
$\quad \Leftrightarrow S_{2}=(0+0i)+{\overrightarrow {S_{0}A}}+{\overrightarrow {AE}}+{\overrightarrow {ES_{2}}}$
$\quad \Leftrightarrow S_{2}=\left(-|S_{0}A|\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)+{\overrightarrow {AE}}+{\overrightarrow {ES_{2}}}\quad$, since ${\overrightarrow {S_{0}A}}$ is collinear to the diagonal of the square $\square {ABCD}$
$\quad \Leftrightarrow S_{2}=\left(-{\frac {{\sqrt {2}}\cdot a_{0}}{2}}\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)+{\overrightarrow {AE}}+{\overrightarrow {ES_{2}}}\quad$, since ${\overrightarrow {S_{0}A}}$ is half the diagonal of the square $\square {ABCD}$
$\quad \Leftrightarrow S_{2}=\left(-{\frac {a_{0}}{2}}\cdot (1+i)\right)+{\overrightarrow {AE}}+{\overrightarrow {ES_{2}}}\quad$, rearranging
$\quad \Leftrightarrow S_{2}=\left(-{\frac {a_{0}}{2}}\cdot (1+i)\right)+\left(|AE|\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)+{\overrightarrow {ES_{2}}}\quad$, since ${\overrightarrow {AE}}$ is the height of the triangle $\triangle {AKH}$
$\quad \Leftrightarrow S_{2}=\left(-{\frac {a_{0}}{2}}\cdot (1+i)\right)+\left({\frac {a_{2}}{\sqrt {2}}}\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)+{\overrightarrow {ES_{2}}}\quad$, since ${\frac {a_{2}}{\sqrt {2}}}$ is the value of the height of the triangle $\triangle {AKH}$
$\quad \Leftrightarrow S_{2}=\left(-{\frac {a_{0}}{2}}\cdot (1+i)\right)+\left({\frac {a_{2}}{2}}\cdot (1+i)\right)+{\overrightarrow {ES_{2}}}\quad$, rearranging
$\quad \Leftrightarrow S_{2}=\left(-{\frac {a_{0}}{2}}\cdot (1+i)\right)+\left({\frac {a_{2}}{2}}\cdot (1+i)\right)+\left(-|ES_{2}|\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)\quad$, since ${\overrightarrow {ES_{2}}}$ is collinear to the diagonal of the square $\square {ABCD}$
$\quad \Leftrightarrow S_{2}=\left(-{\frac {a_{0}}{2}}\cdot (1+i)\right)+\left({\frac {a_{2}}{2}}\cdot (1+i)\right)+\left(-{\frac {1}{3}}|AE|\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)\quad$, since the centroid of a triangle is a third of the height
$\quad \Leftrightarrow S_{2}=\left(-{\frac {a_{0}}{2}}\cdot (1+i)\right)+\left({\frac {a_{2}}{2}}\cdot (1+i)\right)+\left(-{\frac {1}{3}}\cdot {\frac {a_{2}}{\sqrt {2}}}\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)\quad$, since the height of a right isoceles triangle with side length a is: $h={\frac {a}{\sqrt {2}}}$
$\quad \Leftrightarrow S_{2}=\left(-{\frac {a_{0}}{2}}\cdot (1+i)\right)+\left({\frac {a_{2}}{2}}\cdot (1+i)\right)+\left(-{\frac {1}{3}}\cdot {\frac {a_{2}}{2}}\cdot (1+i)\right)\quad$, rearranging
$\quad \Leftrightarrow S_{2}={\frac {1}{2}}\cdot \left(-a_{0}+a_{2}-{\frac {1}{3}}\cdot a_{2}\right)\cdot (1+i)\quad$, factoring out
$\quad \Leftrightarrow S_{2}={\frac {1}{2}}\cdot \left(-a_{0}+{\frac {2}{3}}\cdot a_{2}\right)\cdot (1+i)\quad$, rearranging
$\quad \Leftrightarrow S_{2}={\frac {1}{2}}\cdot \left(-a_{0}+{\frac {2}{3}}\cdot {\frac {2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\right)\cdot (1+i)\quad$, applying result of calculation 1c
$\quad \Leftrightarrow S_{2}={\frac {1}{2}}\cdot \left(-1+{\frac {2}{3}}\cdot {\frac {2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot a_{0}\quad$, factoring out
$\quad \Leftrightarrow S_{2}={\frac {1}{6}}\cdot \left(-3+2\cdot {\frac {2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot a_{0}\quad$, factoring out
$\quad \Leftrightarrow S_{2}={\frac {1}{6}}\cdot \left({\frac {4{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}-3\right)\cdot (1+i)\cdot a_{0}\quad$, rearranging
$\quad \Leftrightarrow S_{2}={\frac {1}{6}}\cdot \left({\frac {4{\sqrt {\pi }}-3({\sqrt {\pi }}+{\sqrt {2}}+1)}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot a_{0}\quad$, having a common denominator
$\quad \Leftrightarrow S_{2}={\frac {1}{6}}\cdot \left({\frac {4{\sqrt {\pi }}-3{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot a_{0}\quad$, multiplying into the bracket
$\quad \Leftrightarrow S_{2}={\frac {1}{6}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot a_{0}\quad \approx (-0.218-0.218i)$, adding

Calculation 4

The centroid of the inscribed circle is calculated in orientated expression:
This will be done by rotating the expression as displayed by $225^{\circ }$ which will be done by a $45^{\circ }$ rotation followed by a $180^{\circ }$ rotation
$SO_{1}=S_{1}\cdot \left({\frac {1}{\sqrt {2}}}+{\frac {i}{\sqrt {2}}}\right)\cdot (i^{2})$
$\quad \Leftrightarrow SO_{1}=S_{1}\cdot \left({\frac {1}{\sqrt {2}}}+{\frac {i}{\sqrt {2}}}\right)\cdot (-1)\quad$, since $i^{2}=-1$
$\quad \Leftrightarrow SO_{1}=S_{1}\cdot {\frac {1}{\sqrt {2}}}\cdot \left(1+i\right)\cdot (-1)\quad$, factoring out
$\quad \Leftrightarrow SO_{1}=S_{1}\cdot {\frac {1}{\sqrt {2}}}\cdot \left(-1-i\right)\quad$, multiplying
$\quad \Leftrightarrow SO_{1}={\frac {1}{2}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (1+i)\cdot a_{0}\cdot {\frac {1}{\sqrt {2}}}\cdot \left(-1-i\right)\quad$, applying Calculation 2
$\quad \Leftrightarrow SO_{1}={\frac {1}{2{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (1+i)\cdot \left(-1-i\right)\cdot a_{0}\quad$, rearranging
$\quad \Leftrightarrow SO_{1}={\frac {1}{2{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (-1-i-i-i^{2})\cdot a_{0}\quad$, multiplying
$\quad \Leftrightarrow SO_{1}={\frac {1}{2{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (-1-i-i+1)\cdot a_{0}\quad$, since $i^{2}=-1$
$\quad \Leftrightarrow SO_{1}={\frac {1}{2{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (0-2i)\cdot a_{0}\quad$, adding up
$\quad \Leftrightarrow SO_{1}={\frac {2}{2{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (0-i)\cdot a_{0}\quad$, factoring out
$\quad \Leftrightarrow SO_{1}={\frac {1}{\sqrt {2}}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (0-i)\cdot a_{0}\quad \approx (0-0.229i)\quad$

Calculation 5

The centroid of the inscribed triangle is calculated in orientated expression:
This will be done by rotating the expression as displayed by $225^{\circ }$ which will be done by a $45^{\circ }$ rotation followed by a $180^{\circ }$ rotation
$SO_{2}=S_{2}\cdot \left({\frac {1}{\sqrt {2}}}+{\frac {i}{\sqrt {2}}}\right)\cdot (i^{2})$
$\quad \Leftrightarrow SO_{2}=S_{2}\cdot \left({\frac {1}{\sqrt {2}}}+{\frac {i}{\sqrt {2}}}\right)\cdot (-1)\quad$, since $i^{2}=-1$
$\quad \Leftrightarrow SO_{2}=S_{2}\cdot {\frac {1}{\sqrt {2}}}\cdot \left(1+i\right)\cdot (-1)\quad$, factoring out
$\quad \Leftrightarrow SO_{2}=S_{2}\cdot {\frac {1}{\sqrt {2}}}\cdot \left(-1-i\right)\quad$, multiplying
$\quad \Leftrightarrow SO_{2}={\frac {1}{6}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot a_{0}\cdot {\frac {1}{\sqrt {2}}}\cdot \left(-1-i\right)\quad$, applying Calculation 3
$\quad \Leftrightarrow SO_{2}={\frac {1}{6{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot \left(-1-i\right)\cdot a_{0}\quad$, rearranging
$\quad \Leftrightarrow SO_{2}={\frac {1}{6{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (-1-i-i-i^{2})\cdot a_{0}\quad$, multiplying
$\quad \Leftrightarrow SO_{2}={\frac {1}{6{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (-1-2i-(-1))\cdot a_{0}\quad$, since $i^{2}=-1$
$\quad \Leftrightarrow SO_{2}={\frac {1}{6{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (0-2i)\cdot a_{0}\quad$, adding
$\quad \Leftrightarrow SO_{2}={\frac {2}{6{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (0-i)\cdot a_{0}\quad$, factoring out
$\quad \Leftrightarrow SO_{2}={\frac {1}{3{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (0-i)\cdot a_{0}\quad \approx (0+0.308i)$, reducing