Find the center of the octagon, complex plane The problem given read as follows,
Two consecutive vertices of a regular convex octagon are $(1,2)$ and $(3,-2)$. Find the center of the octagon. 

I would like to know if my answer is correct. Here's what I have thought:
I've drawn the complex plane and set the affixes given accordingly (I've called them $z_{1} = (1,2)$ and $z_{2}=(3,-2)$). 
Since $z_{1}, z_{2}, ... z_{8}$ compose a regular octagon, the angle between $z_{1}$ and $z_{2}$ must be $\frac{\pi}{4}$. So I have an angular relation at my disposal. 
Say $z_{0} = x+iy$ is the center of the polygon, then I have two vectors, namely $\vec{v} = \vec{z_{0} z_{2}}$ and $\vec{u} =  \vec{z_{0} z_{1}}$ and since I have an angle relation between them, I can write:
$\vec{u} = \vec{v}.e^\frac{i \pi}{4} \iff z_{1}-z_{0} = (z_{2} - z_{0}).cis \left(\frac{\pi}{4} \right) (*)$
Substituting with my known affixes, I get:
$1+2i-z_{0}=(3-2i-z_{0}).e^\frac{i \pi}{4}$
Working out some algebra, I've got:
$z_{0} = \frac{3e^{\frac{i \pi}{4}} - 1}{e^\frac{i \pi}{4} - 1} + i\frac{-2-2 e^\frac{i \pi}{4}}{e^\frac{i \pi}{4} - 1}$
Converting from polar form to trigonometric form and doing a lengthy algebra, I've got:
$\boxed{z_{0} = 2 + i \frac{\sqrt{2}}{2-\sqrt{2}}}$
I believe that my reasoning is correct, although substituting my result at $(*)$ I get:
$\left(1 - \sqrt{2}, \frac{4-3\sqrt{2}}{2-\sqrt{2}}\right) \neq \left(\frac{3 \sqrt{2} - 2}{2}, \frac{1-2 \sqrt{2}}{2-\sqrt{2}}\right)$
I would appreciate if someone could help me figure out my mistake.
Thanks, in advance.

$z_0=x+i y;z_1=1+2 i;z_2=3-2 i$
$z_1-z_0=e^{\frac{i \pi }{4}} (z_2-z_0)$
$-e^{\frac{i \pi }{4}} (-x-i y+(3-2 i))-x-i y+(1+2 i)=0$
$
\left\{
\begin{array}{c}
 \frac{x}{\sqrt{2}}-x-\sqrt{2}-\frac{y}{\sqrt{2}}-\frac{3}{\sqrt{2}}+1=0 \\
 \frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}-y+\sqrt{2}-\frac{3}{\sqrt{2}}+2=0 \\
\end{array}
\right.
$
$x=-2 \sqrt{2},y=-1-\sqrt{2}$
but there is also another possibility, as you can see in the 
picture
$z_2-z_0=e^{\frac{i \pi }{4}} (z_1-z_0)$
which leads to the other solution
$x=2 \left(2+\sqrt{2}\right),y=1+\sqrt{2}$
Hope this is useful

