Say that $c_1 = -i$ and $c_2 = 3$. For this problem, let $z_0$ be an arbitrary complex number. We can rotate $z_0$ around $c_1$ by $\pi/4$ counterclockwise to get $z_1$. Next, we canrotate $z_1$ around $c_2$ by $\pi/4$ counter-clockwise to get $z_2$.
There exists a complex number $c$ where we can get $z_2$ from $z_0$ by rotating around $c$ by $\pi/2$ counterclockwise. What is the sum of $c$'s real and imaginary parts?
This question has already been asked previously on M.SE (Rotations of complex graphs), however, my question is about the answer I directly yielded.
I understand how kmeis reached this end result in the duplicate post: $$ \Re(c) + \Im(c) = \frac{(a_1-b_1+\sqrt{2}a_2) + (a_1+b_1+\sqrt{2}b_2)}{\sqrt{2}}= \frac{a_1+ \sqrt{2}{a_2}+\sqrt{2}b_2}{\sqrt{2}}.$$
If $c_1 = 0-i$ and $c_2 = 3+0i$, then $a_1 = 0, b_1 = -1, a_2 = 3, b_2 = 0$. Plugging this into what I have above, I have $\Re (c) + \Im (c) = \frac{3\sqrt{2}}{\sqrt{2}} = 3$.
However, $3$ is incorrect, and I'm not sure exactly where I made my mistake. If someone could point it out, that would be great. Thanks.
I think the given answer is wrong.
We have $$z_1-c_1=(z_0-c_1)\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)\tag 1$$ $$z_2-c_2=(z_1-c_2)\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)\tag 2$$ $$z_2-c=(z_0-c)\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)\tag 3$$
From $(1)$, we have $$z_0=c_1+\frac{\sqrt 2\ (z_1-c_1)}{1+i}\tag{4}$$ From $(2)$, we have $$z_2=c_2+(z_1-c_2)\cdot\frac{1+i}{\sqrt 2}\tag{5}$$ From $(3)$, we have $$c=\frac{z_2-iz_0}{1-i}\tag{6}$$
Then, from $(4)(5)(6)$, we have $$\begin{align}c&=\frac{1}{1-i}\left(c_2+(z_1-c_2)\cdot\frac{1+i}{\sqrt 2}-i\left(c_1+\frac{\sqrt 2\ (z_1-c_1)}{1+i}\right)\right)\\&=\frac{1+(\sqrt 2-1)i}{2}c_1+\frac{1+(1-\sqrt 2)i}{2}c_2\end{align}$$
So, for $c_1=-i,c_2=3$,
$$c=\frac{\sqrt 2-1+3}{2}+\frac{-1+3-3\sqrt 2}{2}i\Rightarrow \color{red}{\Re(c)+\Im(c)=2-\sqrt 2}.$$