Rotations of complex graphs Let $c_1 = -i$ and $c_2 = 3$. Let $z_0$ be an arbitrary complex number. We rotate $z_0$ around $c_1$ by $\pi/4$ counter-clockwise to get $z_1$. We then rotate $z_1$ around $c_2$ by $\pi/4$ counter-clockwise to get $z_2$.



There exists a complex number $c$ such that we can get $z_2$ from $z_0$ by rotating around $c$ by $\pi/2$ counter-clockwise. Find the sum of the real and imaginary parts of $c$.

I am having some trouble with this problem.  I have tried thinking this problem as if it were to be on the cartesian plane, but I still could not solve it.  Any help is appreciated.
Geometrically, this is simply stating that we wish to define a circle with center $c$ such that the points $z_0$ and $z_2$ lie on this circle. 

In the complex plane, a point can be describe in polar coordinates by $re^{i\theta}+c$ for $0 \leq \theta &lt; 2 \pi$ where $c$ is the center and $r$ is the radius. 

The first point can then be described as $z_1 = z_0e^{i\frac{\pi}{4}} + c_1$ and the second point $z_2 = z_1e^{i\frac{\pi}{4}} + c_2$. We can then describe $z_2$ in terms of a circle based on $z_0$ as:


   $z_2 = \left (z_0 e^{i\frac{\pi}{4}} + c_1\right )e^{i\frac{\pi}{4}} + c_2= z_0e^{i\frac{\pi}{2}} + \underbrace{c_1e^{i\frac{\pi}{4}}+ c_2}_{\text{new center}, c} $


The center of this circle is $c = c_1e^{i\frac{\pi}{4}}+ c_2$.

From Euler's Formula: $re^{i\theta} = r(\cos(\theta) + i\sin(\theta))$. Using this, we can describe $c$ as:


   $$ c = c_1\cos(\frac{\pi}{4}) + ic_1\sin(\frac{\pi}{4}) + c_2 =  \frac{c_1}{\sqrt{2}} + i\frac{c_1}{\sqrt{2}} + c_2 $$ 


Note that in cartesian $c_1 = a_1+ b_1i$ and $c_2 = a_2+ b_2i$. Which implies $ic_1 = -b_1 + a_1 i$. Which leads to:


   $$ c =  \frac{a_1 + b_1i}{\sqrt{2}} + \frac{-b_1 + a_1 i}{\sqrt{2}} + \frac{\sqrt{2}a_2 + \sqrt{2}b_2i}{\sqrt{2}} = \frac{(a_1-b_1+\sqrt{2}a_2)}{\sqrt{2}} + \frac{(a_1+b_1+\sqrt{2}b_2)}{\sqrt{2}}i$$


Then we just need to find $\Re(c) + \Im(c)$:


   $$ \Re(c) + \Im(c) = \frac{(a_1-b_1+\sqrt{2}a_2) + (a_1+b_1+\sqrt{2}b_2)}{\sqrt{2}}= \frac{a_1+ \sqrt{2}{a_2}+\sqrt{2}b_2}{\sqrt{2}}$$


or 


   $$ \Re(c) + \Im(c) = \frac{\Re(c_1) + \sqrt{2}\left ( \Re(c_2)+ \Im{c_2} \right ) }{\sqrt{2}}$$
Geometrically, this is simply stating that we wish to define a circle with center $c$ such that the points $z_0$ and $z_2$ lie on this circle. 

In the complex plane, a point can be describe in polar coordinates by $re^{i\theta}+c$ for $0 \leq \theta &lt; 2 \pi$ where $c$ is the center and $r$ is the radius. 

The first point can then be described as $z_1 = z_0e^{i\frac{\pi}{4}} + c_1$ and the second point $z_2 = z_1e^{i\frac{\pi}{4}} + c_2$. We can then describe $z_2$ in terms of a circle based on $z_0$ as:


   $z_2 = \left (z_0 e^{i\frac{\pi}{4}} + c_1\right )e^{i\frac{\pi}{4}} + c_2= z_0e^{i\frac{\pi}{2}} + \underbrace{c_1e^{i\frac{\pi}{4}}+ c_2}_{\text{new center}, c} $


The center of this circle is $c = c_1e^{i\frac{\pi}{4}}+ c_2$.

From Euler's Formula: $re^{i\theta} = r(\cos(\theta) + i\sin(\theta))$. Using this, we can describe $c$ as:


   $$ c = c_1\cos(\frac{\pi}{4}) + ic_1\sin(\frac{\pi}{4}) + c_2 =  \frac{c_1}{\sqrt{2}} + i\frac{c_1}{\sqrt{2}} + c_2 $$ 


Note that in cartesian $c_1 = a_1+ b_1i$ and $c_2 = a_2+ b_2i$. Which implies $ic_1 = -b_1 + a_1 i$. Which leads to:


   $$ c =  \frac{a_1 + b_1i}{\sqrt{2}} + \frac{-b_1 + a_1 i}{\sqrt{2}} + \frac{\sqrt{2}a_2 + \sqrt{2}b_2i}{\sqrt{2}} = \frac{(a_1-b_1+\sqrt{2}a_2)}{\sqrt{2}} + \frac{(a_1+b_1+\sqrt{2}b_2)}{\sqrt{2}}i$$


Then we just need to find $\Re(c) + \Im(c)$:


   $$ \Re(c) + \Im(c) = \frac{(a_1-b_1+\sqrt{2}a_2) + (a_1+b_1+\sqrt{2}b_2)}{\sqrt{2}}= \frac{a_1+ \sqrt{2}{a_2}+\sqrt{2}b_2}{\sqrt{2}}$$


or 


   $$ \Re(c) + \Im(c) = \frac{\Re(c_1) + \sqrt{2}\left ( \Re(c_2)+ \Im{c_2} \right ) }{\sqrt{2}}$$