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Let $f(z)$ be a rotation by $\pi/2$ counterclockwise around $2+i$, as in the picture below:

Then we can write \[f(z) = az + b\]for some constant complex numbers $a$ and $b$. What is the ordered pair $(a, b)$?

 Jan 18, 2019
 #2
avatar+21338 
+8

Let $f(z)$ be a rotation by $\pi/2$ counterclockwise around $2+i$, as in the picture below:

 

\(\text{Let $z_0 = 2+i$}\)

\(\begin{array}{|rcll|} \hline f(z)-z_0 &=& (z-z_0)\left( \cos\left(\dfrac{\pi}{2}\right) + i\sin\left(\dfrac{\pi}{2} \right) \right) \\ f(z)-z_0 &=& (z-z_0)\left( 0 + i\cdot 1 \right) \\ f(z)-z_0 &=& (z-z_0)i \\ \mathbf{f(z)} & \mathbf{=}& \mathbf{z_0 + (z-z_0)i} \\\\ f(z) &=& z_0 + (z-z_0)i \quad | \quad z_0 = 2+i \\ f(z) &=& 2+i + \Big(z-(2+i)\Big)i \\ f(z) &=& 2+i + (z-2-i)i \\ f(z) &=& 2+i + iz-2i-i^2 \quad | \quad i^2 = -1 \\ f(z) &=& 2+i + iz-2i+1 \\ f(z) &=& iz+3-i \quad | \quad \text{compare with }f(z) = az + b \\ \\ \mathbf{(a,b)} & \mathbf{=} & \mathbf{(i,3-i)} \\ \hline \end{array} \)

 

laugh

 Jan 18, 2019
 #3
avatar+27480 
+3

As follows:

 

 

Edit:  Just noticed heureka beat me to it.  Luckily, we came up with the same result! 

 Jan 18, 2019
edited by Alan  Jan 18, 2019

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