We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+1
162
3
avatar

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+21978 
+9

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+27664 
+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

20 Online Users

avatar
avatar