Is there a formula for the image of z under reflection over the line y = kx?
Thanks
You are referring to the reflection of the complex number z= x+ iy (or point (x,y)) in the line y= kx?
I don't know any formula off the top of my head but you should be able to work it out. The "reflection" of a point in a line is the point, on the opposite side of the line, such that the line connecting them is perpendicular to the given line and the distance from each point to the given line is the same. The line perpendicular to y= kx has slope -1/k so the line through $\displaystyle (x_0, y_0)$ perpendicular to y= kx is $\displaystyle y= -1/k(x- x_0)+ y_0$. That will cross y= kx when $\displaystyle y= kx= -1/k(x- x_0)+ y_0$. Then $\displaystyle (k+ 1/k)x= ((k^2+ 1)/k)x= x_0/k+ 1= (x_0+ ky_0)/k$ so that $\displaystyle x= \frac{x_0+ ky_0}{k^2+ 1}$. Since this point is on y= kx, [tex]y= \frac{k(x_0+ ky_0)}{k^2+ 1}. You can calculate the change in x from $\displaystyle x_0$ to $\displaystyle \frac{x_0+ ky_0}{k^2+ 1}$ as well as the change in y from $\displaystyle y_0$ to $\displaystyle \frac{k(x_0+ ky_0)}{k^2+ 1}$ and add those again to the point $\displaystyle \left(\frac{x_0+ ky_0}{k^2+ 1}, \frac{k(x_0+ ky_0)}{k^2+ 1}\right)$ to get the corresponding point on the other side of the line.