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Under what conditions is the derivative of $$f(z) = (z-(x_1+iy_1))(z-(x_2+iy_2))(z-(x_3+iy_3))$$ equal to $3(z^2-13)$ where $i$ is the imaginary number? When I put the equation in Wolfram it's a huge mess and I am wondering if there's an easier simplification or mathematical point to keep in mind.

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what are you differentiating with respect to? $z$? What are the $x$s and $y$s? – Robert Mastragostino May 26 '12 at 23:39
Yes, with respect to z. Here is one sample solution: and the x's and y's are just integer values (cartesian coordinates) – John Smith May 26 '12 at 23:40
up vote 1 down vote accepted

I would suggest to denote $z_k=x_k+iy_k$, $k=1,2,3$ to make it more concise. Then $$f\left(z\right)=\left(z-z_{1}\right)\left(z-z_{2}\right)\left(z-z_{3}\right)$$ $$f'\left(z\right)=\left(z-z_{2}\right)\left(z-z_{3}\right)+\left(z-z_{1}\right)\left(z-z_{3}\right)+\left(z-z_{1}\right)\left(z-z_{2}\right)$$ $$3z^{2}-2z\left(z_{1}+z_{2}+z_{3}\right)+z_{2}z_{3}+z_{1}z_{3}+z_{1}z_{2}=3z^{2}-39$$ Now comparing coefficients on both sides and separating real and imaginary parts we obtain: $$x_{1}+x_{2}+x_{3}=0$$ $$y_{1}+y_{2}+y_{3}=0$$ $$x_{2}x_{3}-y_{2}y_{3}+x_{1}x_{3}-y_{1}y_{3}+x_{1}x_{2}-y_{1}y_{2}=-39$$ $$x_{2}y_{3}+x_{3}y_{2}+x_{1}y_{3}+x_{3}y_{1}+x_{1}y_{2}+x_{2}y_{1}=0$$ Which leaves $x_3$, $y_3$ arbitrary

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What do you mean by x3/y3 arbitrary? – John Smith May 27 '12 at 0:23
It means that unless I made a slip somewhere, the system is underdetermined: 6 unknowns, 4 equations. – Valentin May 27 '12 at 0:30

For the derivative to be $3z^2-39$, your $f$ will have to be $f(z) = z^3-39z+c$ where $c$ is an arbitrary complex constant. Solve this cubic equation to get your $x_j+iy_j$, but it will be messy.

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