1
$\begingroup$

If I have some function $f = (1 + 7i - x)(7 + 5i - x)(3 + 1i - x)$

How do I find its derivative? I know it is $(13/3 + 11i/3 - x)(3 + 5i - x)$ but I don't know how to find it.

  • 0
    Exactly like usual differentiation. Multiply out and differentiate (unpleasant) or use the product rule twice. The number $i$ is just a constant.2012-05-26
  • 0
    But I thought all constants drop to 0 in differentiation?2012-05-26
  • 0
    Indeed they do. By the way, the answer you give is not correct.2012-05-26
  • 0
    The solution seems to be fine (I verified it on Wolfram) -- it seems that what I am after here are the complex roots of the derivative. Complex roots are 13/3+11i/3 and 3+5i2012-05-26
  • 0
    The roots may be right. But the coefficient of $x^2$ in the derivative should be $-3$, and the coefficient of $x^2$ in the answer you give is $1$.2012-05-26
  • 0
    Please see the picture on the far right of http://en.wikipedia.org/wiki/Steiner_inellipse2012-05-26
  • 0
    A Wikipedia entry is not necessarily to be believed. Of course, neither am I, but the calculation really is totally routine. Maybe the thing missing in the wikipedia entry is a $3$ in front.2012-05-26
  • 0
    In general I am just trying to take three vertices of a triangle and calculate the coordinates of the foci of the steiner inellipse in a systematic way2012-05-26
  • 0
    See the answer by copper.hat. My guess about what's wrong with the Wikipedia entry turned out to be correct.2012-05-26
  • 0
    The roots are not affected by constant multiplication, eg, $x-1$ and $3x-3$ have the same root, but are different functions.2012-05-26

1 Answers 1

2

Expand the function $f$:

$$f(x) = -x^3+(11+13i)x^2+(16-98i)x-138+134i.$$ Then differentiate as usual.

Factoring the derivative gives:

$$f'(x) = (3+5i-x)(3x-13-11i).$$