0
$\begingroup$

"Find all the roots of the polynomial $f(x)=x^2+(3i-2)x-2(1+i)$. Why does the answer not violate the $Conjugate \space Roots \space Theorem \space (CJRT)$"

I tried using the quadratic formula and got to $$x = \frac{-(3i-2) \pm \sqrt{3-4i}}{2}$$ but I'm having trouble with the $\sqrt{3-4i}$.

If I use de Moivre's Theorem I get $$\sqrt{5}\left(\cos\frac{\theta}{2} + i\sin\frac{\theta}{2}\right)$$ but I'm having trouble finding $\theta$

$\endgroup$
  • $\begingroup$ Is that supposed to be $(3i-2)x$ in the polynomial? $\endgroup$ – KSmarts Dec 5 '14 at 22:56
  • $\begingroup$ Yes it is, my bad. $\endgroup$ – Stephen Dec 5 '14 at 22:57
2
$\begingroup$

The answer does not violate the "conjugate roots theorem" (!) because the "theorem" is only valid for polynomial with real coefficients.

For your problem, to find the two values of $\sqrt{3 - 4 i}$, or of a general $\sqrt{x + i y}$, you have to solve the equation $(X+iY)^2 = x+iy$. This is equivalent to $X^2 - Y^2 = x$ and $2XY = y$. To which you can add $(X^2+Y^2)^2 = x^2+y^2$ by taking moduluses in the equation $(X+iY)^2 = x+iy$. So you have $X^2+Y^2 = \sqrt{x^2+y^2}$. Adding (resp. substracting) this to $X^2 - Y^2 = x$ gives you $X^2 = \frac{x + \sqrt{x^2+y^2}}{2}$ and $Y^2 = \frac{-x + \sqrt{x^2+y^2}}{2}$. Both previous right-hand sides (especially the second one) are always positive, so you'll find two (at most) values for $X$ and $Y$ (using the sign of $y$ through the $2XY = y$ equation) which finally give you roots (the plural handle one root with multiplicity $2$) of $x+iy$.

Application : $\sqrt{3 - 4 i} = \pm (2 - i)$.

Proof : $(\pm (2 - i))^2 = (2 - i)^2 = 4 - 1 - 2\times 2 i = 3 - 4 i$.

$\endgroup$
0
$\begingroup$

The point of the "conjugate roots theorem" is that if $P$ is a polynomial with real coefficients, then $P(\overline{z}) = \overline{P(z)}$; thus if $P(z_0) = 0$, then $P(\overline{z_0}) = \overline{P(z_0)} = 0$. Your given polynomial $f$ does not have real coefficients and do not satisfy $\overline{f(z)} = f(\overline{z})$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.