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 September 29th, 2014, 07:19 AM #1 Newbie   Joined: Sep 2014 From: Sweden Posts: 1 Thanks: 0 Solving Polynomials with complex coefficients? Hey guys! Having some trouble here with solving polynomial equations with complex numbers as coefficients. I can't really seem to get a hang of it, though I never have any trouble with real polynomials. Like this one: $\displaystyle z^2 - (4+i)z + (5+5i)$ First I tried to just solve it using the standard formula for quadratic equations, but that left me with a horrible (and incorrect) answer, with fractions and square roots everywhere. Then I tried factoring it completely, which I guess is the best way to go about it, but I can't really seem to get it down to a $\displaystyle (z-a)$ type form/prime factors? How would you go about factoring/solving a polynomial like this? Any tips for dealing with these types of expressions in the future? Thanks a lot in advance!
September 29th, 2014, 07:34 AM   #2
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Quote:
 Originally Posted by ChrisLang Hey guys! Having some trouble here with solving polynomial equations with complex numbers as coefficients. I can't really seem to get a hang of it, though I never have any trouble with real polynomials. Like this one: $\displaystyle z^2 - (4+i)z + (5+5i)$ First I tried to just solve it using the standard formula for quadratic equations, but that left me with a horrible (and incorrect) answer, with fractions and square roots everywhere. Then I tried factoring it completely, which I guess is the best way to go about it, but I can't really seem to get it down to a $\displaystyle (z-a)$ type form/prime factors? How would you go about factoring/solving a polynomial like this? Any tips for dealing with these types of expressions in the future? Thanks a lot in advance!
The reason using the quadratic formula didn't work is that you must have used it incorrectly! In this case it should work out to be quite simple, although you'll have to be able to take the square root of a complex number. But, in this case, the coefficients have been chosen to give you a "nice" number to take the square root of.

You should try again with the formula and take care to get it right!

Last edited by Pero; September 29th, 2014 at 07:36 AM.

September 29th, 2014, 07:41 AM   #3
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Quote:
 Originally Posted by ChrisLang Hey guys! Having some trouble here with solving polynomial equations with complex numbers as coefficients. I can't really seem to get a hang of it, though I never have any trouble with real polynomials. Like this one: $\displaystyle z^2 - (4+i)z + (5+5i)$ First I tried to just solve it using the standard formula for quadratic equations, but that left me with a horrible (and incorrect) answer, with fractions and square roots everywhere. Then I tried factoring it completely, which I guess is the best way to go about it, but I can't really seem to get it down to a $\displaystyle (z-a)$ type form/prime factors? How would you go about factoring/solving a polynomial like this? Any tips for dealing with these types of expressions in the future? Thanks a lot in advance!
Personally I'd choose to complete the square rather than use the quadratic formula here.

-Dan

September 29th, 2014, 08:21 AM   #4
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Quote:
 Originally Posted by ChrisLang $\displaystyle z^2 - (4+i)z + (5+5i)$
There's no difference between solving a complex quadratic and a real one. In this case let's try completing the square (although the quadratic formula works, too).

You want (z - az)^2 to be z^2 - (4 + i)z + something, so you need a = (4 + i)/2 which gives
$$(z-(2+i/2))^2=z^2-(4+i)z+(2+i/2)^2=z^2-(4+i)z+(15/4+2i)$$
so you get
$$z^2-(4+i)z+(15/4+2i)+(5+5i)-(15/4+2i)=0$$
or
$$z^2-(4+i)z+(15/4+2i)=-5/4-3i$$
or
$$(z-(2+i/2))^2=-5/4-3i$$
and maybe you can take it from here?

September 29th, 2014, 09:33 AM   #5
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Quote:
 Originally Posted by topsquark Personally I'd choose to complete the square rather than use the quadratic formula here. -Dan
The quadratic formula is completing the square.

 September 29th, 2014, 10:00 AM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra You can write $z = x + \mathbb I y$, expand the equation and then the real and imaginary parts get you two equations for $x$ and $y$.
September 29th, 2014, 11:14 AM   #7
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Hello, ChrisLang!

Quote:
 $\displaystyle z^2 - (4+i)z + (5+5i)\:=\:0$

$\begin{Bmatrix}a \:=\: 1 \\ b \:=\: \text{-}(4+i) \\ c \:=\: 5(1+i)\end{Bmatrix}$

$z \;=\;\dfrac{(4+i) \pm\sqrt{(4+i)^2 - 20(1+i)}}{2} \;=\;\dfrac{(4+i)\pm\sqrt{16 + 8i - 1 - 20 - 20i}}{2}$

$\quad =\; \dfrac{(4+i) \pm\sqrt{-5-12i}}{2} \;=\;\dfrac{(4+i) \pm\sqrt{-(5+12i)}}{2} \;=\;\dfrac{(4+i)\pm\sqrt{-(3+2i)^2}}{2}$

$\quad =\;\dfrac{(4+i) \pm i(3+2i)}{2} \;=\;\dfrac{(4+i) \pm(3i-2)}{2}$

$z \;=\;\begin{Bmatrix}\dfrac{4+i + 3i-2}{2} &=& \dfrac{2+4i}{2} &=& 1+2i \\ \dfrac{4+i-3i+2}{2} &=& \dfrac{6-2i}{2} &=& 3-i \end{Bmatrix}$

September 29th, 2014, 01:11 PM   #8
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Quote:
 Originally Posted by v8archie The quadratic formula is completing the square.
In the sense that they give the same answer, certainly. But if you want to make sure you know the math you learn by completing the square. It just seems to work better for me, that's all. Personal preference.

-Dan

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