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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 $\displaystyle (za)$ 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  
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  Quote:
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  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,230 Thanks: 908 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  
September 29th, 2014, 08:21 AM  #4 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  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 $$ z^2(4+i)z+(15/4+2i)=5/43i $$ or $$ (z(2+i/2))^2=5/43i $$ and maybe you can take it from here? 
September 29th, 2014, 09:33 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra  
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  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Hello, ChrisLang! The Quadratic Formula should work. Quote:
$\begin{Bmatrix}a \:=\: 1 \\ b \:=\: \text{}(4+i) \\ c \:=\: 5(1+i)\end{Bmatrix}$ $z \;=\;\dfrac{(4+i) \pm\sqrt{(4+i)^2  20(1+i)}}{2} \;=\;\dfrac{(4+i)\pm\sqrt{16 + 8i  1  20  20i}}{2} $ $\quad =\; \dfrac{(4+i) \pm\sqrt{512i}}{2} \;=\;\dfrac{(4+i) \pm\sqrt{(5+12i)}}{2} \;=\;\dfrac{(4+i)\pm\sqrt{(3+2i)^2}}{2}$ $\quad =\;\dfrac{(4+i) \pm i(3+2i)}{2} \;=\;\dfrac{(4+i) \pm(3i2)}{2} $ $z \;=\;\begin{Bmatrix}\dfrac{4+i + 3i2}{2} &=& \dfrac{2+4i}{2} &=& 1+2i \\ \dfrac{4+i3i+2}{2} &=& \dfrac{62i}{2} &=& 3i \end{Bmatrix}$  
September 29th, 2014, 01:11 PM  #8 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,230 Thanks: 908 Math Focus: Wibbly wobbly timeywimey stuff.  

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