August 17th, 2017, 03:12 PM  #1 
Newbie Joined: Aug 2017 From: Luweero, Uganda Posts: 4 Thanks: 0 Math Focus: differentiation  solve the eqn z^2+2(1+i)z+2=0,giving each result in the form a+bi.
Thanks!

August 17th, 2017, 03:47 PM  #2 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,545 Thanks: 518 Math Focus: Yet to find out. 
?... What have you tried 
August 31st, 2017, 05:37 AM  #3 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,301 Thanks: 94 
What principle can you apply? Complex numbers are a field (satisfy same axioms as real numbers) so you can use the quadratic formula. Then you have to do some algebra to put the two answers in a+bi form. In particular, you need to know $\displaystyle z^{\frac{1}{2}}$ 
August 31st, 2017, 02:40 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 18,721 Thanks: 1536 
$z^2 + 2(1 + i)z + 2 = 0$ implies $(z + 1 + i)^2 = (1 + i)^2  2 = 2i  2$. Let $(a + bi)^2 = 2i  2$, where $a$ and $b$ are real. That implies $a^2  b^2 = 2$ and $ab = 1$. Eliminating $b$, $a^2  1/a^2 = 2$ and so $(a^2 + 1)^2 = 2$. Hence $a = \pm\sqrt{1 + √2}$. Similarly, $b = \pm\sqrt{1 + √2}$, where $a$ and $b$ have the same sign. Hence $z = 1 + \sqrt{1 + √2} + (\sqrt{1 + √2}  1)i$ or $1  \sqrt{1 + √2}  (\sqrt{1 + √2} + 1)i$. 
September 1st, 2017, 09:01 AM  #5 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,301 Thanks: 94 
Let z=a+bi, substitute into OP, and set real and imaginary parts to 0 to solve for a and b. Straightforward, no gimmicks. 
September 1st, 2017, 10:22 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 18,721 Thanks: 1536 
No mention of the irrelevant $z^{\frac12}\!$ now. Have you tried to solve the equations?

September 1st, 2017, 11:08 AM  #7  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,301 Thanks: 94  Quote:
z=a+bi and substitute is the most straightforward, but yes, the algebra is also tedious. I quit after a while I don't like wading through algebra, and I always make mistakes because if it's tedious I'm thinking of other things while I do it. Your solution is very nice except for the fact that you had to think of a simplifying substitution which would never occur to me, and which frankly I don't even understand.  
September 1st, 2017, 05:30 PM  #8  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,301 Thanks: 94  Quote:
$\displaystyle z=[2(1+i) \pm \sqrt{4(1+i)^{2}8}]/2$ $\displaystyle \sqrt{4(1+i)^{2}8}=\sqrt{8}\sqrt{i1}$ $\displaystyle i1=\sqrt{2}e^{\frac{3\pi i}{4}+2k\pi i}$ $\displaystyle \sqrt{i1}=2^{\frac{1}{4}} e^{\frac{3\pi i}{8}}, 2^{\frac{1}{4}}i$ $\displaystyle \sqrt{i1}= 2^{\frac{1}{4}}(\cos60+i\sin60), 2^{\frac{1}{4}}i$ $\displaystyle \sqrt{8}\sqrt{i1}=2^{\frac{3}{4}}(1+\sqrt{3}i),2^{\frac{7}{4}}i$ And that's the problem: $\displaystyle z=(b\pm \sqrt{z_{1}})/2\\$ But $\displaystyle \sqrt{z_{1}}$ has two values, so z has four values, i.e., a complex quadratic has four roots. Does anyone have a math program to solve this? Last edited by skipjack; September 2nd, 2017 at 02:51 AM.  
September 2nd, 2017, 02:34 AM  #9 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 128 Thanks: 39 Math Focus: Algebraic Number Theory, Arithmetic Geometry  $\sqrt{.}$ is a function $\mathbb{C} \to \mathbb{C}$, so $\sqrt{z}$ is just a single complex number for any $z \in \mathbb{C}$ (the principle square root of $z$). However, if $v^2 = z_1$ then $v$ is one of the two square roots of $z_1$: $v = \sqrt{z_1}$ or $v = \sqrt{z_1}$. Fortunately, by writing $\pm \sqrt{z_1}$ in your preceding working, you've already accounted for both square roots  there's no need to count them both again by saying $\sqrt{z_1}$ has two values! This is why it's actually a big problem that you've counted the same thing twice! Hopefully, if you take a step back, you'll see why this couldn't be true. How could a polynomial of degree $n$ over a field have more than $n$ roots in that field? Last edited by cjem; September 2nd, 2017 at 02:37 AM. 
September 2nd, 2017, 02:54 AM  #10 
Global Moderator Joined: Dec 2006 Posts: 18,721 Thanks: 1536  

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