September 2nd, 2017, 08:11 AM  #11 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,174 Thanks: 90  cjem, welcome to earth, you got here just in time. You are absolutely right. $\displaystyle z^{2}+bz+c=0$, all complex $\displaystyle z=\frac{b\pm \sqrt{z_{1}}}{2}$ $\displaystyle \sqrt{z_{1}}$ has the roots $\displaystyle re^{i\theta},re^{i(\theta+\pi)}$ and that's where I made my mistake $\displaystyle e^{i\pi}=1$ So the roots are $\displaystyle z=\frac{b\pm re^{i\theta}}{2}$ skipjack, this is as far as I got with the substitution z=a+ib $\displaystyle a^{2}b^{2}+2a2b+2=0$ for the real part and $\displaystyle ab+a+b=0$ for the imaginary part. At that point it started to get messy so I gave up. Last edited by zylo; September 2nd, 2017 at 08:21 AM. 
September 2nd, 2017, 01:01 PM  #12 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,174 Thanks: 90 
Let $\displaystyle z=a+bi$, substitute into OP, and set real and imaginary parts to 0 to solve for $\displaystyle a$ and $\displaystyle b$. 1) $\displaystyle a^{2}b^{2}+2a2b+2=0$ for the real part and 2) $\displaystyle ab+a+b=0$ for the imaginary part. Substituting for $\displaystyle a$ from 2) into 1) gives: $\displaystyle (b^{2}+2b)^{2}=2$ $\displaystyle b^{2}+2b=\pm\sqrt{2}$ $\displaystyle b^{2}+2b=\sqrt{2}$ gives a complex b, invalid. $\displaystyle b^{2}+2b=+\sqrt{2}$ gives $\displaystyle b=1\pm \sqrt{1+\sqrt{2}}$ $\displaystyle b=1+\sqrt{1+\sqrt{2}}$ gives $\displaystyle a=1+\frac{1}{\sqrt{1+\sqrt{2}}}=1+\frac{\sqrt{3}}{3}\sqrt{\sqrt{2}1}$ $\displaystyle b=1\sqrt{1+\sqrt{2}}$ gives $\displaystyle a=1\frac{1}{\sqrt{1+\sqrt{2}}}=1\frac{\sqrt{3}}{3}\sqrt{\sqrt{2}1}$ 
September 3rd, 2017, 03:39 AM  #13 
Global Moderator Joined: Dec 2006 Posts: 18,140 Thanks: 1415 
You slipped up at the end.

September 3rd, 2017, 11:11 AM  #14 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,637 Thanks: 959 Math Focus: Elementary mathematics and beyond  
September 4th, 2017, 08:59 AM  #15 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,174 Thanks: 90  Your solution is neat. I read it through. Frankly, when I saw your factorization, which I wouldn't have thought of, I lost interest. I thought my two standard methods of approach might benefit OP. Could you quote me last calculation and underline where I slipped up? 
September 4th, 2017, 02:52 PM  #16 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,174 Thanks: 90  
September 4th, 2017, 03:42 PM  #17 
Global Moderator Joined: Dec 2006 Posts: 18,140 Thanks: 1415 
Plugging what into what using what calculator?

September 4th, 2017, 07:05 PM  #18  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,174 Thanks: 90  Quote:
 
September 5th, 2017, 01:15 AM  #19 
Global Moderator Joined: Dec 2006 Posts: 18,140 Thanks: 1415 
Which parts of (3)?

September 5th, 2017, 07:29 PM  #20  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,174 Thanks: 90  Quote:
$\displaystyle \frac{\sqrt{3}}{3}$ doesn't belong in the second expression for a, but it's academic now.  

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