My Math Forum solve the eqn z^2+2(1+i)z+2=0,giving each result in the form a+bi.
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September 2nd, 2017, 08:11 AM   #11
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Quote:
 Originally Posted by cjem ...
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

$\displaystyle a^{2}-b^{2}+2a-2b+2=0$ 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: 1,214 Thanks: 91 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) $\displaystyle a^{2}-b^{2}+2a-2b+2=0$ 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: 18,422 Thanks: 1462 You slipped up at the end.
September 3rd, 2017, 11:11 AM   #14
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Quote:
 Originally Posted by zylo ...a simplifying substitution which would never occur to me, and which frankly I don't even understand.
It's quite simple...

$$(z+1+i)^2-2i+2=z^2+2(1+i)z+2=0\implies(z+1+i)^2=2i-2$$

September 4th, 2017, 08:59 AM   #15
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Quote:
 Originally Posted by skipjack You slipped up at the end.
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
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Quote:
 Originally Posted by skipjack You slipped up at the end.
No I didn't, I checked my answer by plugging in.

Perhaps greg1313 can help you find your mistake. It's all "quite simple" for him.

 September 4th, 2017, 03:42 PM #17 Global Moderator   Joined: Dec 2006 Posts: 18,422 Thanks: 1462 Plugging what into what using what calculator?
September 4th, 2017, 07:05 PM   #18
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Quote:
 Originally Posted by skipjack Plugging what into what using what calculator?
Quote:
 Originally Posted by zylo 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) $\displaystyle a^{2}-b^{2}+2a-2b+2=0$ 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 $\displaystyle b=-1\pm \sqrt{1+\sqrt{2}}$ 3) $\displaystyle b=-1+\sqrt{1+\sqrt{2}}$ gives $\displaystyle a=-1+\frac{1}{\sqrt{1+\sqrt{2}}}=-1+\frac{\sqrt{3}}{3}\sqrt{\sqrt{2}-1}$ 4) $\displaystyle b=-1-\sqrt{1+\sqrt{2}}$ gives $\displaystyle a=-1-\frac{1}{\sqrt{1+\sqrt{2}}}=-1-\frac{\sqrt{3}}{3}\sqrt{\sqrt{2}-1}$
3) into 1) using pencil and paper.

 September 5th, 2017, 01:15 AM #19 Global Moderator   Joined: Dec 2006 Posts: 18,422 Thanks: 1462 Which parts of (3)?
September 5th, 2017, 07:29 PM   #20
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Quote:
 Originally Posted by zylo 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) $\displaystyle a^{2}-b^{2}+2a-2b+2=0$ for the real part and 2) $\displaystyle ab+a+b=0$ for the imaginary part. Substituting for $\displaystyle a$ from 2) into 1) gives: 3) $\displaystyle (b^{2}+2b)^{2}=2$ $\displaystyle b^{2}+2b=\pm\sqrt{2}$ $\displaystyle b^{2}+2b=-\sqrt{2}$ 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 $\displaystyle a=-1-\frac{1}{\sqrt{1+\sqrt{2}}}=-1-\frac{\sqrt{3}}{3}\sqrt{\sqrt{2}-1}$
3) is wrong. It's really a quartic in b- that rips it. So b above is wrong. That leaves the standard quadratic formula to solve OP, or you might find a simplifying factorization.

$\displaystyle \frac{\sqrt{3}}{3}$ doesn't belong in the second expression for a, but it's academic now.

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