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September 2nd, 2017, 08:11 AM   #11
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Quote:
Originally Posted by cjem View Post
...
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.
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September 2nd, 2017, 01:01 PM   #12
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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 $\displaystyle a=-1-\frac{1}{\sqrt{1+\sqrt{2}}}=-1-\frac{\sqrt{3}}{3}\sqrt{\sqrt{2}-1}$
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September 3rd, 2017, 03:39 AM   #13
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You slipped up at the end.
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September 3rd, 2017, 11:11 AM   #14
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Quote:
Originally Posted by zylo View Post
...a simplifying substitution which would never occur to me, and which frankly I don't even understand.
It's quite simple...

$$(z+1+i)^2-2i+2=z^2+2(1+i)z+2=0\implies(z+1+i)^2=2i-2$$
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September 4th, 2017, 08:59 AM   #15
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Originally Posted by skipjack View Post
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?
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September 4th, 2017, 02:52 PM   #16
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Quote:
Originally Posted by skipjack View Post
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.
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September 4th, 2017, 03:42 PM   #17
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Plugging what into what using what calculator?
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September 4th, 2017, 07:05 PM   #18
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Quote:
Originally Posted by skipjack View Post
Plugging what into what using what calculator?
Quote:
Originally Posted by zylo View Post
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.
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September 5th, 2017, 01:15 AM   #19
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Which parts of (3)?
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September 5th, 2017, 07:29 PM   #20
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Quote:
Originally Posted by zylo View Post
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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