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September 5th, 2017, 09:49 PM   #21
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How did you obtain $\small\dfrac{√3}{3}$ and supposedly verify it by substitution?
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September 5th, 2017, 10:49 PM   #22
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Quote:
Originally Posted by skipjack View Post
How did you obtain $\small\dfrac{√3}{3}$ and supposedly verify it by substitution?
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) $\displaystyle (b^{2}+2b)^{2}=2$
4)$\displaystyle b^{2}+2b=\pm\sqrt{2}$
$\displaystyle b^{2}+2b=-\sqrt{2}$ gives a complex b, invalid.
$\displaystyle b^{2}+2b=+\sqrt{2}$ gives $\displaystyle b=-1\pm \sqrt{1+\sqrt{2}}$
5) $\displaystyle b=-1+\sqrt{1+\sqrt{2}}$ gives $\displaystyle a=-1+\frac{1}{\sqrt{1+\sqrt{2}}}=-1+\frac{\sqrt{3}}{3}\sqrt{\sqrt{2}-1}$
6) $\displaystyle b=-1-\sqrt{1+\sqrt{2}}$ gives $\displaystyle a=-1-\frac{1}{\sqrt{1+\sqrt{2}}}=-1-\frac{\sqrt{3}}{3}\sqrt{\sqrt{2}-1}$
$\displaystyle (\sqrt{2}+1)(\sqrt{2}-1)$ = 3. Made same mistake every time I checked it.

I substituted b and the first expression for $\displaystyle a$ from 5) into 1), which I was absolutely sure of, and thought it was OK- but missed a 2 but didn't recheck because it was what I wanted.

The biggy was 3). I got it because I got a $\displaystyle -2b^{3}$ and a $\displaystyle +2b^{3}$ after substituting for a from 2) into 1) which cancelled out. Finally rechecked it again and both signs were -, which left me with a quartic and I quit.

Sorry I didn't let you know, but the "quite simple" remark still rankled.

But maybe it wasn't all wasted. I suspect if you substituted z=a+bi into the general complex quadratic you would find you had to solve a quartic which would essentially rule out the method.

I assume your answer is correct. Mathematicians never make mistakes in algebra.

I still think THE general approach is the quadratic formula, which should work out to your answer if I finished the calculation I started, and if not, check by substitution. I would do it in the hope of showing you are wrong because of the "quite simple" remark, but the outcome would be highly in doubt given your algebra is probably impeccable and thus not worth the effort.

Finally, I don't think you answered the OP because, in my opinion, the answer should be how to solve A complex quadratic, not THE complex quadratic, which I attempted to do, and that's not quite so "simple".

Last edited by skipjack; September 6th, 2017 at 01:10 AM.
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September 6th, 2017, 12:53 AM   #23
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The value for $b$ and the first value for $a$ are correct in (5) and (6), so you should have been able to verify them by substituting them into (1) without any problem - I don't understand why "missing a 2" would be relevant or what "2" you missed. Can you explain what you meant, being more specific?

The expression $(√2 + 1)(√2 - 1)$ doesn't seem to arise when you do the above substitutions. How did you obtain √3 from it anyway? Also, as you had obtained correct expressions, I don't understand why you aren't satisfied with the method you used. A slip in your arithmetic doesn't imply that your method is unsound.

I didn't use the expression "quite simply".

I took the first step of completing the square, which is a standard solution method, and the method used to prove the quadratic formula.

Given that x² + 2px + q = 0, adding p² - q to both sides gives x² + 2px + p² = p² - q,
i.e. (x + p)² = p² - q.

For p = 1 + i and q = 2, the right-hand side is (1 + i)² - 2, which is the expression I posted.
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September 6th, 2017, 08:21 AM   #24
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Everything worked out after all. Couldn't find where the 3's came from.

$\displaystyle z=a+bi$, substitute into OP, and set real and imaginary parts to $\displaystyle 0$ 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) $\displaystyle (b^{2}+2b)^{2}=2$
4) $\displaystyle b^{2}+2b=\pm\sqrt{2}$
5) $\displaystyle b^{2}+2b=-\sqrt{2}$ gives a complex b, invalid.
6) $\displaystyle b^{2}+2b=+\sqrt{2}$ gives $\displaystyle b=-1\pm \sqrt{1+\sqrt{2}}$
7) $\displaystyle b=-1+\sqrt{1+\sqrt{2}}$ gives $\displaystyle a=-1+\frac{1}{\sqrt{1+\sqrt{2}}}=-1+\sqrt{\sqrt{2}-1}$
8 ) $\displaystyle b=-1-\sqrt{1+\sqrt{2}}$ gives $\displaystyle a=-1-\frac{1}{\sqrt{1+\sqrt{2}}}=-1-\sqrt{\sqrt{2}-1}$

--------------------------------------------------------------------------------------

Scratch work:

9) $\displaystyle a^{2}-b^{2}+2a-2b+2=0$
10) $\displaystyle (a-b)(a+b)+2(a-b)+2=0$
11) $\displaystyle (a-b)(a+b+2)+2=0$

12) $\displaystyle I(z)=0 \rightarrow ab+a+b=0\rightarrow a=\frac{-b}{b+1}$
substituting 12) into 11) gives, after clearing fractions,
13) $\displaystyle (b^{2}+2b)(b^{2}+2b+2)-2(1+b)^{2}=0$
14) $\displaystyle (b^{2}+2b)^{2}=2$

Solving for real b gives, for case 7)
15) $\displaystyle b=-1+\sqrt{\sqrt{2}+1}$

16) $\displaystyle a=\frac{-b}{b+1}=-1+\frac{1}{\sqrt{\sqrt{2
}+1}}$
17) $\displaystyle a=-1+\frac{\sqrt{\sqrt{2}+1}}{\sqrt{2}+1}=-1+\frac{(\sqrt{2}-1)\sqrt{\sqrt{2}+1}
}{(\sqrt{2}-1)(\sqrt{2}
+1)}=-1+\sqrt{\sqrt{2}-1}\sqrt{\sqrt{2}-1}\sqrt{\sqrt{2}+1}$
18 ) $\displaystyle a=-1+\sqrt{\sqrt{2}-1}$

Once I worked with 9) in the form of 11) everything fell into place.

I still have to think about your factoring, and look at substituting a+bi into the general complex quadratic to see if it leads to a quartic equation.
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September 6th, 2017, 10:35 AM   #25
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That's correct, but when your scratch work is taken into account, I'm not sure that your answer still qualifies as "straightforward, no gimmicks".

Any quadratic equation can be solved by use of the quadratic formula. The difficulty here was obtaining the real and imaginary parts of the solutions (separately).

Before investigating whether that's more difficult for "the general complex quadratic", you would need to state how its coefficients are allowed to be defined.
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September 6th, 2017, 10:59 AM   #26
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It's clearly not more difficult as Skipjack's approach was simply to complete the square - exactly the way the formula is derived. The more complicated part is determining $\sqrt{i-1}$, which is necessary in both solutions.

The quartics in $a$ and $b$ only have 2 valid solutions because they have to be real. Skipjack didn't explicitly point out that he was selecting only those solutions, but he (correctly) did so nonetheless.

Last edited by skipjack; September 6th, 2017 at 11:28 AM.
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September 6th, 2017, 01:53 PM   #27
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Why would you derive the solution of a quadratic equation? Do you do that every time you solve a quadratic equation?

And I still don't understand this post:

Quote:
Originally Posted by skipjack View Post
$z^2 + 2(1 + i)z + 2 = 0$ 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 $ab = 1$.
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 $-1 - \sqrt{-1 + √2} - (\sqrt{1 + √2} + 1)i$.
1) is wrong. It should be: $\displaystyle a^{4} +2a^{2}-1=0$ with solutions (quick calculation)
$\displaystyle a=\pm\sqrt{\sqrt{3}-1}$
$\displaystyle b=\pm\frac{\sqrt{2}}{2}\sqrt{\sqrt{3}+1}$

But that's academic: You have
z+1+i = a+bi, Let z=x+yi then
x+yi+1+i =a+bi so

x+1-a=0
y+1-b=0

Now what? Only greg1313 knows, it's "so simple."

You lead me down the garden path, insult me, and when I finally figure it out after assuming you couldn't be wrong, after 3 pgs you put the thread in HS Algebra. Nice crew. The honorable thing to do would be to apologize and put the thread where it's been all along.

Last edited by skipjack; September 7th, 2017 at 04:33 AM.
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September 6th, 2017, 01:57 PM   #28
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Quote:
Originally Posted by zylo View Post
Why would you derive the solution of a quadratic equation? Do you do that every time you solve a quadratic equation?
doesn't make any sense does it... but... sometimes the nature of the equation leads itself towards completing the square rather than direct application of the quadratic formula.

if whole numbers are involved I tend to complete the square.

if decimal fractions are involved I tend to plug everything into the qf.

that's just me though.
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September 7th, 2017, 02:51 AM   #29
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Quote:
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doesn't make any sense does it... but... sometimes the nature of the equation leads itself towards completing the square rather than direct application of the quadratic formula.

if whole numbers are involved I tend to complete the square.

if decimal fractions are involved I tend to plug everything into the qf.

that's just me though.
You missed the point, intentionally I presume. The question is how was that used to solve OP? Please address questions in my previous post #27, including why the thread was moved here after 26 posts.
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September 7th, 2017, 04:23 AM   #30
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Why do you imagine that one should always stick to a single method? I factorise by hand or complete the square by preference because:
  1. it's good algebra practice;
  2. it gives more insight into the expression;
  3. calculators didn't do complex arithmetic when I was younger;
  4. it gives me the freedom to choose the form of the answer that is most useful to me, rather than just blindly accepting a calculator or formula output;
  5. I like to get answers for myself.
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