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September 7th, 2017, 04:35 AM   #31
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Quote:
Originally Posted by zylo View Post
1) is wrong. It should be: $\displaystyle a^{4} +2a^{2}-1 = 0$ with solutions (quick calculation) . . .
It's right. Adding 2 to both sides gives $a^4 + 2a^2 + 1 = 2$, i.e. $(a^2+ 1)^2 = 2$, which I gave originally.

As this is a basic algebra problem that happens to involve complex numbers, it doesn't belong in the complex analysis section.
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September 7th, 2017, 04:50 AM   #32
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Quote:
Originally Posted by zylo View Post
when I finally figure it out after assuming you couldn't be wrong
He's not wrong. But for some reason you refuse to accept that your method that took several iterations and a heap of working isn't better. This despite the fact that your approach apparently still hasn't produced the right answer - although I haven't been checking on that.

A similar this is/was going on in the recent limit thread.

This is a common thread in your posts: an inability to accept when you are incorrect or to accept that ideas contrasting to your own might be better, or at least no worse.
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September 7th, 2017, 08:32 PM   #33
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Review of skipjack's solution, post #4

1) $\displaystyle z^{2}+2(1+i)z+2=0$
The form suggests a step I wouldn't have taken assuming it just leads to quadratic formula:
2) $\displaystyle [z+(1+i)]^{2}=z^{2}+2(1+i)z+(1+i)^{2}$
Substituting
$\displaystyle z^{2}+2(1+i)z=(z+1+i)^{2}-(1+i)^{2}$ into 1) gives:
$\displaystyle (z+1+i)^{2}=2(i-1)$

Now comes the tricky part. Let
3) $\displaystyle a+bi=z+1+i$. and then
4) $\displaystyle (a+bi)^2=2(i-1)$ and square a+bi rather than trying to find $\displaystyle [2(1-i)]^{\frac{1}{2}}$

Solve for a and b by equating real and imaginary parts of 4). Then find z from 3).
At this point the actual algebra is elementary and, in my opinion, a waste of time once you understand how you got here, but it gives for z:
$\displaystyle z=-1+\sqrt{\sqrt{2}-1}+i\left ( -1+\sqrt{\sqrt{2}+1} \right )$
$\displaystyle z=-1-\sqrt{\sqrt{2}-1}+i\left ( -1-\sqrt{\sqrt{2}+1} \right )$

Comment:
It's nice because it let's you solve for z without having to find $\displaystyle [2(1-i)]^{\frac{1}{2}}$.
The actual calculation is much simpler algebra than letting z=a+bi and substituting into OP. Unfortunately, I missed the "tricky" part and didn't understand this proof, and assumed it couldn't be right because you couldn't do it without taking square root of a complex number somewhere (assuming it didn't factor into a perfect square), but the same procedure works for the general complex quadratic. So it isn't just a matter of elementary algebra, aside from the trick

As for the "simple" comments, no comment.
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September 8th, 2017, 10:12 AM   #34
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If you get to a point where $\displaystyle (z+z_{1})^{^{2}}=z_{2}$, is there any advantage in letting $\displaystyle z+z_{1}=a+bi$ and writing $\displaystyle a^{2}-b^{2}+2bi=z_{2}$ and solving for a and b, over simply writing
$\displaystyle z=-z_{1}+z_{2}^{\frac{1}{2}}$ which is already the answer in an accepted form?
$\displaystyle z_{2}=re^{i \theta }, z_{2}^{\frac{1}{2}}=r^{\frac{1}{2}}e^{i \frac {(\theta+2k\pi)}{2}}, k=0,1
$

What if we had $\displaystyle (z+z_{1})^{3}=z_{2}$? Should you think,
then $\displaystyle (a+bi)^{3}=a^{3}+a^{2}bi-ab^{2}-b^{3}i+2abi-2b^{2}=Rez_{2}+Imz_{2}i$ and solve for a and b, as opposed to
$\displaystyle z_{2}=re^{i\theta}, z_{2}^{\frac{1}{3}}$ $\displaystyle =r^{\frac{1}{3}}e^{i\frac{(\theta+2k\pi)}{3}}, k=0,1,2$

The point being, the "trick" is not a preferred solution to $\displaystyle z^{\frac{1}{2}}$ as implied in the thread, but it is an easier calculation than sustituting a+bi into OP, which turns out to be a very awkward calculation.
In other words, the "trick" is not a solution at all to OP, it is just a clever algebraic shortcut seen in the course of the solution.
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September 8th, 2017, 10:18 AM   #35
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Perhaps you should read the original question properly.

Why are you so desperate to show that your idea is the only correct one? There are always multiple ways to reach a result.

Yes, you can use $\sin\frac38\pi$ and $\cos\frac38\pi$ (or whatever) in your answer, but it's not very informative.

Last edited by v8archie; September 8th, 2017 at 10:23 AM.
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September 8th, 2017, 10:36 AM   #36
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Quote:
Originally Posted by v8archie View Post
Perhaps you should read the original question properly.
Perhaps you should read my post properly, instead of just deprecating it.
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September 8th, 2017, 11:05 AM   #37
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For the posted problem, $\theta$ can be found quite easily, zylo, but you omitted to explain how in your post. You would later need to calculate $\cos(\theta/2)$ and $\sin(\theta/2)$, and you omitted to explain how to do that. These steps are required, as the problem specifies that the answer is to be given in the form a + bi. You could add these explanations, of course, but they considerably lengthen your method, so that it's no longer "simple".
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September 11th, 2017, 05:22 AM   #38
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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 $ab = 1$.
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 $-1 - \sqrt{-1 + √2} - (\sqrt{1 + √2} + 1)i$.
skipjack, just curious if above is a viable general procedure.

1) $\displaystyle z^{2}+k_{1}z+k_{2}=0$ and $\displaystyle z,k_{1},k_{2}$ complex.
"Completing the square"
$\displaystyle (z+k_{1}/2)^{2}=z^{2}+k_{1}z+k_{1}^{2}/4$
Substitute $\displaystyle z^{2}+k_{1}z$ into 1). Then
$\displaystyle (z+k_{1}/2)^{2}=-k_{2}+k_{1}^{2}/4=c+di$
Let $\displaystyle a+bi=z+k_{1}/2$
$\displaystyle (a+bi)^{2}=c+di$
$\displaystyle a^{2}-b^{2}=c$
$\displaystyle 2ab=d,d=b/(2a))$, Then
$\displaystyle 4a^{4}-4a^{2}c-d^{2}=0$
$\displaystyle t^{2}-ct-d^{2}/4=0$
$\displaystyle t=[c\pm\sqrt{c^{2}+d^{2}}]/2$ which is always + or -.
Then $\displaystyle a=\pm\sqrt{t}, b=d/2a, z+k_{1}/2=a+bi$ give both roots.

It's a viable procedure. Can't find it on Google or a text book. Is it in new texts, is it "out there," or did you just "see" it?

skipjack, why not post the procedure in a new thread for general availability?
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September 11th, 2017, 07:06 AM   #39
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You slipped up towards the end.

The procedure can be found by use of Google.

It's described in the textbook Maths: A Student's Survival Guide (subtitled A Self-Help Workbook for Science and Engineering Students) by Jenny Olive (ISBN: 978-0521017077) and it's also described here.
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September 11th, 2017, 09:17 AM   #40
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So why is the general solution for a cubic equation with real coefficients valid when the coefficients are complex? Without duplicating real solution step by step?

https://en.wikipedia.org/wiki/Cubic_function

(Same principle works for OP)

EDIT
Or a quartic.

Last edited by zylo; September 11th, 2017 at 09:20 AM.
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