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May 9th, 2017, 04:13 AM   #1
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Question Any way to get a direction of solving complex numbers equations?

Hey,
I am always having issues with finding the correct "Path" on figuring the correct method (algebra) to solve complex number problems.
No matter what way I choose it always fails, unless someone here on this amazing forum gives me the right direction.
Because no matter what way I choose when trying to solve, I always get an answer (nth roots for example) but they are all incorrect

What to do? How to go and practice if I can't seems to get the right way on every different question I get?

For example, I am now stuck on this:

$\displaystyle
(2Z+1)^6 = -1
$

So I went this approach:

Some notes:
$\displaystyle
\bullet -1 => 1 \cdot e^{i\pi} \\
\bullet Z => r \cdot e^{i\theta}
$

Taking the 6th Root of -1 I'll get:

$\displaystyle
2Z+1 = 1 \cdot e^{ i(\frac{\pi +2 \pi K}{6}) } \Rightarrow
(replace\, for \, Z) \\ \Rightarrow
2r \cdot e^{i\theta} + 1 = 1 \cdot e^{ i(\frac{\pi +2 \pi K}{6}) } \Rightarrow
(subtract\, 1\, from \, both \, sides) \\ \Rightarrow
2r \cdot e^{i\theta} = 1 \cdot e^{ i(\frac{\pi +2 \pi K}{6}) } - 1 \Rightarrow
(replace\, -1\, with\, 1 \cdot e^{i\pi}) \\ \Rightarrow
2r \cdot e^{i\theta} = 1 \cdot e^{ i(\frac{\pi +2 \pi K}{6}) } + 1 \cdot e^{i\pi} \Rightarrow
(add\, same\, terms,\, gcd\,6) \\
\Longrightarrow
2r \cdot e^{i\theta} = 1 \cdot e^{ i(\frac{7\pi +2 \pi K}{6}) }
$

So I get this at the end:

$\displaystyle
2r \cdot e^{i\theta} = 1 \cdot e^{ i(\frac{7\pi +2 \pi K}{6}) } \\ \\
r = \frac{1}{2} \\
\theta = \frac{7\pi +2 \pi K}{6}
$

From here I don't think I get the same roots as Wolfram Alpha, which are:



so in, conclusion, what are, if exists, the ways to go and find the correct path to solve complex numbers equations?
I really find it frustrating not knowing (for example in a test, under time pressure) the right way to solve it

Any help would be welcomed!
Thanks!

Last edited by skipjack; May 9th, 2017 at 04:44 AM.
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May 9th, 2017, 05:06 AM   #2
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Get $2Z = e^{i\pi(2k + 1)/6}$ - 1, then use $e^{i\theta} \equiv \cos(\theta) + i\sin(\theta)$, giving $k$ integer values from 1 to 6.
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May 9th, 2017, 10:23 AM   #3
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Quote:
Originally Posted by skipjack View Post
Get $2Z = e^{i\pi(2k + 1)/6}$ - 1, then use $e^{i\theta} \equiv \cos(\theta) + i\sin(\theta)$, giving $k$ integer values from 1 to 6.
Since:

$\displaystyle
1 = e^{i0} \,\,\,\,\, note:\, \text{cis}(0^\circ) = 1
$

then:

$\displaystyle
2re^{i\theta} = 1 \cdot e^{\frac{i(\pi+2 \pi k)}{6}} - e^{i0}
$

I'll get:

$\displaystyle
2re^{i\theta} = 1 \cdot e^{\frac{i(\pi+2 \pi k)}{6} - 0} \\

r = \frac{1}{2} \\
\theta = e^{\frac{i(\pi+2 \pi k)}{6}}
$

Inputing k from 0 to 5 still won't give me the right roots..
Did I miss anything?

Last edited by skipjack; May 10th, 2017 at 10:32 AM.
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May 10th, 2017, 05:06 AM   #4
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You should have proceeded as follows:

let $k$ = 1, then $2Z = e^{i\pi/2} - 1 = \cos(\pi/2) - 1 + i\sin(\pi/2)$, so $Z = -1/2 + i/2$, etc.

You can, of course, start with $k$ = 0 if you wish.
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May 10th, 2017, 06:33 AM   #5
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OK so I finally found a way to solve it.
It wasn't simple at first, so here it is:

Find all roots for:

$\displaystyle
(2Z+1)^6 = -1
$

Notes:
$\displaystyle
\bullet -1 \equiv 1 \cdot e^{i\pi} \\
\bullet Z \equiv r \cdot e^{i\theta} or\, r \cdot \text{cis}(\theta)
$

Algebra:

$\displaystyle
(2Z+1)^6 = 1 \cdot e^{i\pi} \Rightarrow (6th\, Root\, of\, -1) \\
\Rightarrow 2Z+1 = 1 \cdot e^{ i(\frac{\pi +2 \pi K}{6}) } \Rightarrow
(replace\, for \, Z) \\ \Rightarrow
2r \cdot \text{cis}(\theta) + 1 = 1 \cdot e^{ i(\frac{\pi +2 \pi K}{6}) } \Rightarrow
(subtract\, 1\, from \, both \, sides) \\ \Rightarrow
2r \cdot \text{cis}(\theta) = 1 \cdot e^{ i(\frac{\pi +2 \pi K}{6}) } - 1
$

Let's make terms easier to read:

$\displaystyle
\alpha = \frac{\pi +2 \pi K}{6} \\
2r \cdot \text{cis}(\theta) = \text{cis}(\alpha) - 1
$

From here we will get a system of equations for Re(Z) and Im(Z):

$\displaystyle
\begin{cases} 2r \cdot \cos(\theta) = \cos(\alpha) - 1 \\ 2r \cdot \sin(\theta) = \sin(\alpha) \end{cases}
$

Divide the bottom equation with the first, finding θ, r for k=[0,1,2,3,4,5]:

$\displaystyle
\tan(\theta) = \frac{\sin(\alpha)}{\cos(\alpha)-1} \\
\theta = \arctan(\frac{\sin(\frac{\pi +2 \pi K}{6})}{\cos(\frac{\pi +2 \pi K}{6})-1}) \\
r = \frac{\sin(\alpha)}{2\sin(\theta)}

$

Finding the roots:

For k=0:

$\displaystyle
\theta = \arctan(\frac{\sin(\frac{\pi}{6})}{\cos(\frac{\pi} {6})-1}) \Rightarrow
\theta = \frac{-5\pi}{12} \\
r = \frac{\sin(\frac{\pi}{6})}{2\sin(\frac{-5\pi}{12})} \Rightarrow
r = -\frac{1}{2}\sqrt{2-\sqrt{3}} \\
Root\, 0 = r \cdot \text{cis}(\theta) \Rightarrow -\frac{1}{2}\sqrt{2-\sqrt{3}} \cdot \text{cis}(\frac{-5\pi}{12}) \Rightarrow \\
\frac{1}{4} \cdot (\sqrt{3} + (-2 + i))
$

for k = 1:

$\displaystyle
\theta = \arctan(\frac{\sin(\frac{\pi}{2})}{\cos(\frac{\pi} {2})-1}) \Rightarrow
\theta = -\frac{\pi}{4} \\
r = \frac{\sin(\frac{\pi}{2})}{2\sin(-\frac{\pi}{4})} \Rightarrow
r = -\frac{\sqrt{2}}{2} \\
Root\, 1 = r \cdot \text{cis}(\theta) \Rightarrow -\frac{\sqrt{2}}{2} \cdot \text{cis}(-\frac{\pi}{4}) \Rightarrow \\
-\frac{1}{2} + \frac{i}{2}
$

From here, we continue to find all the rest of the roots (2,3,4,5).
Hope someone learned from this as well as me!

Shany

Last edited by skipjack; May 10th, 2017 at 10:20 AM.
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May 10th, 2017, 10:29 AM   #6
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As you don't need to write the answers in the form $r\cdot\text{cis}(\theta)$, you don't need to use $r$ at all.

You can just evaluate $\text{cis}(\pi(2k + 1)/6)/2 - 1/2$ for $k$ = 0, 1, 2, 3, 4, and 5.
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May 11th, 2017, 04:04 AM   #7
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Always great answers @skipjack! You are an amazing mathematician !
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