My Math Forum Any way to get a direction of solving complex numbers equations?

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 May 9th, 2017, 03:13 AM #1 Newbie   Joined: Jul 2016 From: Israel Posts: 24 Thanks: 0 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 03:44 AM.
 May 9th, 2017, 04:06 AM #2 Global Moderator   Joined: Dec 2006 Posts: 17,513 Thanks: 1318 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.
May 9th, 2017, 09:23 AM   #3
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Quote:
 Originally Posted by skipjack 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 09:32 AM.

 May 10th, 2017, 04:06 AM #4 Global Moderator   Joined: Dec 2006 Posts: 17,513 Thanks: 1318 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. Thanks from shanytc
 May 10th, 2017, 05:33 AM #5 Newbie   Joined: Jul 2016 From: Israel Posts: 24 Thanks: 0 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 09:20 AM.
 May 10th, 2017, 09:29 AM #6 Global Moderator   Joined: Dec 2006 Posts: 17,513 Thanks: 1318 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. Thanks from shanytc
 May 11th, 2017, 03:04 AM #7 Newbie   Joined: Jul 2016 From: Israel Posts: 24 Thanks: 0 Always great answers @skipjack! You are an amazing mathematician !

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