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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: 17,908 Thanks: 1382 
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  
Newbie Joined: Jul 2016 From: Israel Posts: 24 Thanks: 0  Quote:
$\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: 17,908 Thanks: 1382 
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. 
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: 17,908 Thanks: 1382 
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. 
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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complex, direction, equations, number, numbers, solving 
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