Complex Numbers Help

Scrutinize

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Sep 16, 2019
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So I have a final coming up and to be truthful the only thing I'm unsure of is what complex numbers are. I missed that lecture completely, and last time I did them was early high school. If anyone could explain how this problem works and what it's even asking me that would be amazing.

What is a b and c? How do I solve the equation with 4 variables, or even 3 variables. I'm not sure if I'm just looking at it wrong because I'm confused on how complex numbers work.

Thank you very much for any help!
 
You need to please show what you've done on the problem, even if it doesn't work. We can help you better that way.

Here's a start: Notice that [math]-\dfrac{1}{2} - i ~ \dfrac{ \sqrt{3} }{2}[/math] is one of the cube roots of 1. So [math]z^3 = 1[/math]
To see how this is useful note that [math]z^{3a} = \left ( z^3 \right ) ^a = (1)^a = 1[/math]. Does this give you any ideas?

-Dan
 
You need to please show what you've done on the problem, even if it doesn't work. We can help you better that way.

Here's a start: Notice that [math]-\dfrac{1}{2} - i ~ \dfrac{ \sqrt{3} }{2}[/math] is one of the cube roots of 1. So [math]z^3 = 1[/math]
To see how this is useful note that [math]z^{3a} = \left ( z^3 \right ) ^a = (1)^a = 1[/math]. Does this give you any ideas?

-Dan
You need to please show what you've done on the problem, even if it doesn't work. We can help you better that way.

Here's a start: Notice that [math]-\dfrac{1}{2} - i ~ \dfrac{ \sqrt{3} }{2}[/math] is one of the cube roots of 1. So [math]z^3 = 1[/math]
To see how this is useful note that [math]z^{3a} = \left ( z^3 \right ) ^a = (1)^a = 1[/math]. Does this give you any ideas?

-Dan
The thing is, I don't have anything done with the problem because I don't understand it at all. Complex numbers were always confusing to me because I never had a proper teacher. So this question just doesn't really make much sense to me.

Also how is that a cubed root of 1? How did you know that at least?

Thank you!
 
Complex numbers are just another extension of the concept of number, just as real numbers are an extension of the concept of number. The main things to remember with complex numbers is that there is no order property and that

[MATH]i^2 = -\ 1 = (-\ i)^2.[/MATH]
So to start things off. I would notice that [MATH]z^{3a} = (z^3)^a.[/MATH]
I would then calculate:

[MATH]z^3 = \left ( - \dfrac{1}{2} - \dfrac{i\sqrt{3}}{2} \right )^3 = \dfrac{1}{8} * (-\ 1 - i\sqrt{3})^3 = [/MATH]
[MATH]\dfrac{1}{8} * (-\ 1 - i\sqrt{3})(-\ 1 - i\sqrt{3})^2 = \dfrac{1}{8} * (-\ 1 - i\sqrt{3})(1 + 2i\sqrt{3} + 3i^2 ) =[/MATH]
[MATH]\dfrac{1}{8} * (-\ 1 - i\sqrt{3})(1 - 3 + 2i\sqrt{3}) = \dfrac{1}{8} * (-\ 1 - i\sqrt{3})(-\ 2 + 2i\sqrt{3}) =[/MATH]
[MATH]\dfrac{2}{8} * (-\ 1 - i\sqrt{3})(-\ 1 + i\sqrt{3}) = \dfrac{1}{4} * \{(-\ 1)^2 - (i * \sqrt{3})^2\}) =[/MATH]
[MATH]\dfrac{1}{4} * \{1 - 3(i^2)\} = \dfrac{1}{4} * \{1 - 3(-\ 1)\} = \dfrac{1}{4} * 4 = 1. [/MATH]
Now you may have memorized that

[MATH]\sqrt[3]{1 \pm 0 * i} = -\ \dfrac{1}{2} - \dfrac{i\sqrt{3}}{2} = \sqrt[3]{1 \pm 0 * i} = \sqrt[3]{1}.[/MATH]
But why bother memorizing anything, when you calculate it.

And [MATH]z^{3a} = (z^3)^a = 1^a = 1.[/MATH]
Now notice that

[MATH]z^{(3b+1)} = z * z^{3b} = z * (z^3)^b = z * 1^b = z = -\ \dfrac{1}{2} - \dfrac{i\sqrt{3}}{2}.[/MATH]
And [MATH]z^{(3c+2)} = z^2 * z^{3c} = z^2 * 1 = z^2 = \left ( -\ \dfrac{1}{2} - \dfrac{i\sqrt{3}}{2} \right )^2 =[/MATH]
[MATH]\dfrac{1}{4} + 2 * \dfrac{(-\ 1) * (-\ i \sqrt{3})}{2 * 2} + \dfrac{3i^2}{4} = \dfrac{1}{4} - \dfrac{3}{4} + \dfrac{2i\sqrt{3}}{4} = -\ \dfrac{1}{2} + \dfrac{i\sqrt{3}}{2}.[/MATH]
Now what happens when you add all that up?

Complex numbers involve a lot of mechanics, but the mechanics are an old story.
 
Complex numbers are just another extension of the concept of number, just as real numbers are an extension of the concept of number. The main things to remember with complex numbers is that there is no order property and that

[MATH]i^2 = -\ 1 = (-\ i)^2.[/MATH]
So to start things off. I would notice that [MATH]z^{3a} = (z^3)^a.[/MATH]
I would then calculate:

[MATH]z^3 = \left ( - \dfrac{1}{2} - \dfrac{i\sqrt{3}}{2} \right )^3 = \dfrac{1}{8} * (-\ 1 - i\sqrt{3})^3 = [/MATH]
[MATH]\dfrac{1}{8} * (-\ 1 - i\sqrt{3})(-\ 1 - i\sqrt{3})^2 = \dfrac{1}{8} * (-\ 1 - i\sqrt{3})(1 + 2i\sqrt{3} + 3i^2 ) =[/MATH]
[MATH]\dfrac{1}{8} * (-\ 1 - i\sqrt{3})(1 - 3 + 2i\sqrt{3}) = \dfrac{1}{8} * (-\ 1 - i\sqrt{3})(-\ 2 + 2i\sqrt{3}) =[/MATH]
[MATH]\dfrac{2}{8} * (-\ 1 - i\sqrt{3})(-\ 1 + i\sqrt{3}) = \dfrac{1}{4} * \{(-\ 1)^2 - (i * \sqrt{3})^2\}) =[/MATH]
[MATH]\dfrac{1}{4} * \{1 - 3(i^2)\} = \dfrac{1}{4} * \{1 - 3(-\ 1)\} = \dfrac{1}{4} * 4 = 1. [/MATH]
Now you may have memorized that

[MATH]\sqrt[3]{1 \pm 0 * i} = -\ \dfrac{1}{2} - \dfrac{i\sqrt{3}}{2} = \sqrt[3]{1 \pm 0 * i} = \sqrt[3]{1}.[/MATH]
But why bother memorizing anything, when you calculate it.

And [MATH]z^{3a} = (z^3)^a = 1^a = 1.[/MATH]
Now notice that

[MATH]z^{(3b+1)} = z * z^{3b} = z * (z^3)^b = z * 1^b = z = -\ \dfrac{1}{2} - \dfrac{i\sqrt{3}}{2}.[/MATH]
And [MATH]z^{(3c+2)} = z^2 * z^{3c} = z^2 * 1 = z^2 = \left ( -\ \dfrac{1}{2} - \dfrac{i\sqrt{3}}{2} \right )^2 =[/MATH]
[MATH]\dfrac{1}{4} + 2 * \dfrac{(-\ 1) * (-\ i \sqrt{3})}{2 * 2} + \dfrac{3i^2}{4} = \dfrac{1}{4} - \dfrac{3}{4} + \dfrac{2i\sqrt{3}}{4} = -\ \dfrac{1}{2} + \dfrac{i\sqrt{3}}{2}.[/MATH]
Now what happens when you add all that up?

Complex numbers involve a lot of mechanics, but the mechanics are an old story.
Are there a lot of identities like this where I would have needed to know that z^3 in this case was equal to one? Because when I take the test, doing all this math for each 'identity' seems like it might take too long, and there are so many numbers to work with that a mistake could easily be made. If there are a lot of identities I should memorize them right?

[MATH]z^3 = \left ( - \dfrac{1}{2} - \dfrac{i\sqrt{3}}{2} \right )^3 = \dfrac{1}{8} * (-\ 1 - i\sqrt{3})^3 = [/MATH]
[MATH]\dfrac{1}{8} * (-\ 1 - i\sqrt{3})(-\ 1 - i\sqrt{3})^2 = \dfrac{1}{8} * (-\ 1 - i\sqrt{3})(1 + 2i\sqrt{3} + 3i^2 ) =[/MATH]
[MATH]\dfrac{1}{8} * (-\ 1 - i\sqrt{3})(1 - 3 + 2i\sqrt{3}) = \dfrac{1}{8} * (-\ 1 - i\sqrt{3})(-\ 2 + 2i\sqrt{3}) =[/MATH]
[MATH]\dfrac{2}{8} * (-\ 1 - i\sqrt{3})(-\ 1 + i\sqrt{3}) = \dfrac{1}{4} * \{(-\ 1)^2 - (i * \sqrt{3})^2\}) =[/MATH]
[MATH]\dfrac{1}{4} * \{1 - 3(i^2)\} = \dfrac{1}{4} * \{1 - 3(-\ 1)\} = \dfrac{1}{4} * 4 = 1. [/MATH]

Cause that seems like a lot of work for a short question. (On the test I would probably be given around 5 minutes to solve this question or a question like this.)


Thank you very much for your response though, it did really help!
 
Actually, I suspect that 5 minutes is about what it might take to work this problem out if you remembered these identities.

[MATH](a + bi) + (c + di) = (a + c) + (b + d)i.[/MATH]
[MATH](a + bi)(c + di) = (ac - bd) + (ad + bc)i.[/MATH]
From the one above you get [MATH](a + bi)^2 = (a^2 - b^2) + 2abi[/MATH]
[MATH](a + bi)(a - bi) = a^2 + b^2.[/MATH]
[MATH]\dfrac{a + bi}{c + di} = \dfrac{a + bi}{c + di} * \dfrac{c - di}{c - di} = \dfrac{(ac + bd) + (bc - ad)i}{c^2 + d^2}.[/MATH]
So, using those you would quickly have:

[MATH]z = -\ \dfrac{1}{2} - \dfrac{\sqrt{3}}{2} * i = -\ \dfrac{1}{2} * (1 + i \sqrt{3}) \implies[/MATH]
[MATH]z^2 = \dfrac{1}{4} * (1 - 3 + 2 * 1 * i\sqrt{3}) = \dfrac{1}{4} * (-\ 2 + 2i\sqrt{3}) = \dfrac{1}{2} * (-\ 1 + i \sqrt{3}) \implies[/MATH]
[MATH]z^3 = \left \{-\ \dfrac{1}{2} (1 + i \sqrt{3}) \right \} * \left \{ \dfrac{1}{2} * (-\ 1+ i \sqrt{3} \right ) =[/MATH]
[MATH]-\ \dfrac{1}{4} * \{ 1(-\ 1) + (i\sqrt{3})^2 \} = - \ \dfrac{1}{4} * \{-\ 1 + 3(-\ 1)\} = 1.[/MATH]
The calculations are simple so long as you keep your signs straight.
 
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