Using Argument and DeMoivre's theory

[cutecarebear]

Hello everyone,
I'm having a bit of trouble with using the argument function with DeMoivres formula. I have the question:
z⁸= 16
and am meant to find the solution using DeMoivre's formula (zⁿ=rⁿ(cosn(Θ) +isinn(Θ)) ). The problem is, I have no idea what an argument function is or how to find it. I've read around a bit and found that root(a²+b²)= r and that tan^-1(b/a) =arg

but as you can see, I only have a (16) so that b=0, so is the argument then, tan^-1(0)? I have several problems and they all have a but no b. However, in the answer section, they all have arguments. What the heck am I doing wrong?

Homework Equations

I can do this one:
z³=-1 as it has no argument (though I don't know why), and is just a matter of plugging in numbers.
z³=-1
z³=1(cos(Θ)+isin(Θ)
z=₃√(1)(cos(Θ)+isin(Θ))^1/3
z=1(cos(2kπ)+isin(2kπ)^1/3
z=1(cos(2kπ/3)+isin(2kπ/3)

and then, since it has no argument, k can equal 0, +/-1, +/-2, +/-3 etc..., you plug in the k and solve.

The Attempt at a Solution

I haven't gotten very far with this one

z⁸=16
z⁸= 16⁸(cos8(Θ)+isin8(Θ)) or if I do it the other way
z=₈√(16)(cos(2kπ)+isin(2kπ))^1/8
z=₈√(16)(cos(2kπ/8)+isin(2kπ/8))

and I can't insert anything for k, because I don't know the argument. I know I sound like a real novice at math (I am), but I hope someone can help me! Thank you so much in advance!

 

[phyzguy]

Imagine plotting a complex number z = x + i y in the complex plane, with the x coordinate the real part and the y coordinate the imaginary part. The argument is then the angle made between the real axis and a line drawn between the origin and the point z. Any positive real number has an argument of zero, and any negative real number has an argument of $$\pi$$. A positive imaginary number has an argument of $$\pi$$/2. So, don't say that 1 has "no argument", say that it has an argument of zero.

 

[cutecarebear]

Hey, thanks so much for your reply!

Using what you told me, I've managed to get to the next question, but I get stuck here. I'm not sure what I am doing wrong. Here's where I am so far:

z⁶=1+i
z=r^1/6(cosΘ+isinΘ)^1/6
z=₆√2(cos(Θ/6)+isin(Θ/6))
z=₆√2(cos(π/12)+isin(π/12))

and so

z_k=₆√2 e^{i(π+2πk/12)}

The answer I need to get is:

z_k=2^1/12e^{i(π+8πk/24)}

I don't understand at all what happened. Am I still doing the argument incorrectly? Thanks in advance!

 

[phyzguy]

I think you're still doing the argument incorrectly. It looks like you have the right answer for the magnitude, since the sixth root of sqrt(2) = 2^1/12. But what is the argument of 1+i? You should get tan^-1(1/1) = tan^-1(1) = $$\pi$$/4

 

[cutecarebear]

I finally got it! Thanks so much!