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Using Argument and DeMoivre's theory

  • #1
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!
 

Answers and Replies

  • #2
phyzguy
Science Advisor
4,493
1,442
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 [tex]\pi[/tex]. A positive imaginary number has an argument of [tex]\pi[/tex]/2. So, don't say that 1 has "no argument", say that it has an argument of zero.
 
  • #3
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=r1/6(cosΘ+isinΘ)1/6
z=6√2(cos(Θ/6)+isin(Θ/6))
z=6√2(cos(π/12)+isin(π/12))

and so

zk=6√2 ei(π+2πk/12)

The answer I need to get is:

zk=21/12ei(π+8πk/24)

I don't understand at all what happened. Am I still doing the argument incorrectly? Thanks in advance!
 
  • #4
phyzguy
Science Advisor
4,493
1,442
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) = 21/12. But what is the argument of 1+i? You should get tan-1(1/1) = tan-1(1) = [tex]\pi[/tex]/4
 
  • #5
I finally got it! Thanks so much!
 

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