What is the Argument of a Complex Number with a Given Modulus?

kasse
Messages
383
Reaction score
1
z^4= 1/2 + i sqrt(3)/2

I start by transforming into polar form:

z^4 = e^(i*Pi/3)

But then I'm blank.
 
Physics news on Phys.org
kasse said:
z^4= 1/2 + i sqrt(3)/2

I start by transforming into polar form:

z^4 = e^(i*Pi/3)

But then I'm blank.

Have you tried De Moivre's formula ?
 
I believe the complex root is defined by (that is: it's usually continued from the real function by)
\sqrt{r e^{i \phi}} = \sqrt{r} e^{i \phi / 2}
 
CompuChip said:
I believe the complex root is defined by (that is: it's usually continued from the real function by)
\sqrt{r e^{i \phi}} = \sqrt{r} e^{i \phi / 2}

yes, and from De Moivre's formula we get the general:

z^{1/n} = r^{1/n}*exp(i \phi / n)
 
malawi_glenn said:
yes, and from De Moivre's formula we get the general:

z^{1/n} = r^{1/n}*exp(i \phi / n)


So e^((i Pi)/12) is a solution. How about the other three?
 
Remember that the same complex number can be obtained by rotating 2*pi and so on.

i.e arg((1/2) + i (rot3 / 2)) = pi / 3 + 2 pi * n, where n is ... -3,-2,-1,0,1,2,3...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top