complex numbers

[m05]

Hi, I'm trying to solve this excercise:
"Suppose that the complex number z has two m-th roots and that these are conjugated. is it true that z is real?"
I tried with this:
Consider [math]z = p (cos(\alpha)+isin(\alpha))[/math][math]\ = pe^{i(\alpha + 2k\pi)}[/math]Now let [math]w[/math] be the m-th root of [math]z[/math] ,so [math]w = p^{ \frac{1}{m}}e^{\frac{i(\alpha + 2k\pi)}{m}}[/math]Now the conjugated of [math]w[/math] is [math]p^{ \frac{1}{m}}e^{\frac{i(-\alpha + 2k\pi)}{m}}[/math] but we know tha w-conjugated is a m-th root also, so [math]w[/math] conjugated is also equal to [math]w = p^{ \frac{1}{m}}e^{\frac{i(\alpha + 2k\pi)}{m}}[/math].
We get
[math]p^{ \frac{1}{m}}e^{\frac{i(-\alpha + 2k\pi)}{m}} = p^{ \frac{1}{m}}e^{\frac{i(\alpha + 2k\pi)}{m}} \\ e^{\frac{i(-\alpha + 2k\pi)}{m}} = e^{\frac{i(\alpha + 2k\pi)}{m}} \\ \frac{i(-\alpha + 2k\pi)}{m} = \frac{i(\alpha + 2k\pi)}{m} \\ -\alpha + 2k\pi = \alpha + 2k \\ 2\alpha = 0 \\ \alpha = 0[/math]
Now if [math]\alpha = 0[/math] [math]z = p[/math], so [math]z\ is \ real[/math]
What do you think?

 

[pka]

Hi, I'm trying to solve this excercise:
"Suppose that the complex number z has two m-th roots and that these are conjugated. is it true that z is real?"
I tried with this:
Consider [math]z = p (cos(\alpha)+isin(\alpha))[/math][math]\ = pe^{i(\alpha + 2k\pi)}[/math]Now let [math]w[/math] be the m-th root of [math]z[/math] ,so [math]w = p^{ \frac{1}{m}}e^{\frac{i(\alpha + 2k\pi)}{m}}[/math]Now the conjugated of [math]w[/math] is [math]p^{ \frac{1}{m}}e^{\frac{i(-\alpha + 2k\pi)}{m}}[/math] but we know tha w-conjugated is a m-th root also, so [math]w[/math] conjugated is also equal to [math]w = p^{ \frac{1}{m}}e^{\frac{i(\alpha + 2k\pi)}{m}}[/math].

How did you conclude that [imath]w=\overline{~w~}[/imath]???

 

[blamocur]

In your equation (after the "We get" line) you are assuming that [imath]k[/imath] is the same for both [imath]w[/imath] and [imath]\bar w[/imath], but this does not have to be the case. I believe that your final answer (that [imath]z[/imath] is real) is still correct, but the derivation is not clean.

 

[m05]

In your equation (after the "We get" line) you are assuming that [imath]k[/imath] is the same for both [imath]w[/imath] and [imath]\bar w[/imath], but this does not have to be the case. I believe that your final answer (that [imath]z[/imath] is real) is still correct, but the derivation is not clean.

yes, i've noticed that. Could it be so?
[imath]p^{\frac{1}{m}}e^{\frac{(-\alpha+2k\pi)}{m}} = p^{\frac{1}{m}}e^{\frac{(\alpha+2k_1\pi)}{m}} \\ -\alpha+2k\pi = \alpha +2k_1\pi\\ \alpha=(k_1-k)\pi[/imath]
since it is a multiple of pi, the cosine will be 1 or -1 while the sine is always 0 so the imaginary part vanishes in any case and the number z is real.

 

[lex]

[imath]\text{(1)\;}z^m=c[/imath]
[imath]\Rightarrow \overline{z^m}=\bar{c}[/imath]
[imath]\therefore \bar{\,z\,}^{ m}=\bar{c}[/imath]
So, if you have conjugate solutions to equation (1), then [imath]c=\bar{c}[/imath], and c is real.

 

[blamocur]

yes, i've noticed that. Could it be so?
[imath]p^{\frac{1}{m}}e^{\frac{(-\alpha+2k\pi)}{m}} = p^{\frac{1}{m}}e^{\frac{(\alpha+2k_1\pi)}{m}} \\ -\alpha+2k\pi = \alpha +2k_1\pi\\ \alpha=(k_1-k)\pi[/imath]
since it is a multiple of pi, the cosine will be 1 or -1 while the sine is always 0 so the imaginary part vanishes in any case and the number z is real.

Agree. You got a wrong sign for [imath]\alpha[/imath], but it does effect change the answer.