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 $$z = p (cos(\alpha)+isin(\alpha))$$$$\ = pe^{i(\alpha + 2k\pi)}$$Now let $$w$$ be the m-th root of $$z$$ ,so $$w = p^{ \frac{1}{m}}e^{\frac{i(\alpha + 2k\pi)}{m}}$$Now the conjugated of $$w$$ is $$p^{ \frac{1}{m}}e^{\frac{i(-\alpha + 2k\pi)}{m}}$$ but we know tha w-conjugated is a m-th root also, so $$w$$ conjugated is also equal to $$w = p^{ \frac{1}{m}}e^{\frac{i(\alpha + 2k\pi)}{m}}$$.
We get
$$p^{ \frac{1}{m}}e^{\frac{i(-\alpha + 2k\pi)}{m}} = p^{ \frac{1}{m}}e^{\frac{i(\alpha + 2k\pi)}{m}} \\ e^{\frac{i(-\alpha + 2k\pi)}{m}} = e^{\frac{i(\alpha + 2k\pi)}{m}} \\ \frac{i(-\alpha + 2k\pi)}{m} = \frac{i(\alpha + 2k\pi)}{m} \\ -\alpha + 2k\pi = \alpha + 2k \\ 2\alpha = 0 \\ \alpha = 0$$
Now if $$\alpha = 0$$ $$z = p$$, so $$z\ is \ real$$
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 $$z = p (cos(\alpha)+isin(\alpha))$$$$\ = pe^{i(\alpha + 2k\pi)}$$Now let $$w$$ be the m-th root of $$z$$ ,so $$w = p^{ \frac{1}{m}}e^{\frac{i(\alpha + 2k\pi)}{m}}$$Now the conjugated of $$w$$ is $$p^{ \frac{1}{m}}e^{\frac{i(-\alpha + 2k\pi)}{m}}$$ but we know tha w-conjugated is a m-th root also, so $$w$$ conjugated is also equal to $$w = p^{ \frac{1}{m}}e^{\frac{i(\alpha + 2k\pi)}{m}}$$.

How did you conclude that $w=\overline{~w~}$???

 

[blamocur]

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

 

[m05]

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

yes, i've noticed that. Could it be so?
$p^{\frac{1}{m}}e^{\frac{(-\alpha+2k\pi)}{m}} = p^{\frac{1}{m}}e^{\frac{(\alpha+2k_1\pi)}{m}} \\ -\alpha+2k\pi = \alpha +2k_1\pi\\ \alpha=(k_1-k)\pi$
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]

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

 

[blamocur]

yes, i've noticed that. Could it be so?
$p^{\frac{1}{m}}e^{\frac{(-\alpha+2k\pi)}{m}} = p^{\frac{1}{m}}e^{\frac{(\alpha+2k_1\pi)}{m}} \\ -\alpha+2k\pi = \alpha +2k_1\pi\\ \alpha=(k_1-k)\pi$
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 $\alpha$, but it does effect change the answer.