Prove that if $z$ in $\mathbb{C}$ is a solution of $z^5= 1$ then the conjugate of z is also a solution I have to show that $(z^*)^5=1$, if $z^5=1$, but how do I do that?
I tried to show it by inserting $x-yi$ into the equation, so $(x-yi)^5=1$.
After using the binomial formula for a power of $5$, I thought I might be able to see similarities between the imaginary and real parts but I just got a complete mess.
I'd be happy if someone could help me.
 A: We can easily verify(by the definition of complex addition and multiplication) that
$$
\overline{w+z}=\bar{w}+\bar{z} \text { and } \overline{w z}=\bar{w} \bar{z}
$$
Here's a stronger version of your claim.

Theorem(Polynomials with real coefficients have roots in pairs). Suppose $p\in \mathcal P(\mathcal C)$ is a polynomial with real coefficients. If $\lambda \in \mathbb C$ is a root of $p$, then so is $\bar \lambda $.

Proof. Let
$$
p(z)=a_{0}+a_{1} z+\cdots+a_{m} z^{m}
$$
Assume $\lambda$ is a root of the polynomial, then
$$
a_{0}+a_{1} \lambda+\cdots+a_{m} \lambda^{m}=0
$$
Take the complex conjugate of both sides of this equation, obtaining
$$
a_{0}+a_{1} \bar{\lambda}+\cdots+a_{m} \bar{\lambda}^{m}=0
$$
so $\bar \lambda$ is also a root of $p$.
The polynomial mentioned in your case is $z^5=1$, thus for one of its roots $\lambda$, $\bar \lambda$ is also a root of $z^5=0$.
A: Using the polar form we can write the fifth roots of unity as $z = e^{2k\pi i /5}$ with $k$ an integer and $-2 \leq k \leq 2$ which has conjugate $z^* = e^{-2k\pi i /5}$. Then we calculate that $$(z^*)^5 = (e^{-2k\pi i /5})^5 = e^{-2k \pi i} = (e^{2k \pi i })^{-1} = 1^{-1}=1$$ and we are done. Alternatively we can repeatedly use that $x^*y^*=(xy)^*$ by expanding $(z^*)^5$.
A: What you can use is the general fact $$\overline{zw} = \overline z\cdot \overline w.$$ If you want to prove it, just let $z = a+ib$, $w = x + iy$ and calculate both sides.
By induction, it then follows $$\overline {z^n} = \overline z^n$$ for any $n\in\mathbb N.$ You can use it to prove the statement you need.
More generally, if you combine the above with the fact $$\overline{z+w} = \overline z + \overline w,$$ it follows that if $p$ is a polynomial with real coefficients, then $$\overline{p(z)} = p(\overline{z}),$$ which proves that $z$ is a root of $p$ if and only if $\overline{z}$ is. The key is that the coefficients of $p$ have to be real, so they are not affected by complex conjugation.
A: For a general polynomial with real coefficients
$a_0x^0+a_1x^1+a_2x^2+...a_{n-1}x^{n-1}+a_nx^n = 0$
If $z$ is a soltuion for this equation then $z*$ will also be a sloution for the equation.

Proof:
Let $z_1 = a_1+ib_1$ and $z_2 =a_2+ib_2$ be two arbitary complex numbers
Therefore $z_1* = a_1 - ib_1$ and $z_2* = a_2 - ib_2$
$z_1*+ z_2* = (a_1+b_1)-i(b_1+b_2) = (z_1+z_2)*$
$z_1*.z_2* = (a_1a_2-b_1b_2)-i(a_1b_2+a_2b_1)=(z_1.z_2)*$
Thus we have established the fact that
$z_1*+ z_2*=(z_1+z_2)*$ and
$z_1*.z_2* =(z_1.z_2)*$

If $z$ is the solution of $a_0x^0+a_1x^1+a_2x^2+...a_{n-1}x^{n-1}+a_nx^n = 0$,
$a_0z^0+a_1z^1+a_2z^2+...a_{n-1}z^{n-1}+a_nz^n = 0$
$a_0{z*}^0+a_1{z*}^1+a_2{z*}^2+...a_{n-1}{z*}^{n-1}+a_n{z*}^n$
$=(a_0z^0+a_1z^1+a_2z^2+...a_{n-1}z^{n-1}+a_nz^n)*$
$=0$*
$=0$

Thus, 
For a general polynomial with real coefficients 
$a_0x^0+a_1x^1+a_2x^2+...a_{n-1}x^{n-1}+a_nx^n = 0$
If $z$ is a soltuion for this equation then $z*$ will also be a sloution for the equation.
Now, talking specifically about $z^5=1$.
$(z*)(z*)(z*)(z*)(z*)$
$=(z.z.z.z.z)*$
$=(z^5)*$
$=1$

A very important tool here is ${z*}^k = {z^k}*<br>$
