# Proving if $z$ is an n'th root, $\bar z$ is also an n'th root

Let $$n>0$$ be an even number, and let $$z$$ be an $$n$$'th root of a real number. Is $$\bar z$$ also an $$n$$'th root of this number?

My answer is yes. The way I solved this was to consider a complex number $$z = a+bi$$ on the (polar) form $$z = re^{i\theta}$$.

Solving $$z^n = re^{i\theta}$$, I get $$z = \sqrt[n] r e^{{(i\theta+2\pi m)}/n}$$, $$0 \leq m \leq n-1$$.

Since $$n>0$$ is an even number, we will always have an even number of solutions. If we then draw the solutions in the complex plane, we will always have that for each solution that $$z$$ that has an angle $$\phi=\theta$$, we will always find $$\bar z$$, also a solution with $$\phi=-\theta$$ on the complex plane.

However, my answer is more a geometric interpreration/realization than a rigorious algebraic proof.

I wanted to try to prove this in a more general algebraic form: My attempt was to assume $$z=a+bi$$ is an $$n$$'th root of a real number. We now want to show that $$z=a-bi$$ is also an $$n$$'th root of the same real number.

I tried setting up $$(a+bi)^2 = a^2-b^2+2iab$$ and $$(a-bi)^2 = a^2-b^2-2iab$$, noticing that the real part of these numbers are always the same, the imaginary part is different in sign.

Now it's easy to realize that $$(2iab)^n=(-2iab)^n$$ for all $$n>0$$ given that $$n$$ is even.

But I have not proved that if we have a complex number $$z = a+bi$$ (being an $$n$$'th root of a real number, there also exists a $$\bar z = a-bi$$ such that $$\bar z$$ is also an $$n$$'th root of the same real number.

If this were the case, shouldn't $$(a+bi)^n=(a-bi)^n$$ for all even $$n>0$$? This is obviously not the case, I can find $$a$$ and $$b$$ not satisfying this equation.

I guess I'm confused about what I really have to do to prove this.

• A strongest result holds: Complex conjugate root theorem says that if $z$ is a solution for a given real polynomial $P$ then $\bar{z}$ is also a solution for $P$. In your case $P= x^n=a$ with a real number. – ALG Oct 18 '18 at 14:50
• COnjugation is a "field automorphism", which is fancy-speak in particular for $\overline{z+w}=\overline z+\overline w$ and $\overline{ z\cdot w}=\overline z\cdot \overline w$. Therefore $\overline z^n=\overline{z^n}$ and $z^n=a$ implies $\overline z^n=\overline a=a$ – Hagen von Eitzen Oct 18 '18 at 14:54
• Why does $n$ have to be even? – Doug M Dec 19 '18 at 21:01
• No polar form, no Cartesian form, no polynomials, no parity of $n$, just the fact that, for every complex $z$, $$\overline{z^n}=\left(\bar z\right)^n$$ hence, if $z^n=u$ with $u$ real then $\bar u=u$ hence $$\left(\bar z\right)^n=\overline{z^n}=\bar u=u$$ – Did Dec 19 '18 at 21:16

It's really easy to prove it with the comments, but I don't think you're allowed to use these stronger results to resolve your problem.

So I think you should stay with your initial idea of using $$z = re^{i\theta}$$.

So, suppose:

1. $$z = re^{i\theta}$$
2. $$\exists n \in \mathbb Z, z^n = r^ne^{ni\theta} = x$$

Since $$x^n \in \mathbb R$$, you can conclude that $$n\theta \equiv 0 \pmod \pi$$

You also have:

$$\bar{z} = re^{-i\theta}$$

So:

$$\bar{z}^n = r^ne^{-ni\theta}$$

From there, you can use the parity of n to prove $$\frac{z^n}{\bar{z}^n}=\frac{x}{\bar{z}^n}=1$$

Edit: As from [Hagen von Eitzen] comment, you don't even need n to be even

• Polar forms are essentially a pain here. The "strongest" result one needs to solve this is that $\overline{zw}=\bar z\bar w$, and if the OP does not know that... well, then they should learn and understand and be able to use it, asap. – Did Dec 19 '18 at 21:19

Lemma: $$(\overline z)^{2k} = \overline {(z^{2k})}$$

Pf: If $$z = a+bi$$ then $$(a+bi)^{2k} = \sum\limits_{j=0}^{2k}{2k \choose j}a^{j}b^{2k-j}i^{2k-j}$$.

Now if $$j$$ is even, then $$2k-j$$ is even, and $$i^{2k-j}$$ is real, and $${2k \choose j}a^{j}b^{2k-j}i^{2k-j}$$ is real.

If $$j$$ odd, then $$2k-j$$ is odd, and $$i^{2k-j}$$ is purely imaginary and $${2k \choose j}a^{j}b^{2k-j}i^{2k-j}$$ is purely imaginary.

So $$(a+b)^{2k} = \sum_{j=0; jeven}^{2k}({2k \choose j}a^{j}b^{2k-j}i^{2k-j}) + i\sum_{j=0; j odd}^{2k}({2k \choose j}a^{j}b^{2k-j}i^{2k-j-1})$$

And so $$\overline {(z^{2k})} = \sum_{j=0; jeven}^{2k}({2k \choose j}a^{j}b^{2k-j}i^{2k-j}) - i\sum_{j=0; j odd}^{2k}({2k \choose j}a^{j}b^{2k-j}i^{2k-j-1})$$

Now if we expand out $$(\overline z)^{2k}=(a-bi)^{2k}=\sum\limits_{j=0}^{2k}(-1)^j{2k \choose j}a^{j}b^{2k-j}i^{2k-j}$$

Now $$(-1)^j$$ is $$1$$ if even and $$-1$$ if odd and $$i^{2k-j}$$ is real if even and imaginary if odd so:

$$(\overline z)^{2k}=(a-bi)^{2k}=\sum_{j=0; jeven}^{2k}({2k \choose j}a^{j}b^{2k-j}i^{2k-j}) - i\sum_{j=0; j odd}^{2k}({2k \choose j}a^{j}b^{2k-j}i^{2k-j-1})=$$

$$\overline{(z^{2k})}$$

....

And that's it really. If $$z$$ is an $$n$$th root then $$z^n = m\in \mathbb R$$ so $$z^n = m = m + 0i = m - 0i = \overline m = \overline {z^n} = (\overline z)^n$$. So $$\overline z$$ is an $$n$$th root of $$m$$.

..... o Or, if it's not to geometric:

If $$z = re^{i\theta}$$ then $$\overline z = re^{-i\theta}$$ and and $$z^n = r^ne^{in\theta}$$ and $$\overline z^n = r^ne^{-in\theta}$$

If $$z^n \in \mathbb R$$ then $$n\theta = k\pi$$ for some integer $$k$$. And so $$-n\theta = -k\pi$$. Now $$-k\pi \equiv k\pi \pmod {2\pi}$$ so $$(\overline z)^n = z^n$$.