# Show that all roots of a Polynomial must be real numbers

I have a polynomial $$p(z) = (z+i)^{100} - (z-i)^{100}$$ and I am trying to show that all roots should be real. I was thinking I should use the Fundamental Theorem of Algebra which says that every non-constant polynomial has at least one zero in C.

I thought maybe I needed to prove that $$p(z)$$ has no complex coefficients and hence no zeros would be in C. However I am not sure if I am on the right track and if I am, where can I go from there. If the exponents were smaller, I would probably expand the function and check the coefficients like that. The number being too big here, I am guessing I am missing a more elegant solution.

Another idea would be to try to find a root or two and and see if there is a pattern whereby I can show that all roots are real. Is this possible? How would I proceed to find any such root?

$$p(z) = 0$$ implies that $$|z-i|^{100} = |z+i|^{100}$$ and that is equivalent to $$|z-i|^2 = |z+i|^2 \\ \iff (z-i)(\overline {z - i}) = (z+i)(\overline {z + i}) \\ \iff 2i(z - \overline z) = 0$$ so that $$z = \overline z$$, i.e. $$z$$ is a real number.

Geometrically, $$|z-i| = |z+i|$$ is the locus of all points in the complex plane which have the same distance from $$i$$ and $$-i$$, and that is the real line.

Let be more general and denote $$p_n(x)= (z+i)^n-(z-i)^n$$.

If $$z$$ is a root of $$p_n$$, we have $$\frac{z+i}{z-i}=\phi_k$$ where $$e^{\frac{i2k\pi}{n}}=\phi_k$$ is a $$n$$-th unit root. Therefore

$$z=i\frac{1+\phi_k}{1-\phi_k}=-\frac{1}{\tan\left(\frac{2k\pi}{n}\right)}$$

Proving the desired result.

• For the last step — showing that $i \frac{1+\varphi_k}{1-\varphi_k}$ is real — a more geometric method is to diagram the points $0$, $1$, $1+\varphi_k$, $1-\varphi_k$ in the complex plane, and the circle of radius $1$ with center $1$. The angle between ${1+\varphi_k}$ and ${1-\varphi_k}$ is then the angle of an inscribed triangle on a diameter of a circle, so it’s a right angle. – Peter LeFanu Lumsdaine Oct 21 at 14:03

(This answer was found based on the answer by mathcounterexamples.net, but simplifies the argument, and I hope makes it a bit more intuitive.)

The equation $$(z+i)^n - (z-i)^n=0$$ can be rewritten as $$(z+i)^n = (z-i)^n$$. Looking just at magnitude, this implies that $$|z+i| = |z-i|$$.

Looking at this geometrically in the complex plane, it’s now quite intuitive that $$z$$ must be real: if $$z$$ were in the upper half-plane, then $$z+i$$ would have greater magnitude than $$z-i$$, or vice versa if $$z$$ were in the lower half-plane.

To prove this intuition algebraically, write $$z= a +ib$$, with $$a,b$$ real; then expanding out the equation $$|z+i|^2 = |z-i|^2$$ in terms of $$a$$ and $$b$$ gives directly that $$b=0$$.

• @leftaroundabout Thanks for the typo catch! – Peter LeFanu Lumsdaine Oct 21 at 23:23