MHB Can Complex Equation |z|=1 Be Proven for 11z^10+10iz^9+10iz-11=0?

[anemone]

Here is this week's POTW:

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Prove that if $11z^{10}+10iz^9+10iz-11=0$, then ${\left| z \right|}=1$.

(Here $z$ is a complex number and $i^2=-1$)

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[anemone]

Congratulations to the following members for their correct solution:):

1. Olinguito
2. Opalg

Solution from Opalg:

Spoiler

If $11z^{10}+10iz^9+10iz-11=0$ then $z^9(11z + 10i) + (10iz - 11) = 0$, or $$z^9 = \frac{11 - 10iz}{11z + 10i}.$$ Define a function $f$ by $f(z) = \frac{11 - 10iz}{11z + 10i}.$ If $w = f(z)$ then $w(11z + 10i) = 11 - 10iz$, so that $11zw + 10i(z+w) - 11 = 0.$ That relation is symmetric in $z$ and $w$, so that if $w=f(z)$ then $z = f(w)$. Thus the function $f$ is its own inverse. In particular, $f$ maps the complex plane (or more accurately the Riemann sphere) onto itself.

The function $f$ is a Möbius transformation, and it has the additional property that if $|z|=1$ then $|f(z)| = 1$. To see that, notice that if $|z|=1$ then (because a number and its complex conjugate have the same absolute value) $$|11 - 10iz| = |11 + 10i\overline{z}| = |z||11 + 10i\overline{z}| = |11z + 10iz\overline{z}| = |11z + 10i|,$$ so that $$|f(z)| = \left|\frac{11 - 10iz}{11z + 10i}\right| = \frac{|11 - 10iz|}{|11z + 10i|} = 1.$$ That fact, together with the fact that $f$ is its own inverse, says that $f$ maps the unit circle onto itself.

Next, $f(0) = \frac{11}{10i}$, which is a point outside the unit circle. If $z$ is any point inside the unit circle, there is a continuous path from $0$ to $z$ that does not cross the unit circle. The image of that path under the continuous mapping $f$ is a continuous path from $\frac{11}{10i}$ to $f(z)$ that does not cross the unit circle. Therefore $f(z)$ must lie outside the unit circle. In other words, if $|z|<1$ then $|f(z)|>1$. By the symmetry between $f$ and its inverse, the converse is also true: if $|z|>1$ then $|f(z)|<1$.

But if $|z|<1$ then $|z^9|<1$; and if $|z|>1$ then $|z^9|>1$. So the equation $f(z) = z^9$ cannot hold if $|z|<1$ or if $|z|>1$. Therefore all the solutions of the equation $f(z) = z^9$ must satisfy $|z|=1.$