Equation in the complex field $(z+2)^6=z^6$

Question

I'm wondering why if $z$ is a solution of the equation

$$ (z+2)^6=z^6 $$ then we must have $\Re(z)=-1$. I've tried to take the real part of both sides, noticing that $$ \Re((z+2)^6)=\Re([(z+2)^3]^2)=|(z+2)^3|^2 $$ but it doesn't seem to work. Thank you in advance.

Answer 1

If $(z+2)^6 = z^6$, then in particular we must have $|z+2|^6 = |z|^6$. Since the absolute value is non-negative real, there is no problem with taking the sixth root here, and we get $|z+2| = |z|$. Geometrically, $|z+2|$ is the distance in the complex plane from $z$ to the point $-2$, while $|z|$ is the distance from $z$ to the origin.

Thus we conclude that any solution to the equation must be equally far from $-2$ as it is from the origin. Which region of the complex plane has this property?

Answer 2

If $(z+2)^6=z^6$, then $|z+2|=|z|$.

We have $(z+2)^6=z^6 \Rightarrow|z+2|=|z| \iff |z+2|^2=|z|^2 \iff z \overline{z}+2z+2\overline{z}+4 =z\overline{z} \iff z+\overline{z}=2\Re(z)=-2.$

Answer 3

You can calculate $\Re(z)$ directly as follows: $$(z+2)^6=z^6 \Rightarrow \left| 1+\frac{2}{z}\right| = 1$$

Now, use $z\bar z = |z|^2$:

$$1 = \left(1+ \frac{2}{z}\right)\left(1+ \frac{2}{\bar z}\right) = 1 +4\Re\left(\frac{1}{z} \right) + \frac{4}{|z|^2} \Leftrightarrow$$ $$\Re\left(\frac{\bar z}{|z|^2} \right) = -\frac{1}{|z|^2} \Leftrightarrow \Re(z) = \Re(\bar z) = -1.$$

Answer 4

Doing $z = w - 1$: $$0 = (z + 2)^6 - z^6 = 4w(3w^4 + 10w^2 + 3),$$ $w = 0$ ($z = -1$) is obviously solution and the biquadratic factor $3w^4 + 10w^2 + 3$ has only purely imaginary solutions because $w^2 = -3,-1/3$.

Answer 5

For absolute values holds necessarily $$|z+2|=|z|$$ thus $z$ lies on the segment bisector of $AB$ with complex coordinates $z_A=-2, z_B=0.$ One can continue to solve the equation as follows:

Accordingly to the above, set $$z=-1+bi,\;b\in\mathbb{R}.\tag 1$$ The equation $(z+2)^6=z^6$ rewrites

$$(1+bi)^6=(1-bi)^6.\tag 2$$ If $1+bi=re^{i\varphi},$ then $1-bi=re^{-i\varphi}$ and $(2)$ is equivalent to (because $r\neq0$) $$e^{12i\varphi}=1.$$ We obtain $\varphi=\frac{k\pi}{6}, k\in \mathbb{Z}.$ The solutions are coordinates of the intersection points of the line $Re z=-1$ and the lines $\varphi =\frac{k\pi}{6}$ (see picture).

It suffices to take $k\in\{0,1,2,4,5\},$ since for other values of $k$ we would get repeatedly points that we have already obtained. If $k=3$ the lines are parallel, no solution.