Find non-real roots of an equation

Question

The equation is $(z+2)^{12}=z^{12}$ I need to find the 10 non-real roots of the equation. I am not sure what direction should I take. I tried to expand by using binomial theorem and obtained an equation with $z$ to the degree of 11. I'm not sure how to proceed.

Answer 1

As $z\ne0,$

$$\left(1+\dfrac2z\right)^{12}=1=e^{2m\pi i}$$ where $m$ is any integer

$\implies1+\dfrac2z=e^{2m\pi i/12}$ where $m=0,1,2,\cdots,11$

Now $e^{2iy}-1=\cos2y+i\sin2y-1=2i\sin y(\cos y+i\sin y)$

$\dfrac1{e^{2iy}-1}=\dfrac1{2i\sin y(\cos y+i\sin y)}=\dfrac{\cos y-i\sin y}{2i\sin y}$

which will be real if $\cot y=0\implies y=(2n+1)\dfrac\pi2$ where $n$ is any integer

Answer 2

Performing the change of unknown $z = x-1$, the equation becomes $(x+1)^{12} - (x-1)^{12} = 0$, or $8x(3x^{10} + 55x^8 + 198x^6 + 198x^4 + 55x^2 + 3) = 0$. The solution $x=0$ gives $z=-1$ which is real, so it is not interesting.

Letting $t = x^2$ the equation becomes $3t^5 + 55t^4 + 198t^3 + 198t^2 + 55t + 3 = 0$. The symmetry of the coefficients shows that $t = -1$ is a solution, which gives $x = \pm \Bbb i$, which in turn gives $z = -1 \pm \Bbb i$.

Dividing the previous polynomial by $t+1$ gives us the new equation $3t^4 + 52t^3 + 146t^2 + 52t + 3 = 0$ which, because of the symmetry of its coefficients, is called a reciprocal equation. The approach to solving it is well known:

• notice that $0$ is not a root of this equation, therefore we may divide by $t^2$ and get

$$3t^2 + 52t + 146 + 52 \frac 1 t + 3 \frac 1 {t^2} = 0 ;$$

• notice that $t^2 + \dfrac 1 {t^2} = \left( t + \dfrac 1 t \right)^2 - 2$;

• according to the above, the equation can be rewritten as

$$3 \left[ \left( t + \frac 1 t \right)^2 - 2 \right] + 52 \left( t + \frac 1 t \right) + 146 = 0 ;$$

• letting $u = t + \dfrac 1 t$ the above equation becomes $3u^2 + 52u + 140 = 0$, having the roots $u = -14$ and $u = -\dfrac {10} 3$; the corresponding values for $t$ can be found by solving the equation $t^2 - ut + 1 = 0$;

• to $u = -14$ correspond the solutions $t = -7 \pm 4 \sqrt 3$, which in turn lead to $x = \sqrt{-7 \pm 4 \sqrt 3}$ and $x = -\sqrt{-7 \pm 4 \sqrt 3}$, which in turn lead to $z = -1 + \sqrt{-7 \pm 4 \sqrt 3}$ and $z = -1 - \sqrt{-7 \pm 4 \sqrt 3}$;

• to $u = -\dfrac {10} 3$ correspond the solutions $t = -3$ and $t = -\dfrac 1 3$, which in turn lead to $x = \pm \sqrt 3 \Bbb i$ and, respectively, $x = \pm \dfrac 1 {\sqrt 3} \Bbb i$, which in turn lead to $z = -1 \pm \sqrt 3 \Bbb i$ and, respectively, $z = -1 \pm \dfrac 1 {\sqrt 3} \Bbb i$.

Collecting all the results above, the non-real solutions of the given equation are

$$-1 \pm \Bbb i, \qquad -1 + \sqrt{-7 \pm 4 \sqrt 3}, \qquad -1 - \sqrt{-7 \pm 4 \sqrt 3}, \qquad -1 \pm \sqrt 3 \Bbb i, \qquad -1 \pm \dfrac 1 {\sqrt 3} \Bbb i .$$

It is easy to show that $\sqrt {-7 \pm 4\sqrt 3} = (2 \mp \sqrt 3) \Bbb i$, so that the solutions can be rewritten as

$$-1 \pm \Bbb i, \qquad -1 + (2 \mp \sqrt 3) \Bbb i, \qquad -1 - (2 \mp \sqrt 3) \Bbb i, \qquad -1 \pm \sqrt 3 \Bbb i, \qquad -1 \pm \dfrac 1 {\sqrt 3} \Bbb i .$$