# Solve for $z:$ $(1+z)^n + (1-z)^n = 0$

I'm confused about how to approach this problem.

I know that I can rewrite this as

$$\left(\frac{1-z}{1+z}\right)^n = -1.$$

However, I do not know where to go from here. Any help would be greatly appreciated.

Also, apologies in advance if the syntax in this question is off; I'm new to this site.

• Have you tried finding the $n$th roots of $-1$? – user170231 Feb 16 at 23:59

Taking the $$n$$-th root of both sides you get $$\frac{1-z}{1+z}=\sqrt[n]{-1}=\sqrt[n]{e^{i\pi}}=\exp\left(i\frac{\pi+2k\pi}{n}\right)=\exp{\left( \frac{i(2k+1)\pi}{n} \right)}=A_k$$ We can see from here that the equation has $$n$$ solutions $$z_k$$, $$k=0,1,...,n-1$$.

Rearranging the equation we get $$1-z_k=A_k(1+z_k)$$ $$z_k=\frac{1-A_k}{1+A_k}$$

Note that $$z\not=-1$$ since the denominator of $$\frac{1-z}{1+z}$$ must be $$\not=0$$, and it can easily be verified that the solution satisfies this condition.

This solution can be further simplified. Hint: multiply the numerator and denominator by $$\exp{\left( -\frac{i(2k+1)\pi}{2n} \right)}$$

• [+1] : but, concerning the $n$th roots of $-1$, I advise you to change the ambiguous notation $\sqrt[n]{-1}=\sqrt[n]{e^{i\pi}}=\text{cis}\left(\frac{\pi+2k\pi}{n}\right)$ into the non-ambiguous one $\exp(i (2k+1) \pi/n)$ or $e^{i (2k+1) \pi/n}$ as in the solution by @Servaes. – Jean Marie Feb 22 at 8:16

Indeed a good first step is to rewrite the equation into $$\left(\frac{1-z}{1+z}\right)^n = -1,$$ where $$-1=e^{\pi i}$$. This shows that $$\frac{1-z}{1+z}$$ is a $$2n$$-th root of unity, so $$\frac{1-z}{1+z}=e^{\frac{k}{n}\pi i},$$ for some odd integer $$n$$. From here it is not hard to isolate $$z$$ to find that $$z=\frac{1-e^{\frac{k}{n}\pi i}}{1+e^{\frac{k}{n}\pi i}}.$$

When seeing the group $$\dfrac{1-z}{1+z}\quad$$ you may consider a $$\tanh$$ transformation.

Indeed: $$\quad\dfrac{1-\tanh(x)}{1+\tanh(x)}=\dfrac{1-\frac{\sinh(x)}{\cosh(x)}}{1+\frac{\sinh(x)}{\cosh(x)}}=\dfrac{\cosh(x)-\sinh(x)}{\cosh(x)+\sinh(x)}=\dfrac{e^{-x}}{e^x}=e^{-2x}$$

So let set $$z=\tanh(w)$$ then the equation resumes to: $$\ e^{-2nw}=-1$$

Thus $$w=-i\dfrac{(2k+1)\pi}{2n}$$ and for $$\ 0\le k\le n-1$$ then $$z_k=-i\tan\left(\frac{2k+1}{2n}\pi\right)$$