# Show that if $(z+1)^{100} = (z-1)^{100}$, then $z$ is purely imaginary

Let $z$ be a complex number satisfying $$(z+1)^{100} = (z-1)^{100}$$ Show that $z$ is purely imaginary, i.e. that $\Re(z) = 0$.

Rearrange to

$$\left(\frac{z+1}{z-1}\right)^{100} = 1$$

I tried using $z = x+iy$ and trying to multiply numerator and denominator by the conjugate, but I hit a roadblock. I also tried substituting $1 = -e^{i\pi}$, but that also doesn't seem to get me anywhere. How can I prove this? Any help is appreciated, thank you!

## 3 Answers

Hint $$|z+1|^2=|z-1|^2 \\ \left(z+1\right) \overline{\left(z+1\right)}=\left(z-1\right) \overline{\left(z-1\right)} \\ z \bar{z}+z+\bar{z}+1=z\bar{z}-z-\bar{z}+1 \\ \bar{z}=-z$$

• Umm does $\overline z$ stand for the complex conjungate of $z$? Commented Sep 17, 2018 at 23:42
• @MohammadZuhairKhan Yes. Commented Sep 17, 2018 at 23:49

Note that $$|z+1|=|z-1|$$

Let $$z=x+iy$$

Therefore

$$\sqrt {(x+1)^2+y^2}=\sqrt {(x-1)^2+y^2}$$

$$(x+1)^2+y^2=(x-1)^2+y^2$$

$$x^2+2x+1=x^2-2x+1$$

$$4x=0$$

$$\therefore x=0$$

As we have found out, $$z=x+iy\implies z=iy$$ therefore yes, it is purely imaginary.

• So it is true for all $z \in i \mathbb R$? Commented Sep 18, 2018 at 6:15
• Yes, as I just proved that there is no real component. Commented Sep 18, 2018 at 7:42

Take absolute values, $$|z+1|^{100}=|z-1|^{100}$$. Absolute values are always non-negative real numbers, for which root extractions are always defined and one-to-one functions; so take $$100^{th}$$ roots to get $$|z+1|=|z-1|$$. Oops, I have an urgent call and need you to continue from there ;-) .

• is it valid to take the abs value of both sides like that? Commented Sep 18, 2018 at 0:07
• Yup, if two elements are equal so are their absolute vaues. Commented Sep 18, 2018 at 0:23