A dilemma

Imaginary part is the coefficient to $i$ . Therefore we have a solution at $x=1$ .

I believe you meant to say $(1+xi)^{n}$ in your problem statement, Raven Herd. In this case, we have:

$(1+xi)^{n} = [\sqrt{1+x^2} \cdot e^{i\arctan(x)}]^n = (1+x^2)^{n/2} \cdot e^{i \cdot n\arctan(x)} = (1+x^2)^{n/2} \cdot [\cos(n\arctan(x)) + i \sin(n\arctan(x))]$ .

We'd like the real and the imaginary parts to equal each other, which yields:

$(1+x^2)^{n/2} \cos(n\arctan(x)) = (1+x^2)^{n/2} \sin(n\arctan(x))$ ;

or $1 = \frac{\sin(n\arctan(x))}{\cos(n\arctan(x))} = \tan(n\arctan(x));$

or $\frac{\pi}{4} = n\arctan(x);$

or $\boxed{\tan(\frac{\pi}{4n}) = x}$

which satisfies $x \in (0,1], n \in \mathbb{N}.$

The answer is YES.