# Root of a simple equation with imaginary exponent

Let $$i=\sqrt{-1}$$. What is the solution of $$(x-1)^i = 0$$?

EDIT: I try to solve the following boundary value problem for $$a>0$$:

with $$\lambda>0$$ and $$y(a)=0$$. It is corresponding to the radial part of the Klein-Gordon equation of a scalar field inside its own Schwarzschild radius $$a$$. The solution of this equation is (from Maple)

I try to find the (spectrum) values of $$\lambda$$, for which the boundary condition $$y(a)=0$$ is satisfied. Since $$HeunC(...)$$, for $$x=a$$, is constant ($$\neq 0$$), I have to ask for a solution of $$(a-x)^{i\times const}=0$$.

I have not expressed this context in the initial question because I don't think that somebody finds it interesting...

• Apparently $0^{-i}$ is undefined, at least according to Wolfram alpha. – Mohammad Zuhair Khan Oct 27 '18 at 14:48
• Also $z^i$ is undefined.The complex logarithm has many branches. On the other hand we expect that exponential functions, if well-defined, do not vanish. – Jack D'Aurizio Oct 27 '18 at 14:52

Hint: If $$z, \alpha \in {\Bbb C}$$, define $$z^\alpha= \exp(\alpha \log z)$$, where $$\exp$$ is defined in some independent manner such as by a power series.