Can $(-1)^{a+i b}$ be expressed without negative based exponentiation, complex exponentiation, complex logarithms or trigonometric functions?

Question

Can this expression, where $a$ and $b$ are both real, be expressed without negative based exponentiation (i.e. $a^b$ where $a$ is negative), complex exponentiation, complex logarithms or trigonometric functions?

$$(-1)^{a+i b}$$

I've got it to this, but can't work out how to refactor the negative based exponentiation.

$$e^{-b\pi}(-1)^a$$

Answer 1

The closest I've come to an answer works when $a$ is a positive multiple of $1\over4$ $$ e^{-b\pi}(\sqrt {0.5} + i \sqrt {0.5})^{4a} $$ and since simple multiplication can be used like so $$ (a + bi)(c + di) = (ac – bd) + (ad + bc)i. $$ $$ (a+bi)^2 = (a^2-b^2)+(2ab)i $$ this doesn't require complex exponentiation.

For $a$ not a multiple of $1\over4$, other values can be used. For example, slightly more unwieldy is this, which works with values of $a$ multiples of $1\over60$. $$ e^{-b\pi}\left(\frac{1}{16} \left(\sqrt{2}-\sqrt{6}-\sqrt{10}+\sqrt{30}+\sqrt{20+4 \sqrt{5}}+\sqrt{60+12 \sqrt{5}}\right) + i \frac{1}{16} \left(-\sqrt{2}-\sqrt{6}+\sqrt{10}+\sqrt{30}+\sqrt{20+4 \sqrt{5}}-\sqrt{60+12 \sqrt{5}}\right)\right)^{60a} $$

Answer 2

If you write $\exp(-b\pi)$ $[ln(e^{-1})]^a$, that should satisfy all your criteria (if I understand complex exponentiation in a way similar to what you defined as being negative exponentiation). There are plenty of choices for what to put between $[~]$, as long it is a function that gives $-1$ and meets your criteria.

Answer 3

Interesting. $$ \mathrm{Re}\big((-1)^{ib}\big) = \cos(b\pi) $$ to be expressed without trig functions... And without any of the known ways to epxress trig functions.