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

$$e^{-b\pi}(-1)^a$$
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 $1\over60$.
$$
e^{-b\pi}\left(\frac{1}{16} \left(\sqrt{2}-\sqrt{6}-\sqrt{10}+\sqrt{30}+\sqrt{20+4 \sqrt{5}}+\sqrt{60+12 \sqrt{5}}\right) + i \frac{1}{16} \left(-\sqrt{2}-\sqrt{6}+\sqrt{10}+\sqrt{30}+\sqrt{20+4 \sqrt{5}}-\sqrt{60+12 \sqrt{5}}\right)\right)^{60a}
$$
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.