If I have a problem such as $2^i$, I would use the rules:
$$ e^{ix} = \cos{x} + i\sin{x} \\ b^n = e^{n\ln{b}} $$
Applying this to the example $2^i$, I would let $x=\ln{2}$: $$ e^{i\ln{2}} = \cos{(\ln{2})} + i\sin{(\ln{2})} = 0.769238901364+0.638961276314i $$ The calculator agrees with my result. However, is it possible to calculate these imaginary exponents without the use of sine and cosine?
On
Notice, when $z\in\mathbb{C}$ and $x\in\mathbb{R}^+$:
$$z=\Re[z]+\Im[z]i=|z|e^{\arg(z)i}=|z|\cos(\arg(z))+|z|\sin(\arg(z))i$$
Where $|z|=\sqrt{\Re^2[z]+\Im^2[z]}$ and $\arg(z)$ is the complex argument.
So, when $z\in\mathbb{C}$ and $x\in\mathbb{R}^+$:
$$x^z=x^{|z|e^{\arg(z)i}}=x^{|z|\cos(\arg(z))+|z|\sin(\arg(z))i}=$$ $$x^{|z|\cos(\arg(z))}\cdot x^{|z|\sin(\arg(z))i}=x^{|z|\cos(\arg(z))}\cdot e^{\ln\left(x^{|z|\sin(\arg(z))i}\right)}=$$ $$x^{|z|\cos(\arg(z))}\cdot e^{|z|\sin(\arg(z))\ln(x)i}=$$ $$x^{|z|\cos(\arg(z))}\cdot\left(\cos(|z|\sin(\arg(z))\ln(x))+\sin(|z|\sin(\arg(z))\ln(x))i\right)$$
You can proceed with either Taylor's theorem or Binomial expansion theorem.
Taylor's theorem:
$$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}\tag{$n!=1\times2\times3\times\dots\times n,0!=1$}$$
Or, for general bases,
$$a^x=e^{x\ln(a)}=\sum_{n=0}^\infty\frac{[x\ln(a)]^n}{n!}$$
For example,
$$e^i=\sum_{n=0}^\infty\frac{i^n}{n!}=1+i-\frac12-\frac i6+\frac1{24}+\frac i{120}-\dots$$
Doesn't give the exact answer with trig functions, but you can calculate it to any degree of accuracy.
To use binomial expansion theorem, assume $1^i=1$.
General binomial expansion theorem is given as
$$(a+b)^n=\frac1{0!}a^n+\frac n{1!}a^{n-1}b+\frac{n(n-1)}{2!}a^{n-2}b^2+\frac{n(n-1)(n-2)}{3!}a^{n-3}b^3+\dots+\frac{n(n-1)(n-2)\dots(n-k)}{k!}a^{n-k-1}b^{k+1}+\dots$$
For example,
$$2^i=(1+1)^i=1+i+\frac{i(i-1)}2+\frac{i(i-1)(i-2)}6+\dots$$
That'll also get you to the answer you want.
In general,
$$a^i=[(a-1)+1]^i=\dots$$
Note binomial expansion will allow you to deal with $i^i$, whereas Taylor's theorem won't (at least not easily).