Expressing $i^{i - 3}$ in the form $a + ib$

Question

In expressing $i^{i - 3}$ in the form $a + ib$ I have gone as far as simplifying this to

$-i^{i + 1}$

but I'm not sure where to go from here. Are there any hints you can give me that I'm missing out?

Hint: $a^{m+n} = a^m\times a^n$. Or perhaps a better hint is to use the exponential representation for complex numbers. — Edward Evans, Oct 22 '17 at 09:43
@ÍgjøgnumMeg I've broken it down to that but I still can't see where that leads to. Unless this means that a = 0 and b = -i^i? But I found that an unusual answer — Sithe, Oct 22 '17 at 09:46
Related: https://math.stackexchange.com/q/2191338/42969, https://math.stackexchange.com/q/191572/42969, https://math.stackexchange.com/q/1810925/42969 — Martin R, Oct 22 '17 at 09:51

score 1 · Answer 1 · answered Oct 22 '17 at 09:47

Simplify the expression using the hint given in comment by ÍgjøgnumMeg: $$i^{i-3} = i^i\cdot i^{-3} = i\cdot i^{i}$$

Now in principle argument of $i$ is $\frac{\pi}{2}$. Therefore $i = e^{i \pi/2}$. Substitute and you get the value: $0+i e^{-\pi / 2}$

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