All possible values of $i^{-2i}$ - NBHM $2013$

Question

Question is to write down all possible values of $i^{-2i}$

I know that $e^{i\theta}=\cos(\theta)+i\sin (\theta)$

So, I can write $i=e^{i.\frac{\pi}{2}}$ then I would have :

$$i=e^{i.\frac{\pi}{2}}\Rightarrow i^{-2i}=(e^{i.\frac{\pi}{2}})^{-2i}=e^{-2i.i.\frac{\pi}{2}}=e^{\pi}$$

At the same time i can write $i=e^{i.(2\pi+\frac{\pi}{2})}$ then I would have :

$$i=e^{i.(2\pi+\frac{\pi}{2})}=e^{i.(\frac{5\pi}{2})}\Rightarrow i^{-2i}==(e^{i.\frac{5\pi}{2}})^{-2i}=e^{-2i.i.\frac{5\pi}{2}}=e^{5\pi}$$

At the same time i can write $i=e^{i.(4\pi+\frac{\pi}{2})}$ then I would have :

$$i=e^{i.(4\pi+\frac{\pi}{2})}=e^{i.(\frac{9\pi}{2})}\Rightarrow i^{-2i}=(e^{i.\frac{9\pi}{2}})^{-2i}=e^{-2i.i.\frac{9\pi}{2}}=e^{9\pi}$$

More generally I can write $i=e^{i.(2n\pi+\frac{\pi}{2})}$ then I would have :

$$i=e^{i.(2n\pi+\frac{\pi}{2})}=e^{i.(\frac{(4n+1)\pi}{2})}\Rightarrow i^{-2i}==(e^{i.\frac{(4n+1)\pi}{2}})^{-2i}=e^{-2i.i.\frac{(4n+1)\pi}{2}}=e^{(4n+1)\pi}$$

So, Possible values of $i^{-2i}$ are $e^{(4n+1)\pi}$ for $n\in \mathbb{Z}$

I would be thankful if some one can have a look at this and confirm if every thing is perfect and if there are any possible values that i am missing.

Thank you .

Answer 1

We define, for complex numbers $w$ and $z$, the expression $z^w = \exp(w \log z)$, where $a + bi = \log z$ satisfies $$z = e^{a+bi} = e^a e^{bi} = e^a(\cos b + i \sin b).$$ If $z = c + di$, then we obtain the system $$\begin{align*} c &= e^a \cos b, \\ d &= e^a \sin b, \end{align*}$$ from which it is easy to see that $c^2 + d^2 = (e^a)^2 (\cos^2 b + \sin^2 b) = e^{2a}$; hence $a = \log \sqrt{c^2 + d^2} = \log |z|$. Next, we also see that $\frac{d}{c} = \tan b$, hence $b = \tan^{-1} \frac{d}{c}$, in the sense that $b$ is an angle corresponding to the angle formed by $z$ in the complex plane (which is unique up to an integer multiple of $2\pi$). Therefore, $$\log z = a + bi = \log |z| + i \arg(z) + 2 \pi i k, \quad k \in {\mathbb Z}.$$ The complex logarithm thus defined, we then easily compute $$\begin{align*} i^{-2i} &= \exp(-2i \log i) \\ &= \exp(-2i(\log |i| + i \arg(i) + 2\pi i k)) \\ &= \exp(-2i(\log 1 + i \frac{\pi}{2} + 2\pi i k)) \\ &= \exp(\pi(4k + 1)), \quad k \in {\mathbb Z}, \end{align*}$$ which is simply the set $\{\ldots, e^{-7\pi}, e^{-3\pi}, e^{\pi}, e^{5\pi}, e^{9\pi}, \ldots\}$.

Answer 2

I think you are right. Try to think about $1^\frac{1}{n}$. It is the solution of $x^n=1$. If you consider its real solution, you will find $1$. If you consider the complex solution, you will find $n$ solutions. So we call $z^\frac{1}{n}$ a multivalue function on $\mathbb C$.

There many multivalue functions in $\mathbb C$, like $z^{-2i}$ and studying them is important.

Answer 3

You are right! You could have parametrized the solution at step 1 as $$ i=e^{i.\left(\frac{\pi}{2}+2 n\pi\right)}\Rightarrow i^{-2i}=(e^{i.\frac{\pi}{2}})^{-2i}=e^{-2i.i.\left(\frac{\pi}{2}+2 n\pi\right)}=e^{\pi+4 n \pi}=e^{(4n+1)\pi}$$