# How many values does $i^{i^z}$ have?

I came across this question in a written exam:

For which complex $z$ does $i^{i^z}$ have finitely many values?

Is there a standard way in which this expression is defined? I tried to define it as the set $$\{e^{(i \pi/2 + 2 \pi i n)e^{z(i \pi/2 + 2 \pi i m)}}\}_{n,m \in \mathbb{Z}}$$ because of the expression $\log z = \log|z| + i \arg(z) + 2 \pi i k$

Is the question asking me to count the size of this set?

There are some answers which help me define $i^z \equiv \{e^{z i \pi/2 + z 2 \pi i k}\}_{k \in \mathbb{Z}}$ but it's not clear that this will now lead to a definition for $i^{i^z}$ unless we define what it means to exponentiate a set of complex numbers since the exponent is now a set. I think what makes sense is to take $\cup_k i^{e^{z i \pi/2 + z 2 \pi i k}}$ in which case I believe we get back the definition I made up $\{e^{(i \pi/2 + 2 \pi i n)e^{z(i \pi/2 + 2 \pi i m)}}\}_{n,m \in \mathbb{Z}}$

• Consider first functions of the form $f(z)$ that could send $i^{f(z)}$ to only finitely many values. For instance, what if $f(z)$ was such that $f : \mathbb{C} \to \mathbb{Z}$? – Emily Aug 11 '14 at 16:31
• I'm very confused about what $i^z$ means. What exactly is it? – Mark Aug 11 '14 at 16:33
• It means $i$ raised to the power $z$. It's like $2^x$, except with complex numbers. – Emily Aug 11 '14 at 16:34
• But it is a set valued function? How would I compute $i^{0.9}$ for example? Which set would that correspond to? – Mark Aug 11 '14 at 16:35
• $i = e^{i\frac{\pi}{2}+2\pi k}$. Now use rules of exponentiation. – Emily Aug 11 '14 at 16:37

My answer to Complex power of a complex number seems relevant here, because it explains in detail what $w^z$ means when $w$ and $z$ are complex, and also how many values $w^z$ represents. I am reproducing it below.

Let us find in general $w^z$ where $w$ and $z$ are complex. This expression is by definition equal to $$\exp\{z\ln w\}$$ where $\ln w$ is one of the complex logarithms of $w$. That is, $\ln w = w'$ where $$e^{w'} = w.$$ Suppose $w = re^{i\theta}$. Then $$w' = \ln r + i\theta +2ik\pi$$ where $k$ is an arbitrary integer and the $\ln$ is the ordinary real-valued logarithm. (Since $r\ge 0$ this is well-defined everywhere except for $r=0$, in which case we are dealing with $0^z$, which really is ambiguous.)

Putting this back into the original formula we have the answer, that \begin{align} (re^{i\theta})^z & = \exp\{z (\ln r + i\theta + 2ik\pi)\}\tag{\star} \\ & =\exp\{z(\ln r + i\theta)\}\cdot \exp\{2ik\pi\cdot z\} \end{align} where $k$ is an integer.

Now observe that although $(\star)$ seems to list an infinite number of solutions, they are not always distinct. For example, when $z$ is a real integer, the second factor, $\exp\{2ik\pi\cdot z\}$, is $1$ for every choice of $k$, and so can be disregarded.

To find out how many values of $(\star)$ are distinct, one needs to ask about the values of $e^{2ik\pi \cdot z}$. When $z$ has nonzero imaginary part, or is a real irrational, these are all distinct and there are an infinite family of values of $w^z$, given by different choices of $k$. But when $z$ is a real rational number with (lowest-terms) denominator $n$, there are exactly $n$ distinct values.

• And how do you exponentiate a set of complex values, like $i^{i^z} = i^{\{i \pi/2 + 2 \pi i k\}_k}?$ Is it like I defined? – Mark Aug 11 '14 at 16:48
• Yes, I think you are interpreting the question correctly. – MJD Aug 11 '14 at 16:53

If $z = a + ib$ then

$$i^z = e^{i\pi z/2} = e^{\frac{1}{2}(-b\pi + ia\pi)} = \frac{\cos \frac{a\pi}{2} + i\sin \frac{a\pi}{2}}{e^{b\pi/2}}$$

Therefore

$$i^{i^z} = e^{(\cos \frac{a\pi}{2} + i\sin \frac{a\pi}{2})·e^{-b\pi/2}·\frac{i\pi}{2}} = \\e^{-\frac{\pi}{2}e^{-b\pi/2}\sin \frac{a\pi}{2}}·\left(\cos\left( \frac{\pi}{2}\cos\frac{a\pi}{2}e^\frac{-b\pi}{2}\right) + i\sin \left( \frac{\pi}{2}\cos\frac{a\pi}{2}e^\frac{-b\pi}{2}\right)\right)$$

So for $r_\theta$ to be expressed as $i^{i^z}$ then we need:

$$\ln \vert r\vert = -\frac{\pi}{2}e^\frac{\large -b\pi}{2}\sin\frac{a\pi}{2}$$

And

$$\theta = \frac{\pi}{2}e^\frac{\large -b\pi}{2}\cos\frac{a\pi}{2}$$

So given $z$ then the module and argument of $i^{i^z}$ are fixed.

In fact, I think $f(z) = i^{i^z}$ bijects $\mathbb{C} \to \mathbb{C}-\lbrace 0\rbrace$ but I might be wrong.

• This seems to miss the point altogether. – Did Sep 16 '14 at 17:17

Defining $$i^{i^z} \equiv \cup_{k\in\mathbb{Z}} i^{e^{i \pi/2 + 2 \pi i k}}$$ $$= \cup_{k\in\mathbb{Z}} \{e^{(i \pi/2 + 2 \pi i n)e^{z(i \pi/2 + 2 \pi i k)}}\}_{n \in \mathbb{Z}}$$

after plugging in the definition for $i^{e^{i \pi/2 + 2 \pi i k}}$

Fixing some $k$, it's not hard to show that $\{e^{(i \pi/2 + 2 \pi i n)e^{z(i \pi/2 + 2 \pi i k)}}\}_{n \in \mathbb{Z}} \subset i^{e^{i \pi/2 + 2 \pi i k}}$ is infinite cardinality unless $e^{z(i \pi/2 + 2 \pi i k)}$ is a rational number.

Writing $z = x+iy$ we get that $x$ must be even and $e^{-y(\pi/2 + 2 \pi k)}$ must be a rational number. But considering $k=0$ we get that $e^{-y(\pi/2)} = p/q$ for some integers $p,q$ and therefore $y= -2/\pi \log(p/q)$.

Then $e^{z(i \pi/2 + 2 \pi i k)} = \pm e^{-y(\pi/2 + 2 \pi k)} = \pm e^ {\log(p/q) (1 + 4k)} = \pm (p/q)^{1+4k}$

Now we have shown $\{e^{(i \pi/2 + 2 \pi i n)e^{z(i \pi/2 + 2 \pi i k)}}\}_{n \in \mathbb{Z}}$ must be of the form $\{e^{\pm (i \pi/2 + 2 \pi i n)(p/q)^{4k+1}}\}_{n \in \mathbb{Z}}$ if it is to be finite. So we are reduced to searching for conditions on $p,q$ for which the set $\{e^{\pm (i \pi/2 + 2 \pi i n)(p/q)^{4k+1}}\}_{n,k \in \mathbb{Z}}$ is finite.

It is necessary that $|p|>|q|$ and sufficient that $|q|=1$. If I can only show that it is necessary for $|q|=1$ then I am done.