# How can $i^i = e^{-\pi/2}$ !!

I was asked a homework question: find $i^i$. The solution provided was as follows:

Let $A = i^i$.

$\log A = i \log i$.

Now, $\log i = \log e^{i\pi/2} = \frac{i\pi}{2}$.

So, $\log A = -\frac{\pi}{2}$

Thus, $i^i = e^{-\pi/2}$.

I understood how the result was obtained, but it is illogical. I understand that multiplying by $i$ is equivalent to rotating the position vector of the complex number in Argand Plane by $90$ degrees anti-clockwise. How can rotating $i$ anti-clockwise $i$ number of times give $e^{-\pi/2}$?

So can somebody explain to me graphically or more intuitively, how $i^i = e^{-\pi/2}$ ?

• "$a^b$ is $a$ multiplied with itself $b$ times" only works when $b$ is a positive integer. – DHMO Apr 19 '17 at 7:12
• You can watch this video to understand more about exponents. – DHMO Apr 19 '17 at 7:13
• Then go build some other numbers which behave the way you want instead. – mathreadler Apr 19 '17 at 7:13
• If you don't have time, here is the gist of the video: powers are mappings from adders into multipliers. – DHMO Apr 19 '17 at 7:15
• @DHMO I know that $4^3=4\times4\times4.$ That is not "$4$ multiplied by itself three times". – bof Apr 19 '17 at 7:27

By definition $$e^{ix} = \cos x+i\sin x = cisx$$

This definition can be proved by observing the Taylor expansions of both the RHS and LHS. You will find they are both identical.

Letting $x= \frac{\pi}{2}$ we get this:

$$i = e^{\frac{i\pi}{2}}$$

Then just play with the powers.

$$i^i = e^{\frac{i^2\pi}{2}}$$

$$i^2=-1$$ Hence, $$i^i = e^{-\frac{\pi}{2}}$$

Of course $i^i$ attains an infinite elemental set of real values (due to the periodic nature of $cisx$) but since your question was to prove $i^i = e^{-\frac{\pi}{2}}$, I have done so accordingly.

• @Sid, you forgot to take branches. $e^{-\pi/2}$ is only one of the branches. – DHMO Apr 19 '17 at 7:22
• @DHMO what branches? – Sid Apr 19 '17 at 7:26
• @Sid $i=e^{5i\pi/2}$ also. – DHMO Apr 19 '17 at 7:26
• Complex logarithm and exponentiation is multivalued. – DHMO Apr 19 '17 at 7:26
• @DHMO so $i^i= e^{-5\frac{\pi}{2}}$ also? – Sid Apr 19 '17 at 7:32

Like DHMO says in his comment, the complex map $\ln$ is multivalued, so complex exponentiation is a multivalued operation. Accordingly,

$$i^i=\exp(i\ln(i))=\exp(i\cdot (\pi/2+2k\pi)i),k\in\mathbb{Z}$$

Then if you want, you may consider the principal branch of the above for $k=0$, which gives the desired answer. $i^i$ gives a set equality and not a single number.

• thats interesting. so the real number $i^i$ attains an infinite elemental set of real numbers? – Sid Apr 19 '17 at 7:40
• $\ln(i) = (\color{red}i\pi/2+2k\color{red}i\pi)$ – DHMO Apr 19 '17 at 7:41
• @Sid yes, that is true. – DHMO Apr 19 '17 at 7:41
• Yes. Choosing the branch makes the difference. – Yiannis Galidakis Apr 19 '17 at 7:41
• @DHMO: right, typo forgot the i – Yiannis Galidakis Apr 19 '17 at 7:50

The homework question is wrong, and so is the provided answer. Exponentiation $a^b$ is well defined when either $b$ is integer (and $a$ is invertible in case $b<0$) or when $a\in\Bbb R_{>0}$; in the former case the "repeated multiplication" definition of exponentiation applies, and in the latter case the definition $a^b=\exp(b\ln a)$ where the functions $\exp:\Bbb C\to\Bbb C$ and $\ln:\Bbb R_{>0}\to\Bbb R$ are the usual well defined ones. In the case of $\def\ii{{\bf i}}\ii^{\ii}$ however neither of these cases applies, so the expression is not well defined.

Many will try to nevertheless use the formula $\exp(b\ln a)$ to give a value to$~a^b$, as is done (somewhat indirectly) in the answer presented in the question. However, this overlooks that fact that the justification for $a^b=\exp(b\ln a)$, namely $$a^b =(\exp(\ln a))^b =\exp((\ln a)b),$$ uses a rule, namely $(\exp y)^z=\exp(yz)$ (or maybe even more generally $(a^y)^z=a^{yz}$ for $a\in\Bbb R_{>0}$), that simply does not hold for all $y,z\in\Bbb C$ (although it does hold for $y\in\Bbb R$ and $z\in\Bbb C$). For a simple example where the rule fails, take $y=2\pi\ii$ and $z=\pi$, then $$(\exp2\pi\ii)^\pi=1^\pi=1\neq \exp(2\pi^2\ii)\approx 0.629681725+0.77685322\ii .$$ An alternative form of the rule is $\ln(x^y)=y\ln(x)$ that also fails in general when $y\notin\Bbb R$, for instance when $y=2\pi\ii$ and $x=e$, where it would give $0=2\pi\ii$. Your "answer" uses this latter rule at the very beginning with $y=\ii$, which is outside of the range where the rule is valid.

• I think the confusion lies herein: $1^\frac{1}{2}$ is both 1 and -1 because both $1^2$ and ${-1}^2$ are 1. It is just a matter of convention that we chose $1^\frac{1}{2}$ to be 1 instead of -1. – Truth-seek Apr 19 '17 at 8:09
• @MathEnthusiast No, $b^x$ for positive real base $b$ is a perfectly single-valued expression (and positive real whenever $x$ is real), so $1^\frac12=1$ without any ambiguity. By the logic you want to apply $e^{0.5}$ would be doubly valued (one value being negative), and $e^{0.1}$ would be $10$-valued, which is just not what $e^x$ means for real $x$. – Marc van Leeuwen Apr 19 '17 at 8:19
• I saw in many books that $z^w$ is defined as $e^{w\ln z}$. Then, from here, we can get the principal value of $\ln z$. – Masacroso Apr 19 '17 at 8:20