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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 $i^i = e^{-\pi/2}$ ?

Truth-seek
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    "$a^b$ is $a$ multiplied with itself $b$ times" only works when $b$ is a positive integer. – DHMO Apr 19 '17 at 07:12
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    You can watch this video to understand more about exponents. – DHMO Apr 19 '17 at 07:13
  • Then go build some other numbers which behave the way you want instead. – mathreadler Apr 19 '17 at 07:13
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    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 07:15
  • Out of curiosity, what does $a^{\frac{1}{2}}$ mean @MathEnthusiast (I'm sure you've encountered this one and have had no trouble justifying the result and meaning) – frog1944 Apr 19 '17 at 07:21
  • since $e^{i (\pi/2) \alpha}$ is an analytic function of $\alpha$ (well, as long as the angle doesn't exceed $4\pi$) you can just analytically continue from real $\alpha$ to complex $\alpha$ – tired Apr 19 '17 at 07:22
  • @tired I don't think OP would understand what "analytic" means. – DHMO Apr 19 '17 at 07:22
  • Where do you get "$i^i$ is $i$ multiplied by itself $i$ times"? If you multiply $a$ by itself once you get $a^2,$ not $a^1.$ – bof Apr 19 '17 at 07:23
  • I think your confusion comes from missing that it is an imaginary power. What does it mean to rotate something an imaginary number of times, and why should that be anything similar to rotating the vector 90 degrees 2 or 5 times. – Johan Löfberg Apr 19 '17 at 07:23
  • @bof it means $4^3 = \underbrace{4\times4\times4}_{3\text{ terms}}$ – DHMO Apr 19 '17 at 07:23
  • @dmho that is why i just wrote a comment with the purpose to stimulate some own research ;) – tired Apr 19 '17 at 07:24
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    @DHMO I know that $4^3=4\times4\times4.$ That is not "$4$ multiplied by itself three times". – bof Apr 19 '17 at 07:27
  • @bof well that is convenient. – DHMO Apr 19 '17 at 07:29
  • First, you have to define which $z^i$ branch-cut you are considering. With the principal branch, $\mathrm{i} = \exp(\mathrm{i}\pi/2)$. – Felix Marin Oct 03 '20 at 15:03

4 Answers4

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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
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    @Sid, you forgot to take branches. $e^{-\pi/2}$ is only one of the branches. – DHMO Apr 19 '17 at 07:22
  • @DHMO what branches? – Sid Apr 19 '17 at 07:26
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    @Sid $i=e^{5i\pi/2}$ also. – DHMO Apr 19 '17 at 07:26
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    Complex logarithm and exponentiation is multivalued. – DHMO Apr 19 '17 at 07:26
  • @DHMO so $i^i= e^{-5\frac{\pi}{2}}$ also? – Sid Apr 19 '17 at 07:32
  • Yes, and also $e^{2n\pi+3\pi/2}$ for any integer $n \in \Bbb Z$. – DHMO Apr 19 '17 at 07:40
  • @DHMO interesting, thanks for the info. but the calculator is only programmed with $x\in [-\pi,\pi]$ – Sid Apr 19 '17 at 07:42
  • you should know when to add the periods. – DHMO Apr 19 '17 at 07:42
  • can somebody provide a graphical proof?? – Truth-seek Apr 19 '17 at 08:27
  • "Of course $i^i$ attains an infinite elemental set of real values, but since your question was to prove that [it takes one particular value], I have done so accordingly." That is a nice example of doublethink. – Marc van Leeuwen Apr 19 '17 at 10:37
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    Just to repeat what I have been saying in my own answer, passing from the correct $i=e^\frac{i\pi}2$ to the incorrect $i^i=e^\frac{i^2\pi}2$ is an instance of applying a rule $(x^y)^z=x^{yz}$ outside the domain where this rule is valid (namely for $y\in\Bbb R$). – Marc van Leeuwen Apr 19 '17 at 10:47
  • @MarcvanLeeuwen so you are saying $i^i$ does not equal $e^{-\pi/2}$ – Sid Apr 19 '17 at 10:51
  • Calculators just compute what their programmer told them to. There is a widespread belief (which you can also find on Wikipedia) that the real identity $a^b=\exp(b\ln a)$ can be used to define exponentiation of complex numbers, and the programmer just implemented this. But any mathematician who studies the question seriously will see that one cannot adopt that definition without running into inconsistencies. – Marc van Leeuwen Apr 19 '17 at 10:53
  • @Sid Yes, as I state in my answer, $i^i$ is not and should not be defined. – Marc van Leeuwen Apr 19 '17 at 10:54
  • @MarcvanLeeuwen wow. You are correct as you must have had more experience. Can you please help me find one inconsistency so that I can change my answer and disprove the question. Thanks – Sid Apr 19 '17 at 11:17
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    As I said a few comments up, passing from $i=e^\frac{i\pi}2$ to $i^i=e^\frac{i^2\pi}2$ is not justified. – Marc van Leeuwen Apr 19 '17 at 11:24
  • @MarcvanLeeuwen ok it is not justified. But please provide me one example of a reason why it is not justified. – Sid Apr 19 '17 at 11:31
  • @MarcvanLeeuwen I request you to please look at : JEE Advanced. This is an examination. Look at the examination paper 2016 maths part. Solve it and tell me when you have all solutions ready. Thank you for your help – Sid Apr 19 '17 at 13:12
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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.

See also this answer.

Marc van Leeuwen
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  • True. I doubt this was a homework question, just him fidgeting with his calculator. – Sid Apr 19 '17 at 08:09
  • 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 08:09
  • @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 08:19
  • 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 08:20
  • @Marc van Leeuwen Then tell me why is $1^{\frac{1}{2}}$ not equal to -1? (other than being a convention) – Truth-seek Apr 19 '17 at 08:23
  • @Masacroso: As I said, the justification of setting $z^w$ to $e^{w\ln z}$ is based on applying a rule outside the domain where it is valid, namely $w\in\Bbb R$. I don't care how many people have made this mistake, even in print, and taught it to students, they are just wrong. They think it is harmless, or at worst leads to multi-values expressions (which is bad enough), but it leads to actual inconsistencies that cannot be defended other than by "I can choose to give any value I like to your expressions, even if they were perfectly uniquely defined before". – Marc van Leeuwen Apr 19 '17 at 08:23
  • @MathEnthusiast: Because $-1$ is negative, and raising a positive real number to a real power never gives a negative number. – Marc van Leeuwen Apr 19 '17 at 08:23
  • @Marc van Leeuwen You mean, by convention, it was decided that positive real numbers raised to real powers "should not" give negative numbers. – Truth-seek Apr 19 '17 at 08:25
  • @Marc thank you for your comment and clarification. I would like to know some bibliography, if it would be possible, that show what you says about this topic, by now I only see the same version (or similar) about exponentiation of complex numbers. – Masacroso Apr 19 '17 at 09:39
  • @MathEnthusiast Since the exponent $\frac12$ was apparently confusing you, I've changed my example to avoid a rational exponent. – Marc van Leeuwen Apr 19 '17 at 10:30
  • Note that I focussed on the failure of $(\exp y)^z=\exp(yz)$ because it is about complex exponential function which (contrary to the complex logarithm) is a perfectly well defined (and need I add single-valued?) function. Moreover the example only involves a power with a positive real base, which case is single-valued, so we get unique but different values on both sides. Even if one goes for a multi-valued interpretation of $(\exp y)^z$, a multi-valued LHS should not equal a single-valued RHS. However you turn it the rule is just not valid when $y\notin\Bbb R$. – Marc van Leeuwen Apr 19 '17 at 10:30
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    Would you give a few examples of inconsistencies that arise from defining complex power and exponential functions? – Daniel Fischer Apr 19 '17 at 11:23
  • @DanielFischer I should reduce my claim a bit: the mere definition of something without claiming the result satisfies any property does not give inconsistencies (unless it contradicts itself, which $a^b=\exp(b\ln a)$ with principal branch of $\ln$ does not). However powers of complex (or negative) numbers easily lead to paradoxical situations, as in this question I already linked to, or this one, due to application "rules" suggested by the exponential notation. – Marc van Leeuwen Apr 19 '17 at 11:54
  • Yes, there's a great potential for wreaking havoc due to applying "rules" that don't hold. – Daniel Fischer Apr 19 '17 at 12:03
  • ... What is particular about using $a^b=\exp(b\ln a)$ to define complex exponentiation is this. Often one extends a definition guided by the desire that some existing identity should continue to hold in a wider context. But here that identity is based on a rule (for exponentials) that, even though it was used to inspire the definition, at the end of the day fails to hold with that very definition. This is what I was attempting to explain in my answer. – Marc van Leeuwen Apr 19 '17 at 12:07
  • @DanielFischer Yes rules need to be applied respecting the restrictions that come with them. Unfortunately these restrictions are hardly ever taught explicitly, so people can be forgiven to not be aware of them. See moreover the rational exponents question where a serious text tries to lay out precise rules that apply, and gets it wrong (they lead to contradictions). – Marc van Leeuwen Apr 19 '17 at 12:12
  • @ Marc van Leeuwen I think it is no use teaching restrictions under which the rules apply. Instead there must be some way in which all restrictions to a particular rule may be logically deduced. Unless this is the case, no matter how many restrictions you are aware of, there might be one you don't know and that will lead you to contradiction at some point of time. You can't just keep applying rules, hoping they are right in the domain you are working, or guessing if they are wrong. – Truth-seek Apr 19 '17 at 14:48
  • @MathEnthusiast: Come on, there is no need to be that pessimistic. Rules for rewriting expressions are just theorems, that like any theorem have conditions to be satisfied before they can be applied. For exponentiation, the important things to know are that only for positive real base they are well defined, and that rules valid for real exponents cannot in general be applied for complex exponents, with the exception of the rule $a^{x+y}=a^xa^y$ which remains valid. This is a bit conservative, but sufficient for everyday use. – Marc van Leeuwen Apr 19 '17 at 16:39
  • @DanielFischer please Marc van give us examples!! Where are the inconsistencies??? – Sid Apr 21 '17 at 04:27
  • @Sid: I've given an example in my answer and linked to two other questions that are directly about perceived inconsistencies; on the subject of exponentiation this is actually one of the most recurring kind of questions on this site. – Marc van Leeuwen Apr 21 '17 at 16:56
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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.

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If you would like, there is also another solution to this problem. Since we cannot count "i" times, the solution is to find a formula for doing exponents, namely, Euler's formula. When you plug in $e^{xi}$, the output will be $\cos(x)+i\sin(x)$ (in radians), also notated as the $\operatorname{cis}(x)$. Now, we can define $i$ as $e^{π/2i}$. So therefore our goal is $(e^{π/2i})^i$. Luckily, with the way that exponents work, we can simplify this to $e^{π/2i\cdot i}$, which of course, by definition, is simplified to $e^{-π/2}$.

Xander Henderson
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  • $(a^b)^c$ might not equal $a^{bc}$ in the case that we are working with complex numbers. This identity you are using is only guaranteed for positive numbers. – JMoravitz Jan 21 '21 at 20:38