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How to find all roots if power contains imaginary or irrational power of complex number? How do I find all roots of the following complex numbers? $$(1 + i)^i, (1 + i)^e, (1 + i)^{ i\over e}$$

EDIT:: Find the value of $i^i$ and $i^{1\over e}$ in $[0, 4\pi]$?

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By root, do you mean value? – robjohn Sep 6 '12 at 16:00
Slightly related: – Isaac Sep 6 '12 at 16:47
@robjohn yep ... if it's multivalued in the given interval – hasExams Sep 6 '12 at 18:07
up vote 10 down vote accepted

Since you probably mean the multi-valued complex exponential, you can go straight to the definition:

$$ x^y := \exp(y \log x) $$

The multi-valued complex logarithm can be computed as

$$ \log x = \ln |x| + \mathbf{i} \arg(x) $$

where $\arg(x)$ is the polar angle of $x$. This, of course, takes values in the set

$$ \arg(x) = \mathop{\text{Arg}}(x) + 2 \pi n $$

where $n$ is an integer, and $\mathop{\text{Arg}}$ is the "principal value". For your particular question, all you need to know about the idea of a principal value is that $\mathop{\text{Arg}}(x)$ is a particular choice of polar angle for $x$; here it doesn't matter which one you choose.

To demonstrate that this reproduces what you already know, consider the case of $(-8)^{1/3}$. In this case, we have:

  • $\mathop{\text{Arg}}(x) = \pi$
  • $\arg(x) = \pi + 2 \pi n$
  • $\log(x) = \ln(8) + \mathbf{i} \pi (1 + 2n)$
  • $\exp(y \log x) = \exp((1/3) \ln(8) + \mathbf{i} \pi (1 + 2n)/3) = \exp(\ln(2)) \exp(\mathbf{i} \pi (1 + 2n)/3)$
  • $x^y = 2 (\cos(\frac{1+2n}{3} \pi) + \mathbf{i} \sin(\frac{1 + 2n}{3} \pi))$

and so the values for $(-8)^{1/3}$ are all of the numbers $2 \exp(\mathbf{i} \pi \frac{1 + 2n}{3})$ where $n$ is an integer. $n=1$ gives you the familiar value $-2$. $n=0$ and $n=2$ give the other two complex cube roots of $-8$. $n=4$ also gives $-2$, because $3 \pi \mathbf{i}$ and $\pi \mathbf{i}$ are the same angle. In fact, as $n$ ranges over all integers, only three distinct values are produced. This agrees with the familiar fact that nonzero complex numbers have three cube roots.

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So $i^i$ or $i^{1 \over e}$ does not have more than one root in $[0, 2 \pi]$ ?? – hasExams Sep 1 '12 at 17:53
The last words of this answer, “here it doesn't matter which one you choose”, are misleading or incorrect. The expression $i^c$ has one value if $c$ is an integer, finitely many values if $c$ is a nonintegral rational number, and infinitely many values otherwise. And remember, it’s only an equation or a function that may be said to have roots. – Lubin Sep 1 '12 at 18:16
@Lubin: The formula for the set of values of $\arg(x)$ definitely does not depend on which branch you select to define the principal value. But your comment does suggest another tidbit of information that's worth adding. – Hurkyl Sep 1 '12 at 19:00
@testuser: Were you able to compute the set of values of $i^i$ using this methodology? – Hurkyl Sep 1 '12 at 20:20
@Hurkyl i have known this method for quite a while ... just want to know if there is more than one root for this kind of expression – hasExams Sep 2 '12 at 3:29

So we want a general rule for evaluating formal expressions $z^w$, where $z=r\, e^{i\phi}\ne 0$ and $w=u+iv$ are complex numbers.

Depending on circumstances one wants the full infinite set of values for $z^w$, which arises from the multivaluedness of the logarithm function in the complex domain, or one wants only the so called principal value of $z^w$. Denote by $\dot{\mathbb C}$ the punctured compex plane and by ${\mathbb C}^-$ the complex plane slit along the negative real axis. Then $\arg:\ \dot{\mathbb C}\to{\mathbb R}/(2\pi{\mathbb Z})$ is the multivalued argument function and ${\rm Arg}:\ {\mathbb C}^-\to\ ]{-\pi},\pi[\ $ the principal value of the argument. Correspondingly one has two forms of the logarithm: $$\log z:=\log|z| + i\arg(z)\ ,\qquad{\rm Log}z:=\log|z| + i{\rm Arg}(z)\ .$$

In the real domain the "general power function" $a^b$ is defined for $a>0$ by $a^b:=\exp(b\,\log a)$. Transferring this principle into the complex world one arrives at $$z^w:=\exp(w\ \log z)=\exp\bigl((u+iv)(\log r+i\phi +2k\pi i)\bigr)\qquad (r>0)\ ,$$ resp. at $$z^w:=\exp(w\ {\rm Log} z)=\exp\bigl((u+iv)(\log r + i\phi)\bigr)\qquad(r>0, \ |\phi|<\pi)\ .$$

It seems that the OP is interested in the first of the two interpretations. Therefore let's look at the example $i^i$. As $i=e^{i\pi/2}$ one has $r=1$, $\phi={\pi\over2}$. Therefore we get $$i^i=\exp\bigl( (0+i)(\log 1+i{\pi\over2}+2k\pi i)\bigr)=\exp\bigl(-{\pi\over2}-2k\pi \bigr)\ ,$$ where this is to be interpreted as the resulting set when $k$ runs through ${\mathbb Z}$. We see that this set $\{c_k\}_{k\in{\mathbb Z}}$ consists of positive real numbers. For $k\leq-1$ the $c_k$ lies outside the interval $[0,4\pi]$; but for all $k\geq0$ one has $c_k\in[0,4\pi]$, so that, yes: there are infinitely many values of $i^i$ in the given interval.

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A number $x$ of $\mathbb{C} $ is a primitive $n^{th}$ root if $x^n - 1 = 0$, but $x^j -1 \not= 0$ for any $0 \leq j < n$. If we for each natural number $n$ finds the primitive roots of $x^n -1 = 0$ and collects them for all such $n$, then we find all roots once. In number, the primitive $n^{th}$ roots are $\varphi (n)$ where the function is Eulers phi function, counting how many numbers in the interval $[0, n)$ being relative prime to $n$, where $j$ is relative prime to $n$ if $gcd (n,j)=1$ (greatest common divisor).For such $j$ we have a root \begin{equation}\nonumber e^{i\frac{ 2 \pi }{n }j}. \end{equation} If $z$ is a complex number, $z$ can be written $z = |z| e^{2 \pi i( \gamma+ m )} = (|z|^{1/n} e^{2 \pi i (\gamma+ m ) /n})^n$. Therefore an equation $x^n - z = 0$ has an expression \begin{equation}\nonumber \Big(\frac{x}{|z|^{1/n} e^{2 \pi i (\gamma + m)/n}} \Big)^n - 1 = 0 \end{equation} From this, the $n^{\text{th }}$ roots are \begin{equation}\nonumber x =|z|^{1/n}e^{i \frac{2 \pi }{n }\big(\gamma + j + m\big) } \end{equation} If $|z|\not =1$, roots belonging to different $n$ will be different. If $|z|=1$, $m+j$ must be prime to $n$. In that case, we get the contribution from $n$ to be the roots \begin{equation}\nonumber x =|z|^{1/n}e^{i \frac{2 \pi }{n }\big(\gamma + j + m\big) } \qquad \text{where } \gcd(n, j+m)=1. \end{equation} or, what is the same, \begin{equation}\nonumber x =|z|^{1/n}e^{i \frac{2 \pi }{n }\big(\gamma +k\big) } \qquad \text{where } \gcd(n, k)=1, \qquad 1 \leq k < n. \end{equation} For example, when $z$ equals $i^i$, we have a real number, so take $\gamma = 0$ and the real part is $\exp (- \pi /2)$. This should give the $n^{th}$ roots \begin{equation}\nonumber x = e^{-\frac{\pi }{2 n } +i \frac{2 \pi }{n } k} . \qquad \text{where } 1 \leq k < n. \end{equation} When $z$ equals $i^{1/e}$, rewritten ($(e^{i \pi /2})^{1/e})=e^{i \pi /(2 e)}$, $|z|=1$ and $\gamma $ from above is $\pi /(2 e)+ 2 \pi m $. Then the roots which are $n^{th}$ roots but not $k^{th}$ roots for $k < n$ are \begin{equation}\nonumber x = e^{i \frac{2 \pi }{n }\big(\frac{1}{4 e } + k\big) } \qquad \text{where } \gcd(n,k) = 1. \end{equation} Collecting together for all natural numbers $n$ should give all possible roots in the respective cases.

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Let $e^z=N≠0$ ,So, $N=e^z=e^z\cdot e^{2n\pi i}$ as $e^{2n\pi i}=1$ for any integer $n$.

$N=e^{z+2n\pi i}\implies Log N=z+2n\pi i$

If $n=0$ ,we get the principal value $z=logN$, so $LogN=logN+2n\pi i$

Let $A=(a+ib)^{x+iy}$

$Log A=(x+iy)Log(a+ib)$

Let $a=r\cos\theta$ and $b=r\sin\theta\implies a+ib=r(\cos\theta+i\sin\theta)=re^{i\theta}$

So, $r=\sqrt{a^2+b^2}$ and $\theta=tan^{-1}\frac{b}{a}$

then $Log(a+ib)=\frac{1}{2}log(a^2+b^2)+i(2k\pi+tan^{-1}\frac{b}{a})$ where $k$ is any integer.

So, $Log A=(x+iy)Log(a+ib)=(x+iy)(\frac{1}{2}log(a^2+b^2)+i(2k\pi+tan^{-1}\frac{b}{a}))$

So, $Log A=\frac{1}{2}xlog(a^2+b^2)-y(2k\pi+tan^{-1}\frac{b}{a})+i(\frac{1}{2}ylog(a^2+b^2)+x((2k\pi+tan^{-1}\frac{b}{a}))$

So, $$A=(a+ib)^{x+iy}=e^{\frac{1}{2}xlog(a^2+b^2)-y(2k\pi+tan^{-1}\frac{b}{a})}\cdot e^{i({\frac{1}{2}ylog(a^2+b^2)+x(2k\pi+tan^{-1}\frac{b}{a}})}$$

$$=e^{\frac{1}{2}xlog(a^2+b^2)-y(2k\pi+tan^{-1}\frac{b}{a})} cis(({\frac{1}{2}ylog(a^2+b^2)+x(2k\pi+tan^{-1}\frac{b}{a}})) $$ where $cis(P)=\cos P+i\sin P$

(1)For $(1+i)^i$ $a=b=y=1$ and $x=0$

So, $r^2=a^2+b^2=2$ and $\sin\theta=\cos\theta=\frac{1}{\sqrt2}\implies \theta=\frac{\pi}{4}$

$(1+i)^i=e^{-(2k\pi+\frac{\pi}{4})}(\cos(\frac{1}{2}log2)+i\sin(\frac{1}{2}log2))$ where $k$ is any integer.

(2)For $(1+i)^e, a=b=1$ and $x=e,y=0$

So, $r^2=a^2+b^2=2$ and $\sin\theta=\cos\theta=\frac{1}{\sqrt2}\implies \theta=\frac{\pi}{4}$

$(1+i)^e=e^{\frac{1}{2}elog2}\cdot cis(e(2k\pi+\frac{\pi}{4}))$ where $k$ is any integer.

$(1+i)^e=2^{(\frac{e}{2})}\cdot cis(e(2k\pi+\frac{\pi}{4}))$

(3) For $(1+i)^{\frac{i}{e}}, a=b=1$ and $x=0,y=\frac{1}{e}$

$(1+i)^{\frac{i}{e}}=e^{-\frac{1}{e}(2k\pi+\frac{\pi}{4})}\cdot cis(\frac{1}{2e}log2)$ where $k$ is any integer.

(4) For $i^i,a=0,b=1,x=0,y=1\implies r^2=1$ and $\sin\theta=1$ and $\cos\theta=0\implies \theta=\frac{\pi}{2}$

So, the general value of $i^i=e^{-(2k\pi+\frac{\pi}{2})}$ where $k$ is any integer.

Now by the given condition, $0 ≤e^{-(2k\pi+\frac{\pi}{2})} ≤4\pi$

$0 ≤e^{-(2k\pi+\frac{\pi}{2})}$ is true for all real finite $k$.

So, we need check $ -(2k\pi+\frac{\pi}{2})≤ln_e{4\pi}$ .

or $2k\pi+\frac{\pi}{2}≥-ln_e{4\pi}$

Without using calculator, we can observe that this inequality is satisfied by all integer $k≥0$

(5) For $i^{\frac{1}{e}}, a=0,b=1,x=\frac{1}{e},y=0$ $\implies r^2=1$ and $\sin\theta=1$ and $\cos\theta=0\implies \theta=\frac{\pi}{2}$

$i^{\frac{1}{e}}=e^{\frac{1}{2e}log2}\cdot cis(\frac{1}{e}(2k\pi+\frac{\pi}{2}))$

can be further simplified to $2^{\frac{1}{2e}}\cdot cis(\frac{1}{e}(2k\pi+\frac{\pi}{2}))$ where $k$ is any integer.

It does have an imaginary part, so can not have any pure real value in any specified range.

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I think you mean, "all possible meaningful values" You know that $$1 + i = \sqrt{2}e^{i\pi/4 + 2n\pi} = e^{i\pi(2n + 1/4) +\log(2)/2}.$$ If you exponentiate by $i$, you get $$(1 + i)^i = e^{-\pi(2n + 1/4) + i\log(2)/2} $$ You can deal similarly with the other cases.

Exponentiation to a complex power requires restricting of domain; this all boils down to the problem of dealing with the complex logarithm.

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I just want to know how many roots are there are this kind of expression. Like for these types $ \sqrt{i} $ there is two roots ... and thank you very much for response. – hasExams Sep 1 '12 at 17:37
What is the "exact" definition of "all possible meaningful values"? ;-) In math we like if something is "regular", however in complex analysis the singularity at least as important (residue theorem and its many applications). – vesszabo Sep 7 '12 at 9:20

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