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Just a basic question :

If $a=i^i$

Then how would $a$ be expressed in the exponential (Euler form)?

My try: I tried to apply log on both sides but eventually couldn't get to the answer given in my book. Any help would be greatly appreciated.

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$i$ can be written as $\displaystyle e^{i\frac{\pi}{2}}$, since it lies on the positive imaginary axis, with magnitude $1$. So $$i^i = \left(e^{i\frac{\pi}{2}}\right)^i = e^{i^2\frac{\pi}{2}}=e^{-\frac{\pi}{2}}$$


As per Umberto's comment, note that the above is the principal value. Functions of the form $z^w$ are multivalued, since $z^w = e^{w\log z}$ and $\log z$ is multivalued: for $z=re^{i\arg(z)}$ we have $\log z = \log r + i\arg z$ and $\arg z$ is multivalued.

The above calculation is working with the principal value of $i$'s argument: $\pi/2$.

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  • $\begingroup$ You may wish to make a comment about the logarithm. It is equally true that $e^{i \frac \pi 2} = i = e^{i\frac {5\pi}2}$, yet you cannot exponentiate to conclude $$e^{-\frac \pi 2} = i^i = e^{-\frac{5\pi}{2}}.$$ Why is the first exponentiation valid but not the second? $\endgroup$ – Umberto P. Apr 20 '17 at 15:14
  • $\begingroup$ @UmbertoP. Cheers for that, have included a comment about it. $\endgroup$ – Zain Patel Apr 20 '17 at 15:18
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Let assume $i^i=\exp(-\frac{\pi}2)$

Then $(-i)^{(-i)}=(i^3)^{(-i)}=i^{(-3i)}=(i^i)^{(-3)}=\exp(\frac{3\pi}2)$

But it is also $(-i)^{(-i)}=\big(e^{-i\frac{\pi}2}\big)^{-i}=e^{(-i)(-i)\frac\pi2}=\exp(-\frac\pi2)$

And the two values do not agree !


This issue arise for any reciprocal of periodic functions. This is the case for $\ln(z)$ since $e^z$ is $2i\pi$ periodic.

In fact, if we want to keep the morphism relation $\ln(ab)=\ln(a)+\ln(b)$, then we cannot restrict ourselve to the principal value of the logarithm.

Note that generally we define this principal value by having a cut on the semi-axis $\Re(z)<0$, so things get weird when we take values on each side of this line. For instance if we call $\operatorname{Ln}$ this principal determination, one can verify that $\operatorname{Ln}(j^2)\neq2\operatorname{Ln}(j)$.

In our example $i^i=\exp(i\operatorname{Ln}(i))=\exp(i(i\frac\pi2))=\exp(-\frac\pi2)$ we use the principal determination of $\operatorname{Ln}(i)$.

But when considering $(-i)=i^3$ rotating point $(1,0)$ three times by an angle of $\pi/2$ we are crossing the cut, and problems arise.


To avoid these issues, and keep arithmetic relations true (like $(x^a)^b=x^{ab}$ in our case) then one must consider all possible values for the logarithm.

$\ln(z)=\operatorname{Ln}(z)+2ik\pi$ with $k\in\mathbb Z$ but this $\ln(\cdot)$ is not a proper function anymore.

With this definition $i^i=\exp(i\ln(i))=\exp(-\frac\pi2+2k\pi)$

Or if you prefer with set notations : $i^i\in\{\exp(-\frac\pi2+2k\pi)\mid k\in\mathbb Z\}$


When you comply with this methodology, there are no more issues with $(-i)^{(-i)}$.

$(-i)^{(-i)}=\exp((-i)\ln(-i))=\exp((-i)(-i\frac\pi2+2ik\pi))=\exp(-\frac\pi2+2k\pi)=i^i$

which agrees with the two values found above for $k=0$ and $k=1$.


Please note that this issue is not strictly a complex issue. This happens also in the real case as soon as we are facing reciprocal of periodic functions.

Let consider for instance the relation $\arctan(x)+\arctan(y)=\arctan(\frac{x+y}{1-xy})\tag{E}$

If we replace $\arctan$ by it principal determination $\operatorname{Arctan}$ $\in]-\frac\pi2,\frac\pi2[$ then the relation is not always true.

In fact one must consider $\begin{cases} a_n = \operatorname{Arctan}(x) + n\pi\\ b_m = \operatorname{Arctan}(y) + m\pi\\ c_k = \operatorname{Arctan}(\frac{x+y}{1-xy}) + k\pi \end{cases}$

And given this, relation $(E)$ should be interpreted as : $\exists (m,n,k)\in\mathbb Z^3\mid a_n + b_m = c_k$.

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