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Let $b\in\mathbb R.$ Find all the values of $i^{2+ib},$ and show that they are all real.

My attempt:

Let $y = i^{2+ib}.$

$$\ln y = (2+ib)\ln i\\ = (2+ib)\ln e^{i((\frac\pi2)+2n\pi)}\\ = (2i-b)((\frac\pi2)+2n\pi)$$

I am stuck at this step. How do I continue from here?

ryang
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Ming Feng Tay
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  • I mathjax'ed your post, but in the process may have incorrectly edited your working (it was a little hard to read). Click on 'Edit' to change whatever necessary. Welcome to MSE! – ryang Oct 28 '21 at 18:40
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    Do you know how your book defines the quantity $i^{2+ib}$? –  Oct 28 '21 at 18:57

3 Answers3

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Easiest is to forgo any consideration of logarithms.

$\displaystyle i = e^{i\pi/2}.$

Therefore, $~\displaystyle i^{2 + ib} = \left[e^{i\pi/2}\right]^{2 + ib}$.

This equals $~\displaystyle e^{(i\pi/2) \times (2 + ib)}.$

This equals $~\displaystyle e^{[(-b\pi/2) + i(\pi)]}.$

This equals

$$e^{-b\pi/2} \times e^{i(\pi)}. \tag1 $$

Since both factors in (1) above are real numbers, the product is a real number.

Addendum
To a certain extent, I pulled a fast one.

The assertion that $~\displaystyle i = e^{i\pi/2}$ is convention dependent.

More generally, $~\displaystyle i = e^{i[(\pi/2) + 2k\pi]} ~: ~k \in \Bbb{Z}.$

Therefore, $~\displaystyle i^{2 + ib} = \left[e^{i[(\pi/2) + 2k\pi]}\right]^{2 + ib}$.

This equals $~\displaystyle e^{i[(\pi/2) + 2k\pi] \times (2 + ib)}.$

This equals $~\displaystyle e^{(-b)(\pi/2 + 2k\pi) + i(\pi + 4k\pi)]}.$

This equals

$$e^{-b(\pi/2 + 2k\pi)} \times e^{i(\pi + 4k\pi)}. \tag2 $$

Since both factors in (2) above are real numbers, the product is a real number.

user2661923
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  • What would be the magnitude of $i^{2+ib}$? – Asher2211 Oct 28 '21 at 18:24
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    @Asher2211 That is convention dependent. By the typical convention, it would be $e^{-b\pi/2}.$ However, more generally, it could be any expression of form $e^{(-b)(\pi/2 + 2k\pi)} ~: ~k \in \Bbb{Z}.$ – user2661923 Oct 28 '21 at 18:27
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$$\Large i^{2+ib}\\\Large=e^{(2+ib)\log(i)}\\\Large=e^{(2+ib)\;i(\frac\pi2+2k\pi)}\\\Large=e^{i(4k+1)\pi}\:e^{-\frac b2(4k+1)\pi}\\\large\in\mathbb R.$$

ryang
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You can have a look at my general presentation for $z^u$ when both $z$ and $u$ are complex numbers.

https://math.stackexchange.com/a/3729281/399263

So as I indicated express $z=i$ in polar form and write the periodicity explicitly: $$i=\exp\left(i\frac{\pi}2+2ik\pi\right)\text{ for }k\in\mathbb Z$$

Then multiply the inners of the exponential by $u=2+ib$ in cartesian form:

$i^{2+ib}=\exp\left((i\frac{\pi}2+2ik\pi)(2+ib)\right)=\exp\left(-\frac{\pi\,b}2-2k\pi\,b+i(\pi+4k\pi)\right)$

To make the principal value appear, separate what's depends on $k$ and what's not:

$$\begin{cases}z_{[0]}=\exp\left(-\frac{\pi\,b}2+i\pi\right)=-\exp(-\frac{\pi\,b}2)\\\\w^k=\exp\big(-2k\pi\,b+\underbrace{4ik\pi}_{=1}\big)=\exp(-2k\pi\,b)\end{cases}$$

Finally you get the multivalued expression $i^{2+ib}=z_{[0]}w^k\quad k\in\mathbb Z$, where $z_{[0]}$ is called the principal value, and $w$ the multiplicative factor, in this case you notice that the result takes only real values.

zwim
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