Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We had the topic of complex numbers for my senior math team meet this week and i wasn't able to get two of the problems.

1.) $z=i^{\displaystyle \left(i^{\displaystyle \left(i^{(2)}\right)}\right)}$ and $a$ is the real part of $z$, find the lowest positive value of $\ln(a)$ [ i know it comes to $i-i$ but i don't know why that is e^(pi/2)]

2.) $$\left[\cos \left(\frac{2\pi}{7}\right) + \cos \left(\frac{4\pi}{7}\right) + \cos \left(\frac{8\pi}{7}\right)\right]^2$$ [I think i can use de moivre's forumla but i dont know how here]

Its non calculator and the answers are $\frac{\pi}{2}$ and \frac{1}{4}$ respectively. I just want to know how to solve them, thanks.

share|cite|improve this question
This is so unreadable as to be almost-uneditable. – Igor Rivin Dec 13 '13 at 4:44
It also has little to do with (abstract-algebra); please edit the tags as well. – Jonathan Y. Dec 13 '13 at 4:46
what does $e^{\pi/20}$ have to do with anything? – Igor Rivin Dec 13 '13 at 4:57
e^(pi/2)* i forgot to capitalize the 0 into a ) – Prithvi klatka Dec 13 '13 at 5:03
up vote 4 down vote accepted

For the first, it is equal to $i^{-i}.$ So, the log is equal to $-i(\pi i/2 + 2ki\pi) = \pi/2 +2 k \pi.$

The second, before you square, you have the real part of $x=\omega + \omega^2 + \omega^4,$ where $\omega$ is the primitive seventh root of unity. Notice that the conjugate of this expression is $\omega^6 + \omega^5 + \omega^3 = 1-x.$ Since the real part of $x$ is the same as that of $\overline{x},$ we have that the real part of $x$ is $1/2,$ so its square is $1/4.$

share|cite|improve this answer

Euler's formula says $e^{i\pi} = -1$

Take the square root of both sides

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

Raise both sides to the -i power

${(e^{\frac{i\pi}{2}})}^{-i} = e^{\frac{\pi}{2}} = i^{-i}$

share|cite|improve this answer

Here is an approach

$$ i^{-i}=e^{-i\ln i} = e^{-i \left(\ln |i|+i\left(\frac{\pi}{2}+2k\pi \right)\right)}= e^{ \left(\frac{\pi}{2}+2k\pi \right)}.$$

Now, if you take $k=0$, you get $e^{\pi/2}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.