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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.

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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

3 Answers 3

up vote 3 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 $1/4.$

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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

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

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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 $e^{\pi/2}$.

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