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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 solve 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]

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

T J. Kim
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Prithvi klatka
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4 Answers4

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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 $1/4.$

Igor Rivin
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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

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

Steve ODonnell
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Okay, the second one took me some time but I solved it, sorry if I do not know how to write math in this program. I will use 2 pieces of data, the first:

$(e^x)^i = cos(x)+i*sin(x)$ [Euler's identity] The second [I don know the name, but is easy to prove]:

if you add the n nth roots of 1 you get 0. $1+(-1) = 0$

$1+e^{\frac{2\pi i}{3}}+e^{\frac{-2\pi i}{3}}=0$

$1+i-1-i=0$

etc...

Now: $0=e^{\pi i \frac{0}{7}}+e^{\pi i \frac{2}{7}}+e^{\pi i \frac{4}{7}}+e^{\pi i \frac{6}{7}}+e^{\pi i \frac{8}{7}}+e^{\pi i \frac{10}{7}}+e^{\pi i \frac{12}{7}}$

(The 7 roots of 1 added up equal 0)

$S=\cos(\frac{2\pi}{7})+\cos(\frac{4\pi}{7})+\cos(\frac{6\pi}{7})=\cos(\frac{8\pi}{7})+\cos(\frac{10\pi}{7})+\cos(\frac{12\pi}{7})$

$S=\frac{1}{2}\big[\cos(\frac{2\pi}{7})+\cos(\frac{4\pi}{7})+\cos(\frac{6\pi}{7})=\cos(\frac{8\pi}{7})+\cos(\frac{10\pi}{7})+\cos(\frac{12\pi}{7})\big]=$

$\frac{1}{2}\big[\cos(\frac{2\pi}{7})+\cos(\frac{4\pi}{7})+\cos(\frac{6\pi}{7})=\cos(\frac{8\pi}{7})+\cos(\frac{10\pi}{7})+\cos(\frac{12\pi}{7})+\cos(0)-cos(0)\big]$

$=\frac{1}{2} Re(e^{\pi i \frac{0}{7}}+e^{\pi i \frac{2}{7}}+e^{\pi i \frac{4}{7}}+e^{\pi i \frac{6}{7}}+e^{\pi i \frac{8}{7}}+e^{\pi i \frac{10}{7}}+e^{\pi i \frac{12}{7}}+e^{\pi i \frac{0}{7}}-e^{\pi i \frac{0}{7}})$=

$Re(0-e^{\frac{0\pi}{7}})$ [sum of roots equals 0]

$S = \frac{Re(-e^0)}{2} = \frac{-1}{2}$

$S^2 = \frac{1}{4}$

So your answer is $\frac{1}{4}$, the problem is hard but solvable

Rhys Hughes
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user560513
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  • Welcome to Math.SE. +1 for the great answer, but please consult this for proper formatting: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference. I'll edit yours now for you. – Rhys Hughes May 09 '18 at 22:00
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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 $e^{\pi/2}$.

Mhenni Benghorbal
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