Check my stuff, please find z : cos z = 3i

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Check my stuff, please find z : cos z = 3i

Mathematics
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\(cos z = \dfrac{e^{iz}+e^{-iz}}{2}= 3i\) \(e^{iz}+ e^{-iz} -6i =0\) \(e^{2iz}-6ie^{iz} +1=0\) Let t = e^(iz)
\(t^2 -6it +1=0\\t = 3i \pm i\sqrt{10}\)
HI!!

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this looks good, solving a quadratic i think there is another way too, but this should work
let me try a different way
oh
i still want to try a different way
\[\cos(z)=3i\\ \cos^2(z)=-9\\ 1-\sin^2(z)=-9\\ \sin^2(z)=10\\ \sin(z)=\sqrt{10}\]
ok lets skip this maybe i can't do the other one, not sure
\(t = i(3+\sqrt {10})= e^{iz} \\ iz = log(i(3+\sqrt{10}) = log| (i(3+\sqrt{10})| + i (arg (i(3+\sqrt{10}) +2k\pi ~~~k\in \mathbb Z \)
but \(|i(3+ \sqrt{10}| = 3 + \sqrt{10} \) and \(3+\sqrt{10}>0\) hence \(arg (i(3+\sqrt{10}) = \pi/2\) that gives us \(z = \{(pi/2 + 2k\pi) -i log(3+\sqrt{10}\}\)
For \(t = i(3-\sqrt{10})\). since \(3-\sqrt{10} <0\) its argument is -\pi/2 the same process but replace the arg, we have \(z = \{(pi/2 + 2k\pi) -i log(3+\sqrt{10}\} \cup \{(-\pi/2 + 2k\pi) -i log(3-\sqrt{10}\}\)
couldn't you just do \[z=\cos ^{-1}(3i)+2k \pi \]
if it is so, cos^-1 (3i) gives me just one case of the angle of 3i, not the real part
yeah i get you
yeah your first bit looks like your on the right path
brb gimmi 2mins gotta do something
you look like your on the right track
\[z=2k \pi -iln \left[ i(3\pm \sqrt{10}) \right] k \epsilon R \]
mmm
would that suffice?
how?
|dw:1444384172324:dw|
i think you're alright
omg, someone take over, i have dinner calling >.< sorry
I'm ok. Thanks chris00
can we do something with logarithms? lets take log on both sides we get- \[\log_{e}cosz =\log_{e}3i\] \[\log_{e}3i=\log_{e}(3)+i \tan^{-1} 3=log_{e}cosz\]
We need z =..... , not log (3i) Such a nick. hahaha... whenever I call you, your nick makes me miss my death Mom.

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