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Loser66
 one year ago
Check my stuff, please
find z : cos z = 3i
Loser66
 one year ago
Check my stuff, please find z : cos z = 3i

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Loser66
 one year ago
Best ResponseYou've already chosen the best response.0\(cos z = \dfrac{e^{iz}+e^{iz}}{2}= 3i\) \(e^{iz}+ e^{iz} 6i =0\) \(e^{2iz}6ie^{iz} +1=0\) Let t = e^(iz)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0\(t^2 6it +1=0\\t = 3i \pm i\sqrt{10}\)

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1this looks good, solving a quadratic i think there is another way too, but this should work

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1let me try a different way

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1i still want to try a different way

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1\[\cos(z)=3i\\ \cos^2(z)=9\\ 1\sin^2(z)=9\\ \sin^2(z)=10\\ \sin(z)=\sqrt{10}\]

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1ok lets skip this maybe i can't do the other one, not sure

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0\(t = i(3+\sqrt {10})= e^{iz} \\ iz = log(i(3+\sqrt{10}) = log (i(3+\sqrt{10}) + i (arg (i(3+\sqrt{10}) +2k\pi ~~~k\in \mathbb Z \)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0but \(i(3+ \sqrt{10} = 3 + \sqrt{10} \) and \(3+\sqrt{10}>0\) hence \(arg (i(3+\sqrt{10}) = \pi/2\) that gives us \(z = \{(pi/2 + 2k\pi) i log(3+\sqrt{10}\}\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0For \(t = i(3\sqrt{10})\). since \(3\sqrt{10} <0\) its argument is \pi/2 the same process but replace the arg, we have \(z = \{(pi/2 + 2k\pi) i log(3+\sqrt{10}\} \cup \{(\pi/2 + 2k\pi) i log(3\sqrt{10}\}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0couldn't you just do \[z=\cos ^{1}(3i)+2k \pi \]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0if it is so, cos^1 (3i) gives me just one case of the angle of 3i, not the real part

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah your first bit looks like your on the right path

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0brb gimmi 2mins gotta do something

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you look like your on the right track

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[z=2k \pi iln \left[ i(3\pm \sqrt{10}) \right] k \epsilon R \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think you're alright

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0omg, someone take over, i have dinner calling >.< sorry

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I'm ok. Thanks chris00

mom.
 one year ago
Best ResponseYou've already chosen the best response.0can we do something with logarithms? lets take log on both sides we get \[\log_{e}cosz =\log_{e}3i\] \[\log_{e}3i=\log_{e}(3)+i \tan^{1} 3=log_{e}cosz\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0We 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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