Loser66
  • Loser66
Check my stuff, please find z : cos z = 3i
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Loser66
  • Loser66
\(cos z = \dfrac{e^{iz}+e^{-iz}}{2}= 3i\) \(e^{iz}+ e^{-iz} -6i =0\) \(e^{2iz}-6ie^{iz} +1=0\) Let t = e^(iz)
Loser66
  • Loser66
\(t^2 -6it +1=0\\t = 3i \pm i\sqrt{10}\)
misty1212
  • misty1212
HI!!

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

misty1212
  • misty1212
this looks good, solving a quadratic i think there is another way too, but this should work
misty1212
  • misty1212
let me try a different way
misty1212
  • misty1212
oh
misty1212
  • misty1212
i still want to try a different way
misty1212
  • misty1212
\[\cos(z)=3i\\ \cos^2(z)=-9\\ 1-\sin^2(z)=-9\\ \sin^2(z)=10\\ \sin(z)=\sqrt{10}\]
misty1212
  • misty1212
ok lets skip this maybe i can't do the other one, not sure
Loser66
  • Loser66
\(t = i(3+\sqrt {10})= e^{iz} \\ iz = log(i(3+\sqrt{10}) = log| (i(3+\sqrt{10})| + i (arg (i(3+\sqrt{10}) +2k\pi ~~~k\in \mathbb Z \)
Loser66
  • Loser66
but \(|i(3+ \sqrt{10}| = 3 + \sqrt{10} \) and \(3+\sqrt{10}>0\) hence \(arg (i(3+\sqrt{10}) = \pi/2\) that gives us \(z = \{(pi/2 + 2k\pi) -i log(3+\sqrt{10}\}\)
Loser66
  • Loser66
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 \(z = \{(pi/2 + 2k\pi) -i log(3+\sqrt{10}\} \cup \{(-\pi/2 + 2k\pi) -i log(3-\sqrt{10}\}\)
Loser66
  • Loser66
@ganeshie8
anonymous
  • anonymous
couldn't you just do \[z=\cos ^{-1}(3i)+2k \pi \]
Loser66
  • Loser66
if it is so, cos^-1 (3i) gives me just one case of the angle of 3i, not the real part
anonymous
  • anonymous
yeah i get you
anonymous
  • anonymous
yeah your first bit looks like your on the right path
anonymous
  • anonymous
brb gimmi 2mins gotta do something
anonymous
  • anonymous
you look like your on the right track
anonymous
  • anonymous
\[z=2k \pi -iln \left[ i(3\pm \sqrt{10}) \right] k \epsilon R \]
anonymous
  • anonymous
mmm
anonymous
  • anonymous
would that suffice?
Loser66
  • Loser66
how?
Loser66
  • Loser66
|dw:1444384172324:dw|
anonymous
  • anonymous
i think you're alright
anonymous
  • anonymous
omg, someone take over, i have dinner calling >.< sorry
Loser66
  • Loser66
I'm ok. Thanks chris00
mom.
  • mom.
can we do something with logarithms? lets take log on both sides we get- \[\log_{e}cosz =\log_{e}3i\] \[\log_{e}3i=\log_{e}(3)+i \tan^{-1} 3=log_{e}cosz\]
Loser66
  • Loser66
We need z =..... , not log (3i) Such a nick. hahaha... whenever I call you, your nick makes me miss my death Mom.

Looking for something else?

Not the answer you are looking for? Search for more explanations.