Question

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

Answer 1

\(cos z = \dfrac{e^{iz}+e^{-iz}}{2}= 3i\)

\(e^{iz}+ e^{-iz} -6i =0\)

\(e^{2iz}-6ie^{iz} +1=0\)
Let t = e^(iz)

Answer 2

\(t^2 -6it +1=0\\t = 3i \pm i\sqrt{10}\)

Answer 3

HI!!

Answer 4

this looks good, solving a quadratic
i think there is another way too, but this should work

Answer 5

let me try a different way

Answer 6

oh

Answer 7

i still want to try a different way

Answer 8

\[\cos(z)=3i\\
\cos^2(z)=-9\\
1-\sin^2(z)=-9\\
\sin^2(z)=10\\
\sin(z)=\sqrt{10}\]

Answer 9

ok lets skip this
maybe i can't do the other one, not sure

Answer 10

\(t = i(3+\sqrt {10})= e^{iz} \\
iz = log(i(3+\sqrt{10}) = log| (i(3+\sqrt{10})| + i (arg (i(3+\sqrt{10}) +2k\pi ~~~k\in \mathbb Z \)

Answer 11

but \(|i(3+ \sqrt{10}| = 3 + \sqrt{10} \) and \(3+\sqrt{10}>0\) hence \(arg (i(3+\sqrt{10}) = \pi/2\)

that gives us \(z = \{(pi/2 + 2k\pi) -i log(3+\sqrt{10}\}\)

Answer 12

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
\(z = \{(pi/2 + 2k\pi) -i log(3+\sqrt{10}\} \cup \{(-\pi/2 + 2k\pi) -i log(3-\sqrt{10}\}\)

Answer 13

@ganeshie8

Answer 14

couldn't you just do

\[z=\cos ^{-1}(3i)+2k \pi \]

Answer 15

if it is so, cos^-1 (3i) gives me just one case of the angle of 3i, not the real part

Answer 16

yeah i get you

Answer 17

yeah your first bit looks like your on the right path

Answer 18

brb gimmi 2mins gotta do something

Answer 19

you look like your on the right track

Answer 20

\[z=2k \pi -iln \left[ i(3\pm \sqrt{10}) \right] k \epsilon R \]

Answer 21

mmm

Answer 22

would that suffice?

Answer 23

how?

Answer 24

i think you're alright