Question

\[\tan^{-1} z\]
inverses of trig functions

Answer 1

what u need to know about this function?

Answer 2

What is your question?

Answer 3

w(z) = tan−1z
show that it is = -i/2*log(i-z)/(i+z)

Answer 4

@sami-21 @eliassaab

Answer 5

http://www.wolframalpha.com/input/?i=arctan%28a%2Bbi%29

looks like it comes from definition of inverse hyperbolic tangent

Answer 6

ok lets do this
we can write it as
\[z=\tan(w)\]
\[\tan(w)=\frac{\sin(w)}{\cos(w)} \]
and
\[\sin(w)=\frac{1}{i}(e^{iw}-e^{-iw})\]
also
\[\cos(w)=(e^{iw}+e^{-iw})\]
so
it becomes

Answer 7

@SkykhanFalcon are u fallowing the steps?

Answer 8

can you solve it for

??

Answer 9

can u solve this for
\[e^{2iw}\]
?

Answer 10

sorry i wasnt here :/

Answer 11

ok look at the above can you solve this for
\[e^{2iw}\]

Answer 12

okey thanks

Answer 13

what did u get after solving?

Answer 14

just take log of both sides of the above and you will get your result :)

Answer 15

@sami-21 dude i just regonized your solution is full of wrongs

Answer 16

mistakes

Answer 17

@SkykhanFalcon just find one mistake. and let me know

Answer 18

i was asking arctanz

Answer 19

the answer \[(-i/2)*\log(i-z)/(i+z)\]