Question

derive arctan(x+jy)

Answer 1

@ganeshie back to the previous question: derive tan(x+jy)
i have a question
you said :
\[\cosh y = \frac{ e ^{-y}+e ^{y} }{ 2 }\]
\[j \sinh y=\frac{ e^{-y}-e ^{y} }{ 2j }\]

in here it says quite different
http://www.suitcaseofdreams.net/Hyperbolic_Functions.htm

or is it the same?

Answer 2

:)

Answer 3

@ganeshie8

Answer 4

multiply \(j\) both sides

Answer 5

im not so sure about arctan(x+jy) formula derivation
@Kainui

Answer 6

disregard this ques. about arctan :)
my question is
is
\[\sinh y= \frac{ e ^{-y} - e ^{y} }{ 2 }\]
and
\[\sinh y = \frac{ e ^{y}-e ^{-y} }{ 2 }\]
the same?

Answer 7

@ganeshie8

Answer 8

Not quite, here, I fixed it
\[ -\sinh y= \frac{ e ^{-y} - e ^{y} }{ 2 }\]
\[\ \sinh y = \frac{ e ^{y}-e ^{-y} }{ 2 }\]
It is worth noting that sinh is an odd function and cosh is an even function, so by that, cosh(-y)=cosh(y) and sinh(-y)=-sinh(y). Really convenient since sine and cosine on the circle have this same symmetry!

Answer 9

whew! THANK YOU! :DDD

Answer 10

Yeah if you have any more questions about it just ask! It's also nice to note the following\[\Large \cosh^2x-\sinh^2x=1 \\ \Large e^x = \cosh x+\sinh x\]

You can prove these to be true yourself if you just plug in their definitions!

Answer 11

I will! stay tuned! kidding :D thanks for your response. it's much appreciated. :)))