Question

i have a question please someone help me! :)

Answer 1

If \[\sinh (z) = \frac{ e ^{z}-e ^{-z} }{ 2 }\]
then why in http://gyazo.com/1c7c5772dfb16f29a1ae9c062fa5df2e
\[\frac{ e ^{-z}-e ^{y} }{ 2i }= \sinh (z) ?????\]
im so confused!! :((
@ganeshie8

Answer 2

so you want to prove \(\large \frac{ e ^{-z}-e ^{y} }{ 2i }\) equals \(\sinh (z)\)

Answer 3

yes? :)

Answer 4

well you can't prove it because it is worng

Answer 5

below is true however \[ \large \frac{ e ^{-y}-e ^{y} }{ 2i } = i\sinh(y) \]

Answer 6

ohh
but sinh (z) = \[\frac{ e ^{z}-e ^{-z} }{ 2 }\]
?

Answer 7

yes thats right

Answer 8

yes and if z=x+iy you can have another formula

Answer 9

Take left hand side\[ \large \frac{ e ^{-y}-e ^{y} }{ 2i } \]

multiiply top band bottom by \(i\), what do you get ?

Answer 10

eh you are thinking about something different @ganeshie8
what would sinh(x+iy)=?

Answer 11

i don't get it.. multiply what? :/

Answer 12

im guessing the op was asking about the last line attachment..

Answer 13

multiply the imaginary unit \(i\)

Answer 14

oh i see

Answer 15

ahhh i get it already!!!! :DDD THANKS!!!

Answer 16

Take left hand side\[ \large \frac{ e ^{-y}-e ^{y} }{ 2i } \]

multiiply top band bottom by \(i\), you get :
\[ \large \frac{ e ^{-y}-e ^{y} }{ 2i } \times \frac{i}{i} \]

Answer 17

and you know that \(\large i^2 = -1\) so that equals

\[ \large i \frac{ e ^{-y}-e ^{y} }{ 2i^2} \]
\[ \large i \frac{ e ^{-y}-e ^{y} }{ -2} \]
\[ \large i \frac{ e ^{y}-e ^{-y} }{ -2} \]
\[\large i\sinh(y)\]

Answer 18

:))) thanks!