Question

derive tan(x+jy)

Answer 1

*tan(x+jy)

Answer 2

@UnkleRhaukus

Answer 3

\[f(x) = \tan(x+jy)\]let \(z = x+jy\)
\[f(z) = \tan(z)\]

\[\frac{\mathrm d }{\mathrm dx}f(z) = \sec^2(z)\frac{\mathrm d z}{\mathrm dx}\]

Answer 4

sir did you differentiate? :)

Answer 5

i took the derivative with respect to z, ie d/dz tan(z) = sec^2(z),

but to get it in terms of x you have to find dz/dx,
because d/dx tan(z) = sec^2(z) * dz/dx

Answer 6

http://openstudy.com/users/karlakalurky#/updates/5492a868e4b080337c024e12

Answer 7

they taught me the sin(a+b) formula?

Answer 8

The sum/difference formula for tangent is:
\[\tan(a\pm b)=\frac{\tan(a)\pm\tan(b)}{1\mp\tan(a)\tan(b)}\]

but i don't like fractions

Answer 9

from your previous question, i think j is imaginary unit and they might want u put tan(x+jy) in standar complex number form : a+jb

Answer 10

rectangular form.. right :)

Answer 11

am i supposed to do like
sin(x+jy)/cos(x+jy) ???

Answer 12

are our answers for the previous question CORRECT ? :)

Answer 13

\[\cos(x+jy) = \frac{ e ^{j(x+jy)}+e ^{-j(x+jy)} }{ 2 }\]
\[\sin(x+jy)= \frac{ e ^{j(x+jy)}-e ^{-j(x+jy)} }{ 2j }\]
??

Answer 14

they look good!
plug them in and simplify

Answer 15

\[\large \begin{align} \dfrac{~~~~~\frac{ e ^{j(x+jy)}-e ^{-j(x+jy)} }{ 2j }}{~~~~ \frac{ e ^{j(x+jy)}+e ^{-j(x+jy)} }{ 2 }}\end{align}\]

Answer 16

\[\frac{ e ^{j(x+jy)}- e ^{-j(x+jy)} }{ j(e ^{j(x+jy)}+e ^{-j(x+jy)}) }\]

Answer 17

so far so good

Answer 18

is that it? or not

Answer 19

you need to separate real and imaginary parts

Answer 20

final expression should be in below form\[\large \clubsuit + i\spadesuit \]

Answer 21

\[\frac{ e ^{j(x+jy)} }{ je ^{j(x+jy)}+je ^{-j(x+jy)}} -\frac{ e ^{-j(x+jy)} }{ je ^{j(x+jy)}+je ^{-j(x+jy)} }\]
??

Answer 22

wolfram says tan(x+iy) equals below expression
http://gyazo.com/680303c3eb010ace26b61e4dd8cd58f9

Answer 23

i can't see the image.

Answer 24

you have this : \[\large \frac{ e ^{j(x+jy)}- e ^{-j(x+jy)} }{ j(e ^{j(x+jy)}+e ^{-j(x+jy)}) } \]

and you know that j*j = -1, lets see if we can simplify :
\[\large \frac{ e ^{jx-y}- e ^{-jx+y} }{ j(e ^{jx-y}+e ^{-jx+y}) }\]

Answer 25

:)

Answer 26

what can we do next hmm

Answer 27

http://math.stackexchange.com/questions/860947/tan-x-i-y-a-ib-then-tan-x-iy-a-ib
see first reply ?

Answer 28

??

Answer 29

one sec let me grab pen and paper

Answer 30

:)

Answer 31

next divide \(2j\) top and bottom and replace exponentials with hyperbolic trig functions

Answer 32

divide 2j?

Answer 33

one sec there is a mistake in sign

Answer 34

\[\begin{align} \frac{ e ^{jx-y}- e ^{-jx+y} }{ j(e ^{jx-y}+e ^{-jx+y}) } & = \dfrac{e^{-y}[\cos x + j\sin x] - e^{y}[\cos x - j\sin x]}{j(e^{-y}[\cos x + j\sin x] + e^{y}[\cos x - j\sin x])}\\~\\
&= \frac{\cos x[e^{-y}-e^y] +j\sin x[e^{-y}+e^{y}]}{j\cos x[e^{-y}+e^y] -\sin x[e^{-y}-e^{y}]}\\~\\
&= \frac{\cos x[\frac{e^{-y}-e^y}{2j}] +\sin x[\frac{e^{-y}+e^{y}}{2}]}{\cos x[\frac{e^{-y}+e^y}{2}] -\sin x[\frac{e^{-y}+e^{y}}{2j}]}\\~\\
& = \frac{\sin x \cosh y + j \cos x \sinh y}{ \cos x \cosh y-i\sin x\sinh y}
\end{align}\]

Answer 35

see if that makes sense more or less ^

Answer 36

why do we need to divide them by 2j?

Answer 37

thanks for the effort! I truly appreciate it! :DDD

Answer 38

@ganeshie8

Answer 39

because we want to use hyperbolic trig functions :
\[\frac{e^{-y}+e^y}{2} = \cosh y\]

\[\frac{e^{-y}-e^y}{2j} = j\sinh y\]

Answer 40

oh thank you!!! :DD

Answer 41

wait... how bout for tan inverse?

Answer 42

@hartnn