# Hyperbolic sine in Taylor Series

damo03
I am reading through a worked example of the Taylor series expansion of Sinh(z) about z=j*Pi

The example states: sinh(j*Pi)=cos(Pi)*Sinh(0) +jcosh(x)sin(y)

I am unsure of this relation. I understand why the x terms are zero but don't know the relation to expand sinh. Can anyone shed some light on this for me?

Cheers.

Homework Helper
I am reading through a worked example of the Taylor series expansion of Sinh(z) about z=j*Pi

The example states: sinh(j*Pi)=cos(Pi)*Sinh(0) +jcosh(x)sin(y)
I recommend you go back and read the example again. What you write has a constant on the left and a function of y on the right. They cannot be equal. In fact it is not too hard to show that $sinh(j\pi)= 0$.

Do you mean $sinh(j(y+ \pi))$?

I am unsure of this relation. I understand why the x terms are zero but don't know the relation to expand sinh. Can anyone shed some light on this for me?

Cheers.

damo03
The example asks: Calculate directly the first two non-zero terms in the Taylor Series expansion of sinh(z) about z=j*$\pi$

The first step showin is to calculate the two non-zero terms in the Taylor series expansion of sinh(z) about z=j*$\pi$

f(z)=sinh(z) , Z0=j*$\pi$

Sinh(j*$\pi$)

sinh(x+jy)=cos(y)*sinh(x)+jcosh(x)sin(y) (in this case x=0, y=$\pi$)

It's the above line that I don't understand (how they expand sinh)...

The worked example then goes on to show that sinh(x+jy)=cos(Pi)*sinh(0)+jcosh(0)sin(Pi)

which does indeed =-1*0 + j*1*0 = 0

but that one line expansion of sinh is where i am lost.

$$cos(y)sinh(x)+ i cosh(x)sin(y)= \frac{e^{iy}+ e^{-iy}}{2}\frac{e^x- e^{-x}}{2}+ i\frac{e^x+ e^{-x}}{2}\frac{e^{iy}- e^{-iy}}{2i}$$
$$e^xe^{iy}+ e^{x}e^{iy}- e^{-x}e^{iy}- e^{-x}e^{-iy}$$
$$e^xe^{iy}- e^xe^{-iy}+ e^{-x}e^{iy}- e^{-x}e^{-iy}$$