# Hyperbolic sine in Taylor Series

• damo03
The first term in the expansion of sinh(x+jy) will be e^xe^{iy}+ e^xe^{-iy}+ e^{-x}e^{-iy}=0The second term will be e^xe^{-iy}+ e^xe^{-x}=0f

#### 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.

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.

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.

You can go from the left side to the right side by "adding and subtracting" the appropriat things but to see what those things should be, it is simplest to work with the right side.
$$cos(y)sinh(x)+ i cosh(x)sin(y)= \frac{e^{iy}+ e^{-iy}}{2}\frac{e^x- e^{-x}}{2}+ i\frac{e^x+ e^{-x}}{2}\frac{e^{iy}- e^{-iy}}{2i}$$
The "i" in the numerator and the "i" in the denominator in the second term will cancel. (Sorry about writing "i" instead of "j"- I just can't stop myself!)
The denominators of course will be 4. Multiplying out the numerators,
$$e^xe^{iy}+ e^{x}e^{iy}- e^{-x}e^{iy}- e^{-x}e^{-iy}$$
for the first term and
$$e^xe^{iy}- e^xe^{-iy}+ e^{-x}e^{iy}- e^{-x}e^{-iy}$$
for the second term.
Now, observe what terms add and what terms cancel.