# Taylor expansion for complex number:

Determine the Taylor expansion around h = 0 for $\cosh(i\frac{\pi}{2}+ h)$

With regards to the question above:

I calculate that the expansion is as follows: $i(x + \frac{x^3}{3!} + \frac{x^5}{5!}\ldots)$

But I am not sure that can be correct as am confused as to why the expansion is completely imaginary.

Surely we can write $\cosh(i\frac{\pi}{2})$ as the real number $\cos(\frac{\pi}{2})$ which has no imaginary part so i am confused why the expansion is imaginary...

• Your expansion is not correct. Check it over. – Lubin May 4 '17 at 18:32
• Use expansion of $e^x$ in $\cosh x=\frac12(e^{x}+e^{-x})$. – Nosrati May 4 '17 at 18:38
• But if you differentiate the function and plug into it zero we notice a pattern of 0 i 0 i 0 i 0 i and then you plug them into the general expression of the macluarin expansion you get what i get. Where did i go wrong? – David Abraham May 4 '17 at 18:40
• differentiate $\cosh(i\frac{\pi}{2})$.? It's a number. – Nosrati May 4 '17 at 18:45
• no cosh(iπ/2 + h) – David Abraham May 4 '17 at 18:49

## 1 Answer

Your answer is totally correct ...

$$\begin{eqnarray*} \cosh(i\frac\pi 2+x)&=&\cosh(i\frac\pi 2)\cosh(x)+\sinh(i\frac\pi 2)\sinh(x) \\&=& \cos(\frac\pi 2)\cosh(x)+i\sin(\frac\pi 2)sinh(x) \\&=&i\sinh(x) \end{eqnarray*}$$ So it is no surprise that your answer is purely imaginary ( because $\cos(\frac\pi 2)=0$)