Trig functions of complex numbers I was studying complex numbers with the help of Boas textbook. 
    I came about certain problems, which I solved only to find
     that the answers provided in the solution manual to be different. Any help ? 
    express $sin i$ in $x+iy$ form
    since $\sin z = \frac{(e^{iz}) -(e^{-iz})}{2i}$. 
    so on substituting $i$ for $z$ I get 
    $\sin i = \frac{((ie^2)-i)}{2e}$. but the answer shown is $\frac{13}{5}$. Please help. Thank you! Also how do I write equations in a more formatted way here in math.stackexchange?
 A: The text book has an erratum and the solution posted in the question is correct.  We have 
$$\begin{align}
\sin (i) &=\frac{e^{i(i)-}e^{-i(i)}}{2i}\\\\
&=\frac{e^{-1}-e^{1}}{2i}\\\\
&=i\frac{e^{1}-e^{-1}}{2}\\\\
&=i\sinh (1)
\end{align}$$
In fact, we can generalize this to find $\sin (iz)$
$$\begin{align}
\sin (iz) &=\frac{e^{i(iz)-}e^{-i(iz)}}{2i}\\\\
&=\frac{e^{-z}-e^{z}}{2i}\\\\
&=i\frac{e^{z}-e^{-z}}{2}\\\\
&=i\sinh (z)
\end{align}$$
Thus, for all $z$ we have
$$\bbox[5px,border:2px solid #C0A000]{\sin (iz)=i\sinh(z)}$$
A: Express $\sin(i)$ in the form $a+bi$:
We know the following thing:
$$a+bi=|a+bi|e^{\arg(a+bi)i}=|a+bi|(\cos(\arg(a+bi))+\sin(\arg(a+bi))i)$$

$$\sin(i)=$$
$$\sinh(1)i=$$
$$\left|\sinh(1)i\right|e^{\arg\left(\sinh(1)i\right)i}=$$
$$\sqrt{\Re(\sinh(1)i)^2+\Im(\sinh(1)i)^2}e^{\arg\left(\sinh(1)i\right)i}=$$
$$\sqrt{0^2+\sinh(1)^2}e^{\arg\left(\sinh(1)i\right)i}=$$
$$\sqrt{\sinh(1)^2}e^{\arg\left(\sinh(1)i\right)i}=$$
$$\sinh(1)e^{\arg\left(\sinh(1)i\right)i}=$$
$$\sinh(1)e^{\frac{\pi}{2}i}=$$
$$\sinh(1)\left(\cos\left(\frac{\pi}{2}\right)+\sin\left(\frac{\pi}{2}\right)i\right)=$$
$$\sinh(1)\cos\left(\frac{\pi}{2}\right)+\sinh(1)\sin\left(\frac{\pi}{2}\right)i=$$
$$\sinh(1)\cdot 0+\sinh(1)\cdot 1i=$$
$$\sinh(1)\cdot 1i=$$
$$\sinh(1)i$$
So $\sin(i)$ expressed in the form $a+bi$ gives us: $0+\sinh(1)i$
