Express $\sinh(z)$ in $x+yi$ form

Question

How can I express $\sinh(z)$ with $z=i+1$ in $x+yi$ form? I know $\sinh(z)$ is equal to $\frac{e^z-e^{-z}}{2}$ ,the formula for the polar form and Euler's formula but I couldn't get to the answer using only these.

Answer 1

1) employ $\sinh(z)=(e^z+e^{-z})/2$

2) use $e^z=e^{i+1}=e^1e^{1i}=e(\cos(1)+i\sin(1))$ via the Euler formula

3) evaluate $e^{-i-1}$ analogously

4) collect terms

Answer 2

For $x,\,y\in\Bbb R$ we seek $u,\,v\in\Bbb R$ such that $u+vi=\sinh z,\,z:=x+yi$ (if you'll pardon my slight change to requested notation).

Option 1:

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

Option 2:

$$\sinh(x+yi)=\sinh x\cosh yi+\cosh x\sinh yi=\sinh x\cos y+i\cosh x\sin y.$$

Answer 3

Note,

$$\sinh(1+i)= \frac12(e^{1+i}-e^{-1-i}) =\frac12 \left[ e(\cos1+i\sin1)-e^{-1}(\cos1-i\sin1)\right]$$ $$=\frac12 (e-e^{-1})\cos1+\frac i2(e+e^{-1})\sin1=\sinh1\cos1+i\cosh1\sin1$$