Find the following in x + iy form

[Logan Land]

sinh(ln2+ ipi/3)

so I have a general formula of
sinh z= (e^z - e^-z)/2

so I obtained the following

sinh= (e^(ln2 + ipi/3) - e^-(ln2 + ipi/3))/2

sinh= (e^(ln2) e^(ipi/3) - e^(-ln2) e^(ipi/3))/2

e^(ln2)=2
e^(-ln2)=1/2
e^(ipi/3)= (1/2)+(isqrt3/2)
e^(-ipi/3)= (1/2)-(isqrt3/2)

so when I plug that in I seem to have the following

(2((1/2)+(i(sqrt3)/2)) - (1/2)((1/2)-(i(sqrt3)/2)))/2

which I think is

(1+(2i(sqrt3)/2) - (1/4)+((1/2)i(sqrt3)/2))/2

and from there if that is even correct I don't no where to go from there
I did wolfram and it said that the exact answer was

icos((pi/6)+iln2)

but I don't see how that is x +iy form

 

[Prove It]

Using the identities $\displaystyle \begin{align*} \sinh{ \left( \alpha + \beta \right) } \equiv \sinh{\left( \alpha \right) } \cosh{ \left( \beta \right) } + \cosh{\left( \alpha \right) } \sinh{ \left( \beta \right) } \end{align*}$, $\displaystyle \begin{align*} \cosh{ \left( \mathrm{i}\,\alpha \right) } \equiv \cos{ \left( \alpha \right) } \end{align*}$ and $\displaystyle \begin{align*} \sinh{ \left( \mathrm{i}\,\alpha \right) } \equiv \mathrm{i}\,\sin{ \left( \alpha \right) } \end{align*}$ we have

$\displaystyle \begin{align*} \sinh{ \left[ \ln{(2)} + \mathrm{i}\,\frac{\pi}{3} \right] } &= \sinh{ \left[ \ln{(2)} \right] } \cosh{ \left( \mathrm{i}\,\frac{\pi}{3} \right) } + \cosh{ \left[ \ln{(2)} \right] } \sinh{ \left( \mathrm{i}\,\frac{\pi}{3} \right) } \\ &= \sinh{ \left[ \ln{(2)} \right] } \cos{ \left( \frac{\pi}{3} \right) } + \mathrm{i}\,\cosh{ \left[ \ln{(2)} \right] } \sin{ \left( \frac{\pi}{3} \right) } \\ &= \frac{1}{2}\sinh{ \left[ \ln{(2)} \right] } + \mathrm{i}\,\frac{\sqrt{3}}{2} \, \cosh{ \left[ \ln{(2)} \right] } \\ &= \frac{1}{2} \, \left[ \frac{\mathrm{e}^{\ln{(2)}} - \mathrm{e}^{-\ln{(2)}}}{2} \right] + \mathrm{i}\,\frac{\sqrt{3}}{2}\,\left[ \frac{e^{\ln{(2)}} + \mathrm{e}^{-\ln{(2)}}}{2} \right] \\ &= \frac{1}{2} \left[ \frac{2 - \mathrm{e}^{\ln{\left( 2^{-1} \right)}}}{2} \right] + \mathrm{i}\,\frac{\sqrt{3}}{2} \, \left[ \frac{2 + \mathrm{e}^{\ln{ \left( 2^{-1} \right) }}}{2} \right] \\ &= \frac{1}{2} \, \left( \frac{2 - \frac{1}{2}}{2} \right) + \mathrm{i}\,\frac{\sqrt{3}}{2} \, \left( \frac{2 + \frac{1}{2}}{2} \right) \\ &= \frac{3}{8} + \mathrm{i}\,\frac{5\,\sqrt{3}}{8} \end{align*}$

 

[MarkFL]

Another way to proceed is as follows:

\(\displaystyle \exp(z)=\exp\left(\ln(2)+i\frac{\pi}{3}\right)=1+\sqrt{3}i\)

Hence:

\(\displaystyle \sinh(z)=\frac{1+\sqrt{3}i-\frac{1}{1+\sqrt{3}i}}{2}=\frac{-3+2\sqrt{3}i}{2\left(1+\sqrt{3}i\right)}\cdot\frac{1-\sqrt{3}i}{1-\sqrt{3}i}=\frac{3+5\sqrt{3}i}{8}=\frac{3}{8}+\frac{5\sqrt{3}}{8}i\)