Complex Anti-derivative of tan(z) 
Show $tan(z)$ has a complex anti-derivative on $S=\mathbb{C}\backslash((-\infty,-\pi/2]\cup[\pi/2,\infty))$
If F(z) is the complex antiderivative of $tan(z)$ on $S$, find $F(i)$, if $F(0)=0$

I know that for $D\subset \mathbb{C}$ open and starshaped, and if $f:D\rightarrow \mathbb{C}$ is holomorphic, then $f$ has an anti-derivative.
First I prove $S$ is star shaped.
Choose center $z_0=(0,0)\in S$ and consider $\{(x,y)\in S:x\geq0,y\geq0\}$. Trivially $\forall z\in(0,\pi /2)$, $[z_0,z]\subset S$.
Now take $z=(x,y)\in S$, such that $x\geq0$ and $y>0$.  The straight line connecting $z_0$ to $z$ is given by $[z_0,z]=\{z_0+t(z-z_0):t\in[0,1]\}$. We want to show that $[z,z_0]$ never intersects $[\pi /2,\infty)$.
Suppose to the contrary, that is suppose $[z_0,z]$ intersects $[\pi /2,\infty)$.
Note that $z_0+t(z-z_0)=(0,0)+t((x,y)-(0,0))=t(x,y)=(tx,ty)$ for $t\in [0,1]$.
Since by assumption, if $[z_0,z]$ intersects $[\pi /2, \infty)$, then there exists $t\in [0,1]$, such that $tx\in[\pi /2, \infty)$ and $ty=0$. But $y>0$ and thus implies $t=0$. But if $t=0$, that implies $tx=0\notin[\pi /2,\infty)$, and thus a contradiction. That is, $[z_0,z]\subset\{(x,y)\in S:x\geq0,y\geq0\}$
Apply this same argument to all four quadrants of $\mathbb{C}$ which thus shows that $S$ is star-shaped.
Now, since $tan(x)=\frac {sin(z)}{cos(z)}$ and since $sin(z)$ and $cos(z)$ are holomorphic on their domains, then $tan(z)$ is holomorphic on its domain. Thus $tan(z)$ has an antiderivative.
I was wondering if my above logic is correct.
Furthermore, how do we go about finding $F(i)$ if $F(0)=0$
I appreciate any help. Thanks in advance!

Answer to second part: the antiderivative vanishing at $0$ is given by the integral $\int_{[0,z]} tan (\zeta)\, d\zeta$. so $F(i)=\int_{[0,i]} tan (\zeta)\, d\zeta =i\int_0^{1} tan (it)\, dt$. Now $\tan (it)=\frac {i\sinh t} {\cosh (t)}$. Make the substitution $u=\cosh t$ to evaluate the integral. 
