How do I find the value of $\tan\bigl(\frac{\pi}{4}-3i\bigr)$ in terms of elementary functions? The question is to compute $t=\tan \left( {{\pi  \over 4} - 3i} \right)$.
So I change it into $$t={{\sin ({\pi  \over 4} - 3i)} \over {\cos ({\pi  \over 4} - 3i)}}$$
Then $$\eqalign{t=
  & {{{e^{i({\pi  \over 4} - 3i)}} - {e^{ - i({\pi  \over 4} - 3i)}}} \over {2i}} \div {{{e^{i({\pi  \over 4} - 3i)}} + {e^{ - i({\pi  \over 4} - 3i)}}} \over 2}  \cr 
  =& {{{e^{i({\pi  \over 4} - 3i)}} - {e^{ - i({\pi  \over 4} - 3i)}}} \over {2i}} \times {2 \over {{e^{i({\pi  \over 4} - 3i)}} + {e^{ - i({\pi  \over 4} - 3i)}}}}  \cr 
  =& {{{e^{i({\pi  \over 4} - 3i)}} - {e^{ - i({\pi  \over 4} - 3i)}}} \over {{e^{i({\pi  \over 4} - 3i)}} + {e^{ - i({\pi  \over 4} - 3i)}}}} \cr} $$
Are my steps correct and what i should do next?
 A: Yes your steps so far are correct, except that you forgot a factor $i$ in the denominator of the last expression of $t$. To go further, multiply the numerator and denumerator of the fraction $$it=\frac{e^{i(\pi/4 - 3i)} - e^{ - i(\pi/4 - 3i)}}{e^{i(\pi/4 - 3i)} + e^{ - i(\pi/4 - 3i)}}$$ by $e^{i(\pi/4 - 3i)}$. This yields  $$it=\frac{e^{i(\pi/2 - 6i)} - 1}{e^{i(\pi/2 - 6i)} + 1}=\frac{e^{i\pi/2}e^6 - 1}{e^{i\pi/2}e^6 + 1}.$$ One knows that $e^{i\pi/2}=i$ hence $$it=\frac{ie^6 - 1}{ie^6 + 1}=\frac{ie^6 - 1}{ie^6 + 1}\frac{1-ie^6}{1-ie^6 }=\frac{e^{12}-1+2ie^6}{e^{12}+1}.$$ Thus, $$i\tan \left(\pi/4 - 3i\right)=\frac{e^{12}-1}{e^{12}+1}+i\frac{2e^6}{e^{12}+1}=\tanh(6)+i\frac1{\cosh(6)}$$ or, equivalently, $$\tan \left(\pi/4 - 3i\right)=\frac1{\cosh(6)}-i\tanh(6)=\frac{1-i\sinh(6)}{\cosh(6)}.$$
A: Since, in the post, were mentioned elementary functions, what I was trying to suggest in my comments was basically to use the fact that we could develop $$t=\tan(\frac{\pi}{4}-3i)=\frac{\tan(\frac{\pi}{4})-\tan(3i)}{1+\tan(\frac{\pi}{4})\tan(3i)}=\frac{1-\tan(3i)}{1+\tan(3i)}=\frac{1-i\tanh(3)}{1+i\tanh(3)}$$ $$t=\frac{(1-i\tanh(3))^2}{(1+i\tanh(3))(1-i\tanh(3))}=\frac{1-\tanh^2(3)-2i\tanh(3))^2}{1+\tanh^2(3)}$$ $$t=\frac{1-\tanh^2(3)}{1+\tanh^2(3)}-i\frac{2\tanh^2(3)}{1+\tanh^2(3)}=\text{sech}(6)-i \tanh (6)$$ More generally, the same procedure would lead to $$\tan(\frac{\pi}{4}\pm ib)=\text{sech}(2 b) \pm i \tanh (2 b)$$
