According to my teacher, if Z = X+iY and:
$$\tan(z) = \frac{\sin(z)}{\cos(z)}$$
Then $\tan(z)$ will be:
$$\tan(z) = \frac{\sin(2x)+i\sinh(2y)}{\cos(2x)+\cosh(2y)}$$
But I'm stuck expanding the $\sin/\cos$ quotient, specifically in this part:
$$\frac{\sin (z)}{\cos(z)} = \frac{\sin(x)*\cosh(y)+\cos(x)*i\sinh(y)}{\cos(x)*\cosh(y)-\sin(x)*i\sinh(y)}$$
I'd be very grateful for any help on how to solve this, thanks.
On
Remember that $\displaystyle\sin z= \frac{e^{iz}-e^{-iz}}{2i},\ \cos z= \frac{e^{iz}+e^{-iz}}{2}$. Therefore $$\sin(iz)=\frac{e^{-z}-e^{z}}{2i}=i\sinh(z),\quad \cos(iz)=\frac{e^{-z}+e^{z}}{2}=\cosh(z).$$ Then, by letting $z=x+iy$, and by using addition formulas, you obtain $$\sin(z)=\sin x\cos (iy)+\cos x\sin(iy)=\sin x\cosh y+i\cos x\sinh y$$ and $$\cos(z)=\cos x\cos (iy)-\sin x\sin(iy)=\cos x\cosh y-i\sin x\sinh y.$$ If follows that your formula for $\tan z$ is correct. However if you want to separate its real and imaginary parts, you need a few more steps: $$\tan z=\frac{\sin z}{\cos z}=\frac{\sin z\cdot \overline{\cos z}}{|\cos z|^2}=\frac{\sin x \cos x +i \sinh y \cosh y}{\cosh^2 y - \sin^2 x}.$$
In your final answer, the denominator is real. In your expansion of $\sin(z)/\cos(z)$, this is not the case. To eliminate the imaginary component in the denominator, you can multiply both the numerator and denominator by the complex conjugate of the denominator.
When you multiply this complex conjugate, you will end up with terms like $\cos^2(x)$, $\cosh^2(x)$, $\sin^2(x)$, and $\sinh^2(x)$. You can express all of these in terms of $\cos(2x)$ and $\cosh(2x)$ using double angle formulas. (You can find them with a bit of Googling.)
I believe that similar efforts to the numerator (using $\sin(2x)$ and $\sinh(2x)$ formulas) will give you the answer you seek. The identity $\cosh^2(x) - \sinh^2(x) = 1$ might also help.