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This is value of $\tan(z)$ in polar coordinates when I manually calculated it:

this is value of tan(z) in polar coordinates when I manually calculated it

This is the value I got when I wrote the code for it in Maxima:

this is the value I got when I wrote the code for it in maxima

How can I reduce the first form to the second? I can't think of a way to convert the hyperbolic trigonometric function to normal ones. I tried doing this using the normal form of the formula $\tan(x+y)$, and I still don't seem to be gettting this equivalent form. Is there any other way to do it?

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    Double angle identities...? – 2017-02-16
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    Your original expression should have some "$r\cos\theta$"s: $$\frac{\sin\left(r \cos\theta\right) \cosh\left(r\sin\theta\right) + i \sinh\left(r \sin\theta\right)\cos\left(r \cos\theta\right)}{\cos\left(r \cos\theta\right) \cosh\left(r \sin\theta\right) - i \sin\left(r \cos\theta\right) \sinh\left(r \sin\theta\right)}$$ – 2017-02-16
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    *"I can't think of a way to convert the hyperbolic trigonometric function[s] to normal ones."* Conversion uses the definitions in terms of the complex exponential: $$\cosh z := \frac12\left(e^{z} + e^{-z}\right) \qquad \cos z := \frac12\left(e^{iz}+ e^{-iz}\right)$$ $$\sinh z := \frac12\left( e^{z} - e^{-z}\right) \qquad \sin z := \frac{1}{2i}\left( e^{iz} - e^{-iz}\right)$$ From these, we deduce: $$\cosh z = \cos(iz) \qquad \cos z = \cosh(iz)$$ $$\sinh z = - i \sin(i z) \qquad \sin z = - i \sinh(iz)$$ See https://en.wikipedia.org/wiki/Hyperbolic_function . – 2017-02-16
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    I wouldn't try to convert hyperbolic to trig - it doesn't look necessary. Remember that by definition, $\cosh{z} = \frac{1}{2}(e^z + e^{-z})$ and $\sinh(z) = \frac{1}{2}(e^z - e^{-z})$. That gives you all kinds of nice identities - like $\cosh{z}\sinh{z} = \frac{1}{2}\sinh{2z}$. This and similar identities will probably get you the equivalence. – 2017-02-16

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$$\tan(z)=\frac{\sin(x)\cosh(y)+i\cos(x)\sinh(y)}{\cos(x)\cosh(y)-i\sin(x)\sinh(y)}=$$

$\frac{[\sin(x)\cosh(y)+i\cos(x)\sinh(y)][\cos(x)\cosh(y)+i\sin(x)\sinh(y)]}{[\cos(x)\cosh(y)-i\sin(x)\sinh(y)][\cos(x)\cosh(y)+i\sin(x)\sinh(y)]}=$

$\frac{[\sin(x)\cosh(y)\cos(x)\cosh(y)-\cos(x)\sinh(y)\sin(x)\sinh(y)]+i[\cos(x)\sinh(y)\cos(x)\cosh(y)+\sin(x)\cosh(y)\sin(x)\sinh(y)]}{\cos(x)^2\cosh(y)^2+\sin(x)^2\sinh(y)^2}=$

$\frac{\sin(x)\cos(x)(\cosh(y)^2-\sinh(y)^2)+i\sinh(y)\cosh(y)(\cos(x)^2+\sin(x)^2)}{\cos(x)^2\cosh(y)^2+(1-\cos(x)^2)(\cosh(y)^2-1)}=\frac{\sin(x)\cos(x)+i\sinh(y)\cosh(y)}{\cos(x)^2+\cosh(y)^2-1}=$

$\frac{2\sin(x)\cos(x)+2i\sinh(y)\cosh(y)}{2\cos(x)^2+2\cosh(y)^2-2}=\frac{\sin(2x)+i\sinh(2y)}{(2\cos(x)^2-1)+(2\cosh(y)^2-1)}=\displaystyle{\frac{\sin(2x)+i\sinh(2y)}{\cos(2x)+\cosh(2y)}}$