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How can the tangent addition theorem for complex numbers

$$\tan(z+w) = \frac{\tan z + \tan w }{1 - \tan z \tan w}$$

be proved?

For real numbers, the wikipedia page says one can use Euler's formula

$$e^{ix} = \cos x + i\sin x.$$

But I dont see if and how this helps assuming complex numbers too.

So I'd be happy if somebody could give me a hint to get started.

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  • $\begingroup$ This is some kind of a first test questions since I'm not sure if I should dare asking here ... So please just tell me it this question is too stupid, simple, or otherwise inappropriate without shooting me and I'll remove it ;-) $\endgroup$
    – Dilaton
    Sep 1, 2012 at 21:44
  • $\begingroup$ It's a perfectly OK question. $\endgroup$ Sep 1, 2012 at 21:44

3 Answers 3

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the proof is the same as for real numbers. You do need to convince yourself of the two addition formulas over $\mathbb C,$ namely $$ \sin(z + w) = \sin z \; \cos w + \cos z \; \sin w, $$ with $$ \cos(z+w) = \cos z \; \cos w - \sin z \; \sin w. $$ These follow from identities such as $$ \cos z = \frac{e^{iz} +e^{-iz} }{2} $$ and $$ \sin z = \frac{e^{iz} -e^{-iz} }{2i}. $$ Then, same as for the reals, write out the fraction $$ \tan(z+w) = \frac{\sin(z+w)}{\cos(z+w)} $$ and divide numerator and denominator by $\cos z \; \cos w.$

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One way to do it would be to extend the identity from the real case by analyticity. Let

$$ f(z,w) = \tan(z+w) - \frac{\tan z+\tan w}{1-\tan z\tan w}$$

I'll take your word that this is identically zero on $\mathbb R\times\mathbb R$. Temporarily fix $z_0\in\mathbb R$ and consider the function $w\mapsto f(z_0,w)$. This is a meromorphic function of $w$, and since it is identically zero for $w\in \mathbb R$, it is actually zero for all $w$.

So $f$ is zero on $\mathbb R\times \mathbb C$. Now fix $w_0\in\mathbb C$ and consider $z\mapsto f(z,w_0)$ . . .

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  • $\begingroup$ This is elegant and cool, I would not have thought about it this way myself $\endgroup$
    – Dilaton
    Sep 1, 2012 at 22:04
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Hint: $\tan(z+w) = \frac{\sin(z+w)}{\cos(z+w)} = {\frac {-i \left( {{\rm e}^{i \left( z+w \right) }}-{{\rm e}^{-i \left( z+w \right) }} \right) }{{{\rm e}^{i \left( z+w \right) }}+{ {\rm e}^{-i \left( z+w \right) }}}} $ and assume $z=x+iy$ and $w=u+iv$, and work out your proof. It is a good idea to work the left hand side, then the right hand side, and then compare the results. I think it is easier to work with the exponential function than working with sine and cosine functions. Otherwise, you need to use some identities like $\sin(A+B)$ and $\cos(A+B)$, since you have to work out like this function $ \sin((x+u)+i(y+w) ) $. Here is what you should get on both sides,

$$ {\frac {4\,\sin \left( x+u \right) \cos \left( x+u \right) {{\rm e}^{2 \,y+2\,v}}-i+i{{\rm e}^{4\,y+4\,v}}}{ \left( 4\, \left( \cos \left( x+ u \right) \right) ^{2}-2 \right) {{\rm e}^{2\,y+2\,v}}+1+{{\rm e}^{4 \,y+4\,v}}}} \,.$$

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