Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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 ;-) –  Dilaton Sep 1 '12 at 21:44
    
It's a perfectly OK question. –  Raskolnikov Sep 1 '12 at 21:44
add comment

3 Answers

up vote 7 down vote accepted

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 $\cos z \; \cos w.$

share|improve this answer
add comment

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 $z\mapsto f(z,w_0)$ . . .

share|improve this answer
    
This is elegant and cool, I would not have thought about it this way myself –  Dilaton Sep 1 '12 at 22:04
add comment

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,

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.