What is the Complex Tangent Formula Proof for Homework?

AI Thread Summary
The discussion focuses on deriving the complex tangent formula, specifically tan(z) = (tan(a) + i tanh(b)) / (1 - i tan(a) tan(b)). The user initially struggles with the proof, attempting to use the definitions of complex sine and cosine, as well as hyperbolic functions. After several attempts, the user realizes that a misprint in their textbook was causing confusion, as the correct formula is tan(z) = (tan(a) + i tanh(b)) / (1 - i tan(a) tanh(b)). With this clarification and the application of relevant trigonometric identities, the user successfully completes the proof. The discussion highlights the importance of accurate formulas and the use of foundational equations in mathematical proofs.
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Homework Statement



This is an easy one, but keep in mind I'm kind of a newbie, anyway I can't figure out how to get the next formula...
tan(z) = (tan(a)+i tanh(b))/(1 - i tan(a)tan(b))

Homework Equations



This is the third part of an excercise, previous I proof the follow, -all using the definitions, of complex sin,cos, and tan, and definitions of real sinh,cosh, and tanh-...

cos(z)=cos(a)cosh(b)-i sin(a)sinh(b);
sin(z)=sin(a)cosh(b)+icos(a)sinh(b)

The Attempt at a Solution



I tried a lot of things, but couldn't get any way, maybe I'm missing something important, and that's what I fear.
I tried to replace tan(z) = sin(z)/cos(z) with the other equations but, I'm getting nothing.

Sorry for my english, lot of thanks for the help!
 
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\tan(z)=\frac{\sin(z)}{\cos(z)}=\frac{\frac{e^{z}-e^{-z}}{2i}}{\frac{e^{z}+e^{-z}}{2}}=\frac{1}{i}\frac{e^{z}-e^{-z}}{e^{z}+e^{-z}}=\frac{1}{i}\frac{e^{a+ib}-e^{-(a+ib)}}{e^{a+ib}+e^{-(a+ib)}}
Now write out the expression for e^{a+ib}...
 
Svein said:
\tan(z)=\frac{\sin(z)}{\cos(z)}=\frac{\frac{e^{z}-e^{-z}}{2i}}{\frac{e^{z}+e^{-z}}{2}}=\frac{1}{i}\frac{e^{z}-e^{-z}}{e^{z}+e^{-z}}=\frac{1}{i}\frac{e^{a+ib}-e^{-(a+ib)}}{e^{a+ib}+e^{-(a+ib)}}
Now write out the expression for e^{a+ib}...
Thanks for that!, but, I will have to ask for even more help. Because, I already knew that formula, and I can't find how to pass to the one I'm asking for...

tan(z) = (tan(a)+i tanh(b))/(1 - i tan(a)tan(b))
 
I've solved it! and I wanted to tell you. The biggest problem was that the damn book have a cute problem tan(z)=(tan(a)+itanh(b))/(1−itan(a)tan(b)) is really tan(z)=(tan(a)+itanh(b))/(1−itan(a)tanh(b)) having that in mind and with some equations it's really simple to get. I mean, using cosh2(x)- sinh2(x)=1,cos2(x)+sin2(x)=1, and tan(x)=sin(x)/cos(x), ...and the most important, after the two other equations I posted at first... (and the simplest one) sin(x)=cos(x)tan(x)
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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