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Why can't Mathematica solve

$\quad\quad\frac{1-i}{\sqrt{2}}=e^{i \alpha } \tan \left(\frac{\beta }{2}\right)$

with the restrictions $\alpha \in [0, 2 \, \pi)$ and $\beta \in [0,\pi]$:

Solve[(1 - I)/Sqrt[2] == E^(I alpha) *Tan[beta/2] &&  0 <= beta <= π && 0 <= alpha < 2 π]

Solve::nsmet: This system cannot be solved with the methods available to Solve.

The solution is $\beta=\frac{\pi}{2}$ and $\alpha=\frac{7 \cdot \pi}{4}$.

If I give Mathematica the value for $\beta$,

Solve[((1 - I)/Sqrt[2] == E^(I alpha) *Tan[beta/2] /. 
  beta -> π/2) && 0 <= beta <= π && 0 <= alpha < 2 π]

I get the solution

(*{{alpha ->  ConditionalExpression[(7 \[Pi])/4, 0 < beta < \[Pi]]}}*)

What is the problem here?

I also tried the other functions like FindRoot, Reduce, etc.

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Closely related Symbolically solve and Plot equation involving Tan –  Artes Aug 18 at 17:14
    
Wow, Thanks! Do you know why Mathematica can solve it with $\cos[\alpha]+I*\sin[\alpha]$ but not with $\exp[I \alpha]$? –  Fabian Aug 18 at 17:37

2 Answers 2

up vote 4 down vote accepted

You may use the following:

Reduce[(1 - I)/Sqrt[2] == ExpToTrig[Exp[I alpha]]*Tan[beta] &&  0 < beta < Pi && 0 < alpha < 2 Pi];

{ToRules[%]} // FullSimplify
{{alpha -> (7 π)/4, beta -> π/4}, {alpha -> (3 π)/4, beta -> (3 π)/4}}

Although I'm not sure why it doesn't work without the ExpToTrig[] thing

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I hope you don't mind my edit. I think it makes the validity of your solution more clear. –  Mr.Wizard Aug 18 at 23:32
    
@Mr.Wizard Of course. Thanks. –  belisarius Aug 22 at 16:11
Solve[{(1-I)/Sqrt[2]==Exp[I alpha] Tan[beta],0<beta<Pi,0<alpha<2Pi}//ComplexExpand,
{alpha,beta},Method->Reduce]
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