Question

Find the value of $\phi$, If $\tan \left( {\theta + i\phi } \right) = {e^{i\alpha }}$,
A. $\dfrac{1}{2}\log \tan \left( {\dfrac{\pi }{4} + \dfrac{\alpha }{2}} \right)$
B. $- \dfrac{1}{2}\log \tan \left( {\dfrac{\pi }{4} + \dfrac{\alpha }{2}} \right)$
C. $\dfrac{1}{2}\log \tan \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)$
D. $- \dfrac{1}{2}\log \tan \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)$

Answer 1

We need to convert the exponential function ${e^{i\alpha }}$ to $\cos \alpha + i\sin \alpha$ and making ${e^{ - i\alpha }} = \cos \alpha - i\sin \alpha$ equation to solve and get tangent ratio. Using trigonometric ratios we need to solve the above equation. In between we might need to use the hyperbolic functions as well.

Complete step-by-step answer:
Given $\tan \left( {\theta + i\phi } \right) = {e^{i\alpha }}$
Trigonometric values are based on three major trigonometric ratios, Sine, Cosine and Tangent.
Sine or $\sin \theta$= Side opposite to $\theta$ is to Hypotenuse.
Cosines or $\operatorname{Cos} \theta$= Adjacent side to $\theta$ is to Hypotenuse.
Tangent or $\tan \theta$= Side opposite to $\theta$ is to Adjacent side to$\theta$.
We know that ${e^{i\alpha }} = \cos \alpha + i\sin \alpha$
According to Euler’s formula, the fundamental relationship between the trigonometric functions and the complex exponential function is given by ${e^{ix}} = \cos x + i\sin x$ where x is a real number and i is imaginary number
Therefore $\cos \alpha + i\sin \alpha = \tan \left( {\theta + i\phi } \right)$… (1)
$\Rightarrow \cos \alpha - i\sin \alpha = \tan \left( {\theta - i\phi } \right)$… (2)
From equations (1) and (2)
$\theta + i\phi = {\tan ^{ - 1}}{e^{i\alpha }}$… (3)
$\theta - i\phi = {\tan ^{ - 1}}{e^{ - i\alpha }}$… (4)
Subtracting (4) equation from (3) equation.
$\Rightarrow 2i\phi = {\tan ^{ - 1}}\left( {\dfrac{{{e^{i\alpha }} - {e^{ - i\alpha }}}}{{1 + {e^{i\alpha }}{e^{ - i\alpha }}}}} \right)$
$\Rightarrow 2i\phi = {\tan ^{ - 1}}\left( {\dfrac{{\cos \alpha + i{\text{sin}}\alpha - (\cos \alpha - i{\text{sin}}\alpha )}}{2}} \right)$
$\Rightarrow 2i\phi = {\tan ^{ - 1}}\left( {\dfrac{{\cos \alpha + i{\text{sin}}\alpha - \cos \alpha + i{\text{sin}}\alpha }}{2}} \right)$
$\Rightarrow 2i\phi = {\tan ^{ - 1}}\left( {\dfrac{{2i{\text{sin}}\alpha }}{2}} \right)$
$\Rightarrow 2i\phi = {\tan ^{ - 1}}\left( {i{\text{sin}}\alpha } \right)$
$\Rightarrow \tan \left( {2i\phi } \right) = i{\text{sin}}\alpha$
Hyperbolic functions are analogs of the ordinary trigonometric functions defined for the hyperbola rather than on the circle. It is a function of an angle expressed as a relationship between the distances from a point on a hyperbola to the origin and to the coordinate axes, as hyperbolic sine or hyperbolic cosine: often expressed as combinations of exponential functions.
We know that $\tan \left( {i\theta } \right) = i\tanh \theta$
Therefore,
$\Rightarrow i\tanh \left( {2\phi } \right) = i{\text{sin}}\alpha$
$\Rightarrow \tanh \left( {2\phi } \right) = {\text{sin}}\alpha$
$\Rightarrow 2\phi = {\tanh ^{ - 1}}\left( {{\text{sin}}\alpha } \right)$
We know that${\tanh ^{ - 1}}x = \dfrac{1}{2}\log \left( {\dfrac{{1 + x}}{{1 - x}}} \right)$
Therefore,
$\Rightarrow 2\phi = \dfrac{1}{2}\log \left( {\dfrac{{{\text{1 + sin}}\alpha }}{{1 - {\text{sin}}\alpha }}} \right)$
Since $\sin 2\alpha = \dfrac{{2\tan \alpha }}{{1 + {{\tan }^2}\alpha }}$
$\Rightarrow 2\phi = \dfrac{1}{2}\log \left( {\dfrac{{{\text{1 + }}\dfrac{{2\tan \dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}}}{{1 - \dfrac{{2\tan \dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}}}} \right)$
$\Rightarrow 2\phi = \dfrac{1}{2}\log \left( {\dfrac{{\dfrac{{1 + {{\tan }^2}\dfrac{\alpha }{2} + 2\tan \dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}}}{{\dfrac{{1 + {{\tan }^2}\dfrac{\alpha }{2} - 2\tan \dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}}}} \right)$
$\Rightarrow 2\phi = \dfrac{1}{2}\log \left[ {\left( {\dfrac{{1 + {{\tan }^2}\dfrac{\alpha }{2} + 2\tan \dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}} \right)\left( {\dfrac{{1 + {{\tan }^2}\dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2} - 2\tan \dfrac{\alpha }{2}}}} \right)} \right]$
$\Rightarrow 2\phi = \dfrac{1}{2}\log \left( {\dfrac{{1 + {{\tan }^2}\dfrac{\alpha }{2} + 2\tan \dfrac{\alpha }{2}}}{{1 + {{\tan }^2}\dfrac{\alpha }{2} - 2\tan \dfrac{\alpha }{2}}}} \right)$
We used ${a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}$and ${a^2} - 2ab + {b^2} = {\left( {a - b} \right)^2}$
$\Rightarrow 2\phi = \dfrac{1}{2}\log {\left( {\dfrac{{1 + \tan \dfrac{\alpha }{2}}}{{1 - \tan \dfrac{\alpha }{2}}}} \right)^2}$
$\Rightarrow 2\phi = \log \left( {\dfrac{{1 + \tan \dfrac{\alpha }{2}}}{{1 - \tan \dfrac{\alpha }{2}}}} \right)$
Since $\tan \dfrac{\pi }{4} = 1$
$\Rightarrow 2\phi = \log \left( {\dfrac{{\tan \dfrac{\pi }{4} + \tan \dfrac{\alpha }{2}}}{{1 - \tan \dfrac{\pi }{4}\tan \dfrac{\alpha }{2}}}} \right)$
$\Rightarrow \phi = \dfrac{1}{2}\log \left[ {\tan \left( {\dfrac{\pi }{4} + \dfrac{\alpha }{2}} \right)} \right]$
Therefore, If $\tan \left( {\theta + i\phi } \right) = {e^{i\alpha }}$then $\phi = \dfrac{1}{2}\log \left[ {\tan \left( {\dfrac{\pi }{4} + \dfrac{\alpha }{2}} \right)} \right]$

Note: Trigonometric identities are equalities that involve trigonometric functions and are true for every value of the occurring variables where both sides of the equality are defined. Geometrically, these are identities involving certain functions of one or more angles. To solve the given type of problems we have to use the trigonometric identities.