Complex in Logarithm

Note: For this problem to have a unique solution, we need to assume the principal root of negative 1 $(e^{i \pi / 2})$

Evaluate the argument of the logarithm:

$\large{\frac{2 \sqrt{1/3} + (2/3)i}{2 \sqrt{1/3} - (2/3)i} \times 1 \\ = \frac{2 \sqrt{1/3} + (2/3)i}{2 \sqrt{1/3} - (2/3)i} \times \frac{2 \sqrt{1/3} + (2/3)i}{2 \sqrt{1/3} + (2/3)i} \\ = \frac{1 + \sqrt{3} i}{2} \\ = e^{i \pi / 3} }$

The function therefore simplifies to:

$\large{F = i^{2 i + 3} \,\, \frac{\pi}{3} i \\ = \frac{\pi}{3} \, i^{2 i + 4} \\ = \frac{\pi}{3} e^{(i \pi / 2)(2 i + 4)} \\ = \frac{\pi}{3} e^{- \pi} \, e^{i 2 \pi} \\ = \frac{\pi}{3 \, e^{\pi}}}$

Let us first simplify the the given function i.e.

Let G(x) = $ln{ \left( \frac { { { i }^{ 3 } }{ x }^{ 2 }+{ 2x }+{ i } }{ { i }{ x }^{ 2 }+{ 2x }+{ i }^{ 3 } } \right) }\quad =\quad ln{ \left( \frac { { 2x }+{ i }\left( { 1 }-{ x }^{ 2 } \right) }{ { 2x }-{ i }\left( { 1 }-{ x }^{ 2 } \right) } \right) }$ (by taking i as common)

Thus the above expression can be defined as the complex number (z) with real part 2x and imaginary part $\left( { 1 }-{ x }^{ 2 } \right)$

therefore let us assume that magnitude of (z) is r and principal argument $\theta$ , therefore above expression can be written as

$ln{ \left( \frac { r\left( { \cos { \theta } }+{ i }{ \sin { \theta } } \right) }{ r\left( { \cos { \theta } }-{ i }{ \sin { \theta } } \right) } \right) }$

thus changing into euler form we get

$ln{ \left( \frac { { e }^{ i\theta } }{ { e }^{ -i\theta } } \right) }=ln{ \left( { e }^{ 2i\theta } \right) }=\quad 2i\theta$ and on more simplification

we get $\theta =\tan ^{ -1 }{ \left( \frac { { 1 }-{ x }^{ 2 } }{ 2x } \right) } =\cot ^{ -1 }{ \left( \frac { 2x }{ { 1 }-{ x }^{ 2 } } \right) } =\frac { \pi }{ 2 } -\tan ^{ -1 }{ \left( \frac { 2x }{ { 1 }-{ x }^{ 2 } } \right) } =\frac { \pi }{ 2 } -2\tan ^{ -1 }{ x }$ since x $\epsilon \left( 0,1 \right)$

therefore G(x) = $2i\theta =i(\pi -4\tan ^{ -1 }{ x) }$

Let H(x) = ${ i }^{ 2i+3 }$

therefore on simplification we get

${ i }^{ 2i+3 }=\left( -i \right) { i }^{ 2i }=\left( -i \right) { \left( { e }^{ i\pi /2 } \right) }^{ 2i }=\left( -i \right) \left( { e }^{ -\pi } \right)$

Combining H(x) & G(x) we get

F(x) = $\left( { e }^{ -\pi } \right) \left( { \pi }-{ 4\tan ^{ -1 }{ x } } \right)$

therefore now required question can be solved

Solution inspired by @Ayush Mishra

$\begin{aligned} F(x) & = i^{2i+\color{#3D99F6}3} \ln \left(\frac {{\color{#3D99F6}i^3}x^2+2x+i}{ix^2+2x+{\color{#3D99F6}i^3}}\right) & \small \color{#3D99F6} \text{Note that } i^3 = (i^2)i = -i \\ & = {\color{#D61F06}i}^{2i}{\color{#3D99F6}(-i)} \ln \left(\frac {{\color{#3D99F6}-i}x^2+2x+i}{ix^2+2x\color{#3D99F6}-i}\right) & \small \color{#D61F06} \text{By Euler's formula }e^{i\theta} = \cos \theta + i \sin \theta \\ & = {\color{#D61F06}e}^{{\color{#D61F06}i\frac \pi 2}\times 2i}(-i) \ln \left(\frac {2x+i(1-x^2)}{2x-i(1-x^2)}\right) \\ & = -ie^{-\pi} \ln \left( {\color{#3D99F6} \frac {\frac {2x}{1+x^2}+i\frac {1-x^2}{1+x^2}}{\frac {2x}{1+x^2}-i\frac {1-x^2}{1+x^2}}}\right) & \small \color{#3D99F6} \text{By half-angle tangent substitution} \\ & = -ie^{-\pi} \ln \left( {\color{#3D99F6} \frac {\sin \theta +i\cos \theta}{\sin \theta -i\cos \theta}}\right) & \small \color{#3D99F6} \text{where }x = \tan \frac \theta 2 \\ & = -ie^{-\pi} \ln \left( {\color{#3D99F6} \frac {\cos \left(\frac \pi 2 - \theta\right) +i\sin \left(\frac \pi 2 - \theta\right)}{\cos \left(\frac \pi 2 - \theta\right) +i\sin \left(\frac \pi 2 - \theta\right)}} \right) \\ & = -ie^{-\pi} \ln \left( {\color{#3D99F6} \frac {e^{i\left(\frac \pi 2 - \theta\right)}}{e^{-i\left(\frac \pi 2 - \theta\right)}}}\right) & \small \color{#3D99F6} \text{By Euler's formula }e^{i\phi} = \cos \phi + i \sin \phi \\ & = -ie^{-\pi} \ln \left(e^{2i\left(\frac \pi 2 - \theta\right)}\right) \\ & = -ie^{-\pi}(2i)\left(\frac \pi 2 - {\color{#3D99F6}\theta}\right) & \small \color{#3D99F6} \text{Since }x = \tan \frac \theta 2 \implies \theta = 2 \tan^{-1} x \\ & = \frac {\pi - 2\cdot{\color{#3D99F6}2\tan^{-1}x}}{e^\pi} \end{aligned}$

$\implies F\left(\sqrt{\frac 13}\right) = \dfrac {\pi - 4\tan^{-1}\left(\sqrt{\frac 13}\right)}{e^\pi} = \dfrac {\pi - 4\cdot \frac \pi 6}{e^\pi} = \dfrac \pi{3e^\pi}$ . Therefore, $\alpha + \beta = 1+3 = \boxed{4}$ .

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Previous solution

$\begin{aligned} F(x) & = i^{2i+\color{#3D99F6}3} \ln \left(\frac {{\color{#3D99F6}i^3}x^2+2x+i}{ix^2+2x+{\color{#3D99F6}i^3}}\right) & \small \color{#3D99F6} \text{Note that } i^3 = (i^2)i = -i \\ & = {\color{#D61F06}i}^{2i}{\color{#3D99F6}(-i)} \ln \left(\frac {{\color{#3D99F6}-i}x^2+2x+i}{ix^2+2x\color{#3D99F6}-i}\right) & \small \color{#D61F06} \text{By Euler's formula }e^{i\theta} = \cos \theta + i \sin \theta \\ F\left(\sqrt{\frac 13}\right) & = {\color{#D61F06}e}^{{\color{#D61F06}i\frac \pi 2}\times 2i}(-i) \ln \left(\frac {-i\frac 13 + \frac 2{\sqrt 3}+i}{i\frac 13+\frac 2{\sqrt 3}-i}\right) \\ & = e^{-\pi}(-i) \ln \left(\frac {\frac 2{\sqrt 3}+i\frac 23}{\frac 2{\sqrt 3}-i\frac 23}\right) \\ & = -i e^{-\pi}\ln \left(\frac {\frac 43 \color{#D61F06}\left(\frac {\sqrt 3}2+i\frac 12\right)}{\frac 43\color{#D61F06} \left(\frac {\sqrt 3}2-i\frac 12\right)} \right) & \small \color{#D61F06} \text{By Euler's formula again} \\ & = -i e^{-\pi}\ln \left(\frac {\color{#D61F06}e^{\frac \pi 6 i}}{\color{#D61F06}e^{-\frac \pi 6 i}} \right) \\ & = -i e^{-\pi}\ln \left(e^{\frac \pi 3 i} \right) \\ & = -i e^{-\pi}\times \frac \pi 3 i = \frac {\pi}{3e^\pi} \end{aligned}$

Therefore, $\alpha + \beta = 1+3 = \boxed{4}$ .