Complex on Complex

Let $x$ be that expression, so using the exponential function we have:

$x=e^{(1+i)\ln\left(\sqrt{2} + i\frac{\sqrt{2}}{3}\right)}$

Now, use the formula for the principal logarithm of a complex number:

$\ln(z)=\ln(|z|)+i\text{arg}(z) \\ \ln\left(\sqrt{2} + i\dfrac{\sqrt{2}}{3}\right)=\ln\left(\dfrac{2\sqrt{5}}{3}\right)+i \arctan\left(\dfrac{1}{3}\right)$

So, we have now:

$x=e^{(1+i)\left(\ln\left(\frac{2\sqrt{5}}{3}\right)+i \arctan\left(\frac{1}{3}\right)\right)}$

Expand:

$x=e^{\ln\left(\frac{2\sqrt{5}}{3}\right)-\arctan\left(\frac{1}{3}\right)+i\left(\ln\left(\frac{2\sqrt{5}}{3}\right)+\arctan\left(\frac{1}{3}\right)\right)} \\ x=e^{\ln\left(\frac{2\sqrt{5}}{3}\right)-\arctan\left(\frac{1}{3}\right)} \left[\cos\left(\ln\left(\frac{2\sqrt{5}}{3}\right)+\arctan\left(\frac{1}{3}\right)\right)+i\sin \left(\ln\left(\frac{2\sqrt{5}}{3}\right)+\arctan\left(\frac{1}{3}\right)\right)\right]$

To evaluate that expression, we need to use radians:

$x \approx 1.08058579159(0.75114306241+0.6601394548i) \\ x \approx 0.81167452069+0.71333731532i$

Comparing we get $\lfloor 1000\alpha \rfloor=811$ and $\lfloor 1000\beta \rfloor=713$ , so the final answer is $811+713=\boxed{1524}$ .

First we write $(\sqrt {2} +i \dfrac { \sqrt {2} } {3} )$ in form $k(\cos{a}+i\sin{a})$ .

$X={ ( \sqrt {2} +i \dfrac { \sqrt {2} } {3} ) }^{ 1+i }={ (\frac { 2\sqrt { 5 } }{ 3 } ) }^{ 1+i }{ (\frac { 3 }{ 2 } \sqrt { \frac { 2 }{ 5 } } +i\frac { 1 }{ \sqrt { 10 } } ) }^{ 1+i }$

Now we can write both $(\frac { 2\sqrt { 5 } }{ 3 } )$ and $(\frac { 3 }{ 2 } \sqrt { \frac { 2 }{ 5 } } +i\frac { 1 }{ \sqrt { 10 } } )$ in their exponential form. By separating real and imaginary numbers in exponent we can get one purely real number and one complex.

$\large X={ e }^{ (1+i)\ln { \frac { 2\sqrt { 5 } }{ 3 } } }{ e }^{ (1+i)i\arcsin { \frac { 1 }{ \sqrt { 10 } } } }={ e }^{ \ln { \frac { 2\sqrt { 5 } }{ 3 } } -\arcsin{ \frac { 1 }{ \sqrt { 10 } } } }{ e }^{ i(\ln { \frac { 2\sqrt { 5 } }{ 3 } } +\arcsin { \frac { 1 }{ \sqrt { 10 } } } ) }$

This can easily be transformed to the form of $\alpha+i\beta$ .

$X={ e }^{ \ln { \frac { 2\sqrt { 5 } }{ 3 } } -\arcsin { \frac { 1 }{ \sqrt { 10 } } } } \quad[ \cos { (\ln { \frac { 2\sqrt { 5 } }{ 3 } } +\arcsin { \frac { 1 }{ \sqrt { 10 } } )+i\sin { (\ln { \frac { 2\sqrt { 5 } }{ 3 } } +\arcsin{ \frac { 1 }{ \sqrt { 10 } } ) } } } }]$

$\alpha={ e }^{ \ln { \frac { 2\sqrt { 5 } }{ 3 } } -\arcsin { \frac { 1 }{ \sqrt { 10 } } } } \quad \cos { (\ln { \frac { 2\sqrt { 5 } }{ 3 } } +\arcsin { \frac { 1 }{ \sqrt { 10 } } ) } }\approx0,811674$

$\beta={ e }^{ \ln { \frac { 2\sqrt { 5 } }{ 3 } } -\arcsin { \frac { 1 }{ \sqrt { 10 } } } } \quad \sin{ (\ln { \frac { 2\sqrt { 5 } }{ 3 } } +\arcsin { \frac { 1 }{ \sqrt { 10 } } ) } }\approx0,713337$