To the power of pi

Recall Euler's Formula: ${ e }^{ \theta i }=\cos { (\theta ) } +i\sin { (\theta ) }$ .

We can manipulate Euler's formula to look like the equation in the problem. But first, we'll have to get rid of the $i\sin { (\theta ) }$ by using the identity, $\sin { (\theta ) } =\mp \sqrt { 1-{ \cos { (\theta ) } }^{ 2 } }$ (which can be derived from the Pythagorean trigonometric identity).

The equation can now be rewritten as ${ e }^{ \theta i }=\cos { (\theta ) } \mp i\sqrt { 1-{ \cos { (\theta ) } }^{ 2 } }$ . After moving the real part to the left then squaring both sides, it can then be rewritten as ${ ({ e }^{ \theta i }) }^{ 2 }-2{ e }^{ \theta i }\cos { (\theta ) } +{ \cos { (\theta ) } }^{ 2 }=-1+{ \cos { (\theta ) } }^{ 2 }$ , or ${ ({ e }^{ \theta i }) }^{ 2 }-2{ e }^{ \theta i }\cos { (\theta ) } =-1$ .

Now, we want to isolate the cosine. Moving ${ ({ e }^{ \theta i }) }^{ 2 }$ to the right and dividing both sides by -1 results in the equation $2{ e }^{ \theta i }\cos { (\theta ) } =1+{ ({ e }^{ \theta i }) }^{ 2 }$ . Now all we need to do is divide both sides by $2{ e }^{ \theta i }$ in order to get $\cos { (\theta ) } =\frac { 1 }{ 2{ e }^{ \theta i } } +{ \frac { { e }^{ \theta i } }{ 2 } }$ . Finally, taking the arccosine of both sides results in $\arccos { (\frac { { e }^{ \theta i } }{ 2 } +\frac { 1 }{ 2{ e }^{ \theta i } } ) } =\theta$ .

It now becomes clear that $x$ is equivalent to ${ e }^{ \theta i }$ . Replacing $x$ with ${ e }^{ \theta i }$ , we get the equation $\arccos { (\frac { { e }^{ \theta i } }{ 2 } +\frac { 1 }{ 2{ e }^{ \theta i } } ) } =-{ e }^{ \theta \pi i }$ . However, we already know that $\arccos { (\frac { { e }^{ \theta i } }{ 2 } +\frac { 1 }{ 2{ e }^{ \theta i } } ) } =\theta$ , meaning that $\theta$ must be equal to $-{ e }^{ \theta \pi i }$ . The only value of $\theta$ that would make that true is if $\theta$ was equal to 1, as ${ e }^{ \pi i }=-1$ . This means that $x$ must be equal to ${ e }^{ i }$ , or approximately $0.540+0.841i$ .

The real part of $0.540+0.841i$ is 0.54, and the imaginary part is 0.841. Summing the two values together and rounding to two decimal points, we get our final answer, $\boxed { 1.38 }$ .