1
$\begingroup$

The variable x in this case is real...not complex. At first blush $\frac{i}{(x+e^{ix})}$ wouldn't seem to have any real part. It looks like i divided by some stuff. However, plotting this with Mathematica shows there is definitely a real part. How does one go about finding the real and imaginary parts of this function?

$\endgroup$
  • $\begingroup$ Write $e^{ix}$ as $\cos(x) + i\sin(x)$ and then use the usual technique for realifying the denominator (multiplying and dividing by the conjugate). $\endgroup$ – Cameron Williams Aug 7 '17 at 18:12
4
$\begingroup$

$$\frac { i }{ (x+e^{ ix }) } =\frac { i }{ x+\cos { x+i\sin { x } } } =\frac { i\left( x+\cos { x-i\sin { x } } \right) }{ \left( x+\cos { x+i\sin { x } } \right) \left( x+\cos { x-i\sin { x } } \right) } =\\ =\frac { \sin { x } }{ \left( x+\cos { x } \right) ^{ 2 }+\sin ^{ 2 }{ x } } +i\frac { x+\cos { x } }{ \left( x+\cos { x } \right) ^{ 2 }+\sin ^{ 2 }{ x } } $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.