To simplify the expression $\arg(i + e^{2ix})$, let's break it down step by step.

1. Start with $e^{2ix}$:
   Using Euler's formula, we can rewrite the exponential as a combination of sine and cosine: $e^{2ix} = \cos(2x) + i\sin(2x)$ So, the original expression becomes: $i + e^{2ix} = i + \cos(2x) + i\sin(2x)$ Rearrange the terms: $i + e^{2ix} = (\cos(2x)) + i(1 + \sin(2x))$

2. Find the argument:
   The argument of a complex number $z = a + bi$ is given by $\arg(z) = \tan^{-1}\left( \frac{b}{a} \right)$.

   Here, $a = \cos(2x)$ and $b = 1 + \sin(2x)$.

   Therefore, the argument is: $\arg(i + e^{2ix}) = \tan^{-1}\left( \frac{1 + \sin(2x)}{\cos(2x)} \right)$

3. Simplify the argument:
   We can further simplify the expression inside the arctangent: $\arg(i + e^{2ix}) = \tan^{-1}\left( \frac{1 + \sin(2x)}{\cos(2x)} \right)$ This is the simplified form of the argument.

Final Answer:

$\arg(i + e^{2ix}) = \tan^{-1}\left( \frac{1 + \sin(2x)}{\cos(2x)} \right)$

Would you like more details, or do you have any questions?

Here are 5 related questions:

1. How do you simplify arguments of complex numbers in polar form?
2. Can you explain how Euler’s formula relates complex numbers to trigonometry?
3. What is the geometric interpretation of the argument of a complex number?
4. How can we use trigonometric identities to simplify complex number expressions?
5. What is the difference between the modulus and argument of a complex number?

Tip: The argument of a complex number describes its angle relative to the positive real axis.