# What is the argument and module of the complex number sin(i+3) ?

Mar 6, 2017

See below.

#### Explanation:

Using de Moivre's identity

$\sin x = \frac{{e}^{i x} - {e}^{- i x}}{2 i}$ we have

$\sin \left(i + 3\right) = \frac{{e}^{i \left(i + 3\right)} - {e}^{- i \left(i + 3\right)}}{2 i} = - \frac{1}{2} i \left({e}^{- 1 + 3 i} - {e}^{1 - 3 i}\right) =$

$= \left(\frac{1}{2 e} + \frac{1}{2} e\right) \sin \left(3\right) + i \left(\frac{1}{2} e - \frac{1}{2 e}\right) \cos \left(3\right) =$

$= \frac{\sqrt{1 + {e}^{4} - 2 {e}^{2} \cos \left(6\right)}}{2 e} {e}^{i \phi}$

where

$\phi = {\tan}^{-} 1 \left(\frac{{e}^{2} - 1}{{e}^{2} + 1} \cot \left(3\right)\right)$