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

Question

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

1 Answer

Impact of this question

1280 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-06T15:46:28

See below.

Explanation:

Using de Moivre's identity

$sin x = \frac{e^{i x} - e^{- i x}}{2 i}$ $sin x=(e^(ix)-e^(-ix))/(2i)$ we have

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

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

$= \frac{\sqrt{1 + e^{4} - 2 e^{2} cos (6)}}{2 e} e^{i ϕ}$ $=sqrt[1 + e^4 - 2 e^2 cos(6)]/(2 e) e^(iphi)$

where

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

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question