Respuesta :

Lanuel

Based on the calculations, for any number n > 1, [tex]|(\frac{1}{2} + \frac{3i}{4} )^n|[/tex] is less than one but not greater than or equal to one.

How to solve the equation?

First of all, we would simplify the equation by using the lowest common multiple (LCM) of the fractions as follows:

[tex]|(\frac{2\;+\;3i}{4} )^n|[/tex]

By applying exponent rule, we have:

[tex](a.b)^n=a^n \times b^n\\\\|(\frac{(2\;+\;3i)^n}{4^n} )|\\\\|(\frac{(2\;+\;3i)^n}{(2^2)^n} )|[/tex]

By applying exponent power rule, we have:

[tex]|(\frac{(2\;+\;3i)^n}{2^{2n}} )|[/tex]

Assuming n = 2, we have:

[tex]|(\frac{(2\;+\;3i)^2}{2^{2\times 2}} )|\\\\|(\frac{(2\;+\;3i)^2}{2^{4}} )|\\\\|(\frac{-5\;+\;12i}{16} )|[/tex]

Therefore, for any number n > 1, [tex]|(\frac{1}{2} + \frac{3i}{4} )^n|[/tex] is less than one but not greater than or equal to one.

Read more on complex equations here: https://brainly.com/question/15007300

#SPJ1

ACCESS MORE
EDU ACCESS
Universidad de Mexico