Complex Basics

Algebra Level 2

$\left (\dfrac {1+i}{1-i}\right)^{x}=1,$

For $i = \sqrt{-1}$ , if $x$ satisfy the equation above, which of these answer choices must be true?

x=4n+1

,where

n

is any positive integer.

x=4n

,where

n

is any positive integer.

x=2n

,where

n

is any positive integer.

x=2n+1

where

n

is any positive integer.

2 solutions

Hjalmar Orellana Soto
Oct 24, 2015

$(\frac{1+i}{1-i})^{x}=1$ $=>(\frac{1+i}{1-i}\cdot \frac{1+i}{1+i})^{x}=1$ $=>(\frac{2i}{2})^{x}=1$ $=>i^{x}=1$ $=>x=4n$ Such that $n$ is an integral number

Edwin Gray
Feb 23, 2019

(1 + i)/(1 - i) = [(1 + i)^2]/[(1 + i)(1 - i) = (1 + 2i + i^2)/2 = i. But i^4 = 1, so i^(4n) =1.

Complex Basics

2 solutions

0 pending reports