Question

Find the least positive values of n if


( 1 + i / 1 - i)ⁿ = 1

It's I'd complex numbers
​

Answer 1

Hey!

_____

____________________________________________________________

=> Given that ,

→ ( 1 + i / 1 - i )ⁿ = 1

➖❄The first step is to rationalize the denominator to reduce it into the form of ( a + ib)

=> We have ,

$= > \frac{(1 + i)}{(1 - i)} \times \frac{(1 + i)}{(1 + i)} \\ \\ \\ = > \frac{(1 + i) {}^{2} }{1 - i {}^{2} }$

[ identities : ⭐ ( a + b)² = a² + b² + 2ab.

⭐ a² - b² = ( a + b ) ( a - b) ]

$= \frac{1 + i {}^{2} + 2i }{1 + 1} \\ \\ (as \: \: i {}^{2} = - 1) \\ \\ = > \frac{2i}{2}$

◼The expression gives us => ( i )ⁿ

•°• ( i)ⁿ = 1 [ Given ]

Now we know that the value of i raised to some power is equal to one if n is a multiple of 4.

Hence the least value of n = 4 ✔

$Thanks!!$

_______________________________________________________________

Answer 2

NOW,

(1+i\1-i)ⁿ = (1+i) (1+i)\(1-i)(1+i)..................(MULTIPLYING WITH (1+I))

⇒ (1+i)²\1-i² ....................................(IN NUMERATOR, THE IDENTITY (A+B)² AND IN

                                                   THE DENOMINATOR, IDENTITY: A²-B²)

⇒1+2i-1\2..........................................( AS THE VALUE OF, n²=(-1) AND IT IS A

                                                     REAL NUMBER ACTUALLY))

⇒2i\2

⇒i

NOW, iⁿ=(-1), WILL GIVE THE VALUE OF n = 2, WHICH IS LEAST POSITIVE VALUE OF N.......................

HOPE IT HELPED AND MARK AS THE BRAINLIEST ANSWER.............