Question

(1+I)ⁿ/(1-i) ⁿ=1 then ⁿ=
​

Answer 1

Answer:

Answer: Least value of n = 4

Proof:

To find the least value of n for which

[(1+i)/(1-i)]ⁿ = 1 …………………………………………………………………………….(1) ,

we first simplify the expression within square brackets on left of (1) by rationalizing the denominator. To do that we multiply its numerator and denominator by (1+i), and then use the algebraic identities:(a+b)²=a²+2ab+b², (a+b)(a-b)=a²-b². Thus,

(1+i)/(1-i) = (1+i)(1+i)/(1+i)(1-i) = (1+i)²/(1² - i²) = (1² + 2.1.i + i²)/(1 - i²)

= (1² + 2.1.i + i²)/(1 - i²) = (1 + 2i + i²)/(1 - i²) ………………………………………(2)

Here i is the basic imaginary unit and has the properties:

i = √-1 and i² = √-1 x √-1 = -1

Substituting for i² = -1 in (2),

(1+i)/(1-i) = ((1 + 2i - 1)/(1 - (-1)) = 2i/(1+1) = 2i/2 = i .

∴ Equation (1)→

iⁿ = 1

We now evaluate the quantity on L.H.S. for different integral values of n.

n = 0, iⁿ = i⁰ = 1 (Since a⁰ = 1, for all a≠0)

n = 1, iⁿ= i¹ = i

n = 2, iⁿ= i² =-1

n = 3, iⁿ= i³ =i².i¹ = i².i = -1.i = -i

n = 4, iⁿ= i⁴ =i².i² = (-1) (-1) =+1 = 1

n = 5, iⁿ= i⁵ =i⁴.i = 1.i = i

n = 6, iⁿ= i⁶ =i⁵.i = i.i = i² = -1

n = 7, iⁿ= i⁷ =i⁶.i = (-1)i = -i

n = 8, iⁿ= i⁸ =i⁶.i² = (-1) (-1) = 1

..…………………………………….

In like manner, it can be shown that

i¹² = i¹⁶ = i²⁰ = ……..=1

We see that for n=0, n=4, n=8, n =12=… , that is for all integral values of 4, the Equation (1) is satisfied. But we discard n=0 solution as it is trivial. It is trivial because any quantity (≠ 0) raised to the power 0, the result is 1. Therefore,the nontrivial solutions are n=4, n= 8, n =12……. and the lowest of the solution-sequence is easily seen to be the number 4.

∴ The least value of n = 4 .