Question

if {(1+i)/(1-i)}^n=1 then the least value of n is

Answer 1

Answer:

(1+i/1-i)^n=1

(1+i/1-i×1+i/1+I)^n=1

((1+i)²/1-i²)^n=1

(1+i²+2i/2)=1

(1-1+2i/2)^n=1

(2i/2)^n=1

i^n=1

n=4 answer. (i⁴=1)

Step-by-step explanation:

first rationalise the denominator

expand( 1+I)²

put the value of i²=-1

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Answer 2

SOLUTION

GIVEN

$\displaystyle \sf{ { \bigg( \frac{1 + i}{1 - i} \bigg)}^{n} = 1}$

TO DETERMINE

The least value of n

EVALUATION

Here the given equation is

$\displaystyle \sf{ { \bigg( \frac{1 + i}{1 - i} \bigg)}^{n} = 1} \: \: \: \: - - - - (1)$

Now

$\displaystyle \sf{ \frac{1 + i}{1 - i}}$

$\displaystyle \sf{ = \frac{ - {i}^{2} + i}{1 - i}}$

$\displaystyle \sf{ = \frac{i( - i + 1)}{1 - i}}$

$\displaystyle \sf{ = \frac{i(1 - i)}{1 - i}}$

$\displaystyle \sf{ =i}$

From Equation 1 we get

$\displaystyle \sf{ {i}^{n} = 1 }$

$\displaystyle \sf{ \implies {i}^{n} = {( - 1)}^{2} }$

$\displaystyle \sf{ \implies {i}^{n} = {( {i}^{2} )}^{2} }$

$\displaystyle \sf{ \implies {i}^{n} = {i}^{4} }$

$\displaystyle \sf{ \implies n = 4 }$

FINAL ANSWER

Hence the required least value of n = 4

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