Question

If (1+i/1-i)^2 = 1 then find least positive integral value of m
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Answer 1

Step-by-step explanation:

Given:-

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

To find:-

If (1+i/1-i)^2 = 1 then find least positive integral value of m

Solution:-

Given that (1+i/1-i)²=1

On taking LHS,

(1+i / 1-i)²

The denominator =1-i

Rationalising factor of 1-i=1+i

On Rationalising the denominator then

=>[(1+i)(1+i)]/[(1-i)(1+i)]²

=>(1+i)²/(1-i)(1+i)

Since (a+b)²=a²+2ab+b²

(a+b)(a-b)=a²-b²

Here, a= 1 and b= i

=>[{1²+2(1)(i)+(i)²}/{1²-(i)²}]²

=>[(1+2i+i²)/(1-i²)]²

we know that i²= -1

=>[(1+2i-1)/(1+1)]²

=>(2i/2)²

=>(i)²

Now

If LHS becomes 1 then (i ²)²=1

so,

(i ²)²

i⁴=1

now,

LHS=RHS

So the value of m=4

Answer:-

The value of m=4 for the given problem.