Question

if (1+i÷1-i)m = 1, then find the least positive integral value of m​

Answer 1

Step-by-step explanation:

Consider (

1−i

1+i

)

Rationalize the denominator we get,

⇒(

1−i

1+i

)=

(1−i)(1+1)

(1+i)(1+i)

=

(1−i)(1+i)

(1+i)

2

Here we see that denominator is in the form of (a+b)(a−b)=a

2

−b

2

⇒

1−i

1+i

=

1

2

−i

2

(1+i)

2

⇒

1−i

1+i

=

1−i

2

(1+i)

2

we know that (a+b)

2

=a

2

+2ab+b

2

⇒

1−i

1+i

=

1−i

2

1

2

+(2×1×i)+i

2

⇒

1−i

1+i

=

1−i

2

1

2

+2i+i

2

We know that i

2

=−1 , substituting this we get,

⇒

1−i

1+i

=

1−(−1)

1+2i+(−1)

⇒

1−i

1+i

=

1+1

1+2i−1

⇒

1−i

1+i

=

2

2i

⇒

1−i

1+i

=i

Given that ⇒(

1−i

1+i

)

m

=1

⇒(

1−i

1+i

)

m

=i

m

So, i

m

=1

We know that i

2

=−1

Squaring on both sides we get,

i

4

=(−1)

2

i

4

=1

Therefore the smallest value of m for which (

1−i

1+i

)

m

=1 is m=4