Math, asked by beenaaditya2015, 14 days ago

The least positive integral value of n for which (1+i/1-i) to the power n=1 is (a)2 (b)4 (c)3 (d) none of them​

Answers

Answered by user0888
45

\rm\large\underline{\text{Given question}}

What is the least positive integral value of \rm n for which \rm\left(\dfrac{1+i}{1-i}\right)^n=1?

\rm\large\underline{\text{Main idea}}

Let's rationalize the given fraction.

\rm\cdots\longrightarrow\dfrac{1+i}{1-i}=\dfrac{(1+i)^2}{(1+i)(1-i)}=\dfrac{1+2i-1}{1^2+1^2}

\rm\cdots\longrightarrow\dfrac{1+2i-1}{1^2+1^2}=\dfrac{2i}{2}=i

It simplifies to \rm i.

Now, the question given to us is, "what is the value of \rm n such that \rm i^{n}=1?"

\rm\large\underline{\text{Solution}}

We know that \rm i^2=-1 is the definition of the imaginary unit.

If we take \rm i to the power of 3

\rm\cdots\rightarrow i^3=i^{2}\times i=(-1)\times i=-i

If we take \rm i to the power of 4

\rm\cdots\rightarrow i^4=(i^2)^2=(-1)^2=\underline{1}

So, the least positive integral value of \rm n is 4. Option (B) is right.

\rm\large\underline{\text{More questions like this}}

What is the least positive integer \rm n that \rm x^2-x+1=i^nx has two equal solutions?

The solution is as follows.

\rm\cdots\longrightarrow x^2-(i^{n}+1)x+1=0

Quadratic discriminant is

\rm\cdots\longrightarrow D=(i^n+1)^2-4

To have two equal solutions

\cdots\longrightarrow\rm(i^n+1)^2-4=0

\rm\cdots\longrightarrow(i^n+1)^2=2^2

This provides that

\rm\cdots\longrightarrow i^n+1=2\text{ or }i^n+1=-2

But the last equation cannot be true.

\rm\cdots\longrightarrow i^n+1=2

\rm\cdots\longrightarrow i^n=1

Hence, the least positive integer \rm n is 4.

Answered by singhsuryanshu341
47

Step-by-step explanation:

given :

  • least positive integral value of n for which (1+i/1-i) to the power n=1

to find :

  • least positive integral value of n = ?

solution :

  • please check the attached file

extra information :

  • i value = √-1

  • i value we called as = iota

Attachments:
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