Find the value of \frac{(1+i)^{n}}{(1-i)^{n-2}}

  • Option 1)

    i^{n+1}

  • Option 2)

    i^{n-1}

  • Option 3)

    2(i)^{n-1}

  • Option 4)

    None of the above

 

Answers (1)

As learnt in concept

Power of i in Complex Numbers -

i^{4n}=1,i^{4n+1}=i, i^{4n+2}=-1,i^{4n+3}=-i

- wherein

n\epsilon Integer

 

 

\frac{(1+i)^{n}}{(1-i)^{n-2}}= (1+i)^{2}\left ( \frac{1+i}{1-i} \right )^{n-2}

= (1+i^2+2i)(i)^{n-2}

=2(i)^{n-1}

 


Option 1)

i^{n+1}

Incorrect option

Option 2)

i^{n-1}

Incorrect option

Option 3)

2(i)^{n-1}

Correct option

Option 4)

None of the above

Incorrect option

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