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    If  Im,n=12πi∮Czmz−n dzwhere∣z∣=1I_{m,n} = \frac{1}{2\pi i} \oint_{C} z^m z^{-n} \, dz \quad \text{where} \quad |z| = 1Im,n​=2πi1​∮C​zmz−ndz
    Question

    If  Im,n=12πiCzmzn dzwherez=1I_{m,n} = \frac{1}{2\pi i} \oint_{C} z^m z^{-n} \, dz \quad \text{where} \quad |z| = 1 ,

    Which of the following statements is necessarily true ?

    A.

    Im(m,n)=1if mn\text{Im}(m,n) = 1 \quad \text{if} \, m \geq n​​

    B.

    Im(m,n)=1if m+1=n\text{Im}(m,n) = 1 \quad \text{if} \, m + 1 = n​​

    C.

    Im(m,n)=1if m=n+1\text{Im}(m,n) = 1 \quad \text{if} \, m = n + 1​​

    D.

    N/A

    Correct option is B

    Let, Im,n=12πiCzmzn dzwherez=1I_{m,n} = \frac{1}{2\pi i} \oint_{C} z^m z^{-n} \, dz \quad \text{where} \quad |z| = 1​​

    Option 1.

    Let m \geq​ n  take m = n = 2 Then,

    Im,n=12πiCz2z2 dz=12πic1dz=01I_{m,n} = \frac{1}{2\pi i} \oint_{C} z^2 z^{-2} \, dz=\frac{1}{2\pi i} \oint_{c}1dz =0\neq1

    Option 1 is incorrect.\implies \textbf{Option 1 is incorrect.}

    Option 2.

    Let m+1=n , Then

    Im,n=12πiCzn1zn dz=12πic1zdz=12πi2πi=1 ( Using cauchy integral formulaI_{m,n} = \frac{1}{2\pi i} \oint_{C} z^{n-1} z^{-n} \, dz=\frac{1}{2\pi i} \oint_{c}\frac{1}{z}dz =\frac{1}{2\pi i} 2\pi i=1 \text{ ( Using cauchy integral formula})

    Option 2 is correct.\implies \textbf{Option 2 is correct.}​​

    ​​

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