The absolute value of residue of the function zcos 1/z at z = 0 is (Answer up to one decimal place)Correct answer is '0.5'. Can you explain this answer?

Question

The absolute value of residue of the function zcos 1/z  at z = 0 is (Answer up to one decimal place)Correct answer is '0.5'. Can you explain this answer?

Answer

Explanation:

To find the residue of zcos(1/z) at z=0, we need to expand the function as a Laurent series around z=0.

The Laurent series of cos(1/z) can be given as:

cos(1/z) = 1 - (1/z)^2/2! + (1/z)^4/4! - (1/z)^6/6! + ...

Therefore, zcos(1/z) = z - z(1/z)^2/2! + z(1/z)^4/4! - z(1/z)^6/6! + ...

On simplifying, we get:

zcos(1/z) = z - z/2! + z/4! - z/6! + ...

The residue of the function at z=0 can be given as the coefficient of (1/z) in the Laurent series.

Therefore, the residue can be calculated as:

Res(z=0) = -1/2! + 1/4! - 1/6! + ...

On simplifying, we get:

Res(z=0) = -1/2 + 1/24 - 1/720 + ...

As we only need to find the absolute value of the residue, we can ignore the negative sign.

Therefore, the absolute value of the residue is:

|Res(z=0)| = 1/2! - 1/4! + 1/6! - ...

On simplifying, we get:

|Res(z=0)| = 0.5

Therefore, the absolute value of the residue of the function zcos(1/z) at z=0 is 0.5.

Answer

We have f(z) = zcos1/z.

Residue of f(z) at z = 0 is the coefficient of 1/z i.e -1/2 or - 0.5.

Absolute value = 0.5

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