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Given f(z) = Then
  • a)
    z = ia is a simple pole and ia/2 is a residue at z = ia of f(z)
  • b)
    z = ia is a simple pole ia is a residue at z = ia of f(z)
  • c)
    z = ia is a simple pole and −ia/2 is a residue at z = ia of f(z)
  • d)
    none of the above
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Givenf(z) = Thena)z = ia is a simple pole andia/2 is a residue at z = ...
Concept:
Pole:
The value for which f(z) fails to exists i.e. the value at which the denominator of the function f(z) = 0.
When the order of a pole is 1, it is known as a simple pole.
Residue:
If f(z) has a simple pole at z = a, then

If f(z) has a pole of order n at z = a, then

Calculate:
Given:

For calculating pole:
z2 + a2 = 0
∴ (z + ia)(z - ai) = 0
∴ z = ai, -ai.
∴ z has simple pole at z = ai and -ai.
Residue:
If f(z) has a simple pole at z = a, then

For pole at z = ai

For pole at z = -ai

∴ z has a simple pole at z = ai and  is a residue at z = ia of f(z)
Free Test
Community Answer
Givenf(z) = Thena)z = ia is a simple pole andia/2 is a residue at z = ...
Concept:
Pole:
The value for which f(z) fails to exists i.e. the value at which the denominator of the function f(z) = 0.
When the order of a pole is 1, it is known as a simple pole.
Residue:
If f(z) has a simple pole at z = a, then

If f(z) has a pole of order n at z = a, then

Calculate:
Given:

For calculating pole:
z2 + a2 = 0
∴ (z + ia)(z - ai) = 0
∴ z = ai, -ai.
∴ z has simple pole at z = ai and -ai.
Residue:
If f(z) has a simple pole at z = a, then

For pole at z = ai

For pole at z = -ai

∴ z has a simple pole at z = ai and  is a residue at z = ia of f(z)
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Givenf(z) = Thena)z = ia is a simple pole andia/2 is a residue at z = ia of f(z)b)z = ia is a simple pole ia is a residue at z = ia of f(z)c)z = ia is a simple pole and−ia/2 is a residue at z = ia of f(z)d)none of the aboveCorrect answer is option 'A'. Can you explain this answer?
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