Question

Residue of \(\tan z\) at \(z=\dfrac{\pi}{2}\) is

• A. \(-1\)
• B. \(0\)
• C. \(1\)
• D. \(2\)

Hint:

Remember: \[ \boxed{ \tan z=\frac{\sin z}{\cos z} } \] Residue at a simple pole \[ = \frac{f(a)}{g'(a)}. \]

Answer 1

The Correct Option is A

Concept: The residue of \[ \frac{f(z)}{g(z)} \] at a simple pole \(z=a\) is \[ \boxed{ \operatorname{Res} = \frac{f(a)}{g'(a)} } \] provided \(g(a)=0\).

Step 1: Express \(\tan z\). \[ \tan z = \frac{\sin z}{\cos z}. \] Here, \[ f(z)=\sin z, \qquad g(z)=\cos z. \]

Step 2: Evaluate at \[ z=\frac{\pi}{2}. \] We have \[ \sin\frac{\pi}{2}=1, \] and \[ g'(z) = -\sin z. \] Therefore, \[ g'\!\left(\frac{\pi}{2}\right) = -1. \] Hence, \[ \operatorname{Res} = \frac{1}{-1} = -1. \] Thus, \[ \boxed{\operatorname{Res}\left(\tan z,\frac{\pi}{2}\right)=-1.} \] Therefore, the correct option is \[ \boxed{(A).} \]