Q26·Section II (Mathematics)·26 of 180Complex analysis — Cauchy's integral formulaIf γ={z∈C:∣z−π2∣=1}\gamma=\{z\in\mathbb{C}:|z-\frac\pi2|=1\}γ={z∈C:∣z−2π∣=1}, then the value of ∫γsinz dz(z−π2) z\displaystyle\int_\gamma\frac{\sin z\,dz}{(z-\frac\pi2)\,z}∫γ(z−2π)zsinzdz is: OptionsA0B2πi2\pi i2πiC4i4i4iD−2πi-2\pi i−2πiPick an option to check your answer.Show answerShow solution Report a problem with this question← Q25All questionsQ27 →Keyboard: A–D answer · S solution · ←/→ previous/next