I am trying to solve given integral: $$ \oint_{|z+i| = 1} \frac{z}{(z-1)(z^2+2)}\;\; dz $$ Rewriting it I get this: $$ \oint_{|z+i| = 1} \frac{z}{(z-1)(z-i\sqrt{2})(z+i\sqrt{2})}\;\; dz $$ Area $|z+i|=1$ is circle with radius $1$ and center at $-i$ right?
So only point in area is $z=-i\sqrt{2}$?
Then implementing Cauchy's integral formula: $$ 2\pi i \; f(-i\sqrt{2}) = \oint_{|z+i| = 1} \frac{\frac{z}{(z-1)(z-i\sqrt{2})}}{(z+i\sqrt{2})}\;\; dz $$ Where $f$ is: $$ f(z)=\frac{z}{(z-1)(z-i\sqrt{2})} $$
I get that result of starting integral is $$ -\frac{\pi}{3}(\sqrt{2} + i) $$
But my textbook is saying that result is $$ -\frac{\pi}{2}(1 + i) $$
Can you point me out if I am missing something? Would appreciate any help. Thanks!
I get $2\pi i f(-i\sqrt 2)=2\pi i\dfrac {-i\sqrt 2}{(-i\sqrt 2-1)(-2i\sqrt 2)}=\pi i\dfrac 1{(-i\sqrt 2-1)}=-\pi i\dfrac {(1-\sqrt 2 i)}3=-\dfrac \pi 3(i+\sqrt 2)$.