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1-$$\int_{|z|=\frac{3}{2}}\frac{z^{3}\cos z}{(z-1)(z-2)}dz$$ My Attempt

$\int_{|z|=\frac{3}{2}}\frac{z^{3}\cos z}{(z-1)(z-2)}dz=2\pi i\frac{1^{3}\cos 1}{(1-2)}=-2\pi i\,\cos 1$

2-$$\int_{|z|=3}\frac{\cos z}{(z-1)(z-2)}dz$$ My Attempt

$\int_{|z|=3}\frac{\cos z}{(z-1)(z-2)}dz=\int_{|z|=3}\frac{\cos z}{(z-2)}dz-\int_{|z|=3}\frac{\cos z}{(z-1)}dz=2\pi i{\,\cos 2}-2\pi i{\,\cos 1}=2\pi i({\cos 2}-{\cos 1})$

3-$$\int_{|z-i|=4}\frac{e^z\cos z}{(z^2-2)}dz$$ My Attempt

$\int_{|z-i|=4}\frac{e^z\cos z}{(z^2-2)}dz=-\frac{1}{2\sqrt{2}}\int_{|z-i|=4}\frac{e^z\cos z}{(z-\sqrt{2})}dz-\frac{1}{2\sqrt{2}}\int_{|z-i|=4}\frac{e^z\cos z}{(z+\sqrt{2})}dz=-\frac{2\pi i}{2\sqrt{2}}e^{\sqrt{2}}\cos (\sqrt{2})-\frac{2\pi i}{2\sqrt{2}}e^{-\sqrt{2}}\cos (-\sqrt{2})$

4-$$\int_{|z-i|=1}\frac{e^z}{(z^2+1)}dz$$ My Attempt

$\int_{|z-i|=1}\frac{e^z}{(z^2+1)}dz=\int_{|z-i|=1}\frac{e^z}{(z+i)(z-i)}dz=\frac{2\pi i e^z}{(i+i)}=\pi e^i$

Are the above solutions correct?

Also, can the integral $\int_{|z|=1}\frac{e^z}{(z-1)}dz$ be solved using contour integration? I am confused because $z=1$ lies on the boundary of $|z|=1$.

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    In most cases, when the pole lies on the boundary, you can still solve with residue theorem. Just use $\pi i$ rather than $2 \pi i$ – 2017-01-06
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    @Kaynex, so the answer will be $\pi i e$. Is that right? – 2017-01-06
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    @Kaynex, what about the other integrals are they correct? – 2017-01-06
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    Did not check in minute detail but everything looks right at first glance. – 2017-01-06

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