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Im getting really confused looking at past exam style questions evaluating contour integrals... Can anyone help me in the right direction to solve these..

(i) $\displaystyle\int_\gamma \frac{\sin z}{z^4}dz$ where $\gamma(t)=e^{it}$ and $0\leq t\leq2\pi $

(ii) $\displaystyle\int_\gamma \frac{1}{(z-1)(z+3)}dz$ where $\gamma(t)=2e^{it}$ and $0\leq t\leq 2\pi$

Let $\gamma$ be the unit circle in $\mathbb{C}$ transversed in the anti-clockwise direction.

(iii) $\displaystyle\int_\gamma \frac{\cos^2 z}{z^2}dz$

(iv) $\displaystyle\int_\gamma \frac{1}{(2z+1)(z+3)}dz$

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    $\begingroup$ Use Cauchy's theorem. $\endgroup$ – Ishan Banerjee May 1 '13 at 16:45
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    $\begingroup$ Or the residue theorem. $\endgroup$ – Librecoin May 1 '13 at 16:50
  • $\begingroup$ Wenn you just write out function names like that, $\TeX$ interprets that as a juxtaposition of variable names and formats it accordingly. To get the appropriate font and spacing, you can use predefined commands like \cos, or, if you need an operator name for which there isn't a predefined command, you can use \operatorname{name}. $\endgroup$ – joriki May 1 '13 at 17:06
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Hints (look for the residue(s) at the pole(s) within the integration contour!):

$$\color{red}{(1)}\;\;\;\frac{\sin z}{z^4}=\frac1{z^4}\left(z-\frac{z^3}{3!}+\frac{z^5}{5!}-\ldots\right)=\frac1{z^3}-\frac1{6z}+\ldots$$

$$\color{red}{(2)}\;\;\;\frac1{(z+3)(z-1)}=\frac1{4(z-1)}\frac1{1+\frac{z-1}4}=\frac1{4(z-1)}\left(1-\frac{z-1}4+\frac{(z-1)^2}{16}+\ldots\right)=\ldots$$

$$\color{red}{(3)}\;\;\;\frac{\cos^2z}{z^2}=\frac1{z^2}\left(1-\frac{z^2}{2!}+\ldots\right)^2=\frac{1}{z^2}\left(1-z^2+\ldots\right)=\ldots$$

$$\color{red}{(4)}\;\;\;\frac{1}{(2z+1)(z+3)}=\frac12\frac1{z+\frac12}\cdot\frac25\frac1{1+\frac25\left(z+\frac12\right)}=$$

$$=\frac1{5\left(z+\frac12\right)}+\left(1-\frac{2}{5}\left(z+\frac12\right)+\frac4{25}\left(z+\frac12\right)^2-\ldots\right)=\ldots$$

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  • $\begingroup$ Yucks! The sine and cosine cables short-circuited in my mind. Thanks $\endgroup$ – DonAntonio May 1 '13 at 18:13

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