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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2004

Paper 1

Factor the denominator as $z^2(1+z^2)$. Near zero,
$$
\frac1{z^2(1+z^2)}=z^{-2}-1+z^2-\cdots,
$$
so zero is a <double pole> with <residue> zero. At $z=\pm i$ the <poles> are simple and the denominator derivative is $2z+4z^3=-2z$. Therefore
$$
\boxed{\operatorname{Res}_{0}=0,\qquad\operatorname{Res}_{i}=\frac i2,\qquad\operatorname{Res}_{-i}=-\frac i2.}
$$
These are all the <poles>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 1 / 5A / i / Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 1 / 5A / i / Solution

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Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 1 / 5A / i / Solution

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