Question

please help
find singular part at the Laurant series at z=0 for cos z/ (sin z)^ 3

Answer 1

the singular part is also known as the principal part, and corresponds to negative powers of \(z\). so: $$f(z)=\frac{\cos z}{\sin^3 z}$$

now take power series of these: $$\cos z=1-\frac12 x^2+\frac1{24}x^4-\frac1{720}x^6+\dots\\\sin z=x-\frac16x^3+\frac1{120}x^5-\frac1{5040}x^7+\dots\\\sin^2 z=\left(x-\frac16x^3+\frac1{120}x^5-\frac1{5040}x^7+\dots\right)\left(x-\frac16x^3+\frac1{120}x^5-\frac1{5040}x^7+\dots\right)\\\qquad=x^2-\frac13x^4+\frac2{45}x^6-\frac1{315}x^8+\dots\\\sin^3 z=\left(x-\frac16x^3+\frac1{120}x^5-\frac1{5040}x^7+\dots\right)\left(x^2-\frac13x^4+\frac2{45}x^6-\frac1{315}x^8+\dots\right)\\\qquad=x^3-\frac12x^5+\frac{13}{120}x^7-\frac{41}{3024}x^9+\dots$$ and use long division to find teh first couple terms of \(\frac{\cos z}{\sin^3 z}\)

Answer 2

so \(x^3-\frac12 x^5+\frac{13}{120}x^7-\frac{41}{3024}x^9+\dots\) goes into \(1\) a total of \(\frac1{x^3}\) times, giving a remainder of $$-\frac1{15}x^4+\frac{23}{1890}x^6+\dots$$. now, we go into this a total of \(x\) times, giving some remainder with higher powers of \(x\) and so on; the only term with a negative power is the one we've already found, \(\frac1{x^3}\)

Answer 3

and so the principal part of the Laurent series of \(\frac{\cos z}{\sin^3 z}\) is simply \(1/x^3\)

Answer 4

oops, everywhere I wrote \(z\) I meant \(x\)

Answer 5

alternatively, consider: $$\lim_{z\to0} z^n\cdot\frac{\cos z}{\sin^3 z}$$
for integers \(n>0\).
if \(n=1\), we get \(z/\sin^3 z\to \infty\) and the whole thing blows up.
if \(n=2\), we get \(z^2/\sin^3 z\to\infty\) still and the whole thing still blows up.
if \(n=3\), then we get \(z^3/\sin^3 z\to 1\) and \(\cos z\to 1\).
so we know the least power term in the series is \(1/z^3\)

Answer 6

and then you subtract that and try again for \(n=2\): $$\lim_{z\to0}z^2\left(\frac{\cos z}{\sin^3 z}-\frac1{z^3}\right)=0$$ and likewise for \(n=0\), so there are no terms with \(1/z^2\) and \(1/z\) and the principal part is simply \(1/z^3\)

Answer 7

thanks